some problems you should be able to do · find vector from point pto point q. find a unit vector in...

SOME PROBLEMS YOU SHOULD BE ABLE TO DO

I’ve attempted to make a list of the main calculations you should be ready for on theexam, and included a handful of the more important formulas. There are no examples here:for that, check my lecture worksheets accompanying the corresponding sections. Please letme know if you notice any mistakes or omissions!

I don’t recommend studying exclusively from this list – it is inevitable that I have leftsomething out. Be sure to look at old exams, problem sets, etc. as well.

A few topics we covered only briefly are in italics. It couldn’t hurt to look, but don’t worryabout these too much.

Chapter 11

11.1 and 11.2: Vectors in the plane and vectors in three dimensions.

� Compute basic vector operations: addition/subtraction, scalar multiplication, mag-nitude.

� Know how to interpret these both algebraically and geometrically.� Find vector from point P to point Q.� Find a unit vector in the direction of a given vector (or a vector with some other

specified length)� Midpoint of a line segment.� Equation of a sphere.� Statics problems (balancing forces).

11.3: Dot products.

� Compute the dot product of two vectors algebraically (a1b1 + a2b2 + a3b3) and geo-metrically (|a| |b| cos θ)

� Compute the angle between two vectors� Compute and interpret projv u and scalv(u):

projv u =(u · v

v · v

)v, scalv u = |u| cos θ =

u · v|v|

.

� Work done by a constant force.� Parallel and normal components of a force.

11.4: Cross products.

� Compute the cross product of two vectors.� Find the area of a triangle with given vertices.� Find the area of a parallelogram with given vertices.

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2 SOME PROBLEMS YOU SHOULD BE ABLE TO DO

11.5: Lines and curves in space.

� Parametrize a straight line� You always need to know two things: a point r0 that the line goes through, and a

vector v in the direction of the line (often you will need to do some work to find v,depending on what information is given to you!). Then use the formula r(t) = r0+tv(and think about the bounds)� between two given points� through a point and perpendicular to a given plane� through a point and perpendicular to two given lines� tangent to a curve r(t) at t = a� given as the intersection of two planes

� Parametrize other simple curves (circles)� Check whether lines intersect� Take a limit (by taking the limit of each component)

11.6: Calculus of vector-valued functions.

� Find the derivative r′(t) = drdt

� Find tangent vector to a curve at time t� Find unit tangent vector to a curve� Integrate a vector-valued function (by integrating each component)

11.7: Motion in space.

� Find velocity, acceleration, and speed.� Find position from velocity and velocity from acceleration, given an initial condition

v(0) or a(0).� Movement in a gravitational field.

11.7: Length of curves.

� The arc length of r(t) = 〈f(t), g(t), h(t)〉 is

ˆ b

a

√f ′(t)2 + g′(t)2 + h′(t)2 dt =

ˆ b

a

|r′(t)| dt.

This is the distance traveled by a particle moving along the curve from t = a to t = b.� Find arc length in polar.� Check whether a path is parametrized by arc length.

Chapter 12

12.1: Planes and surfaces.

� Find the equation for a plane with normal vector 〈a, b, c〉 passing through (x0, y0, z0).� Find the equation for a plane through three given points� Find the equation for a line given as the intersection of two planes.� Check whether two planes are orthogonal.� Equations for cylinders

SOME PROBLEMS YOU SHOULD BE ABLE TO DO 3

12.2: Quadric surfaces.

� Sketch the graph of a quadric surface by drawing the xy-, xz-, and yz-traces, or someother traces parallel to the coordinate planes.

� Find the intersection of a line with a quadric surface.

12.3: Limits of functions of several variables.

� Compute the limit of a function of two variables.� Use the two-path test to show that a limit does not exist.

12.4 & 12.5: Partial derivatives.

� Compute the partial derivative of a function f(x, y)� Compute the four second-order partial derivatives of a function� Use the chain rule to compute partial derivatives of functions of two or three variables.� Apply implicit differentiation to an expression F (x, y) = 0

12.6: Directional derivatives and the gradient.

� Compute (and interpret) the directional derivative Duf� Find the gradient ∇f .� Find direction of fastest ascent/descent and the rate of fastest ascent/descent for a

function f(x, y).� Find directions of 0 increase.� Sketch level curves of a function, and tangent directions to level curves.

12.7: tangent planes and linear approximation.

� Find tangent plane to implicit surface F (x, y, z) = 0 at a point (x0, y0, z0).� Find tangent plane to explicit surface z = f(x, y) at a point (x0, y0, z0).� Find the linear approximation to f(x, y) near a point (a, b) and use this to approxi-

mate values of f .� Work with differentials.

12.8: Maxima and minima.

� Find the critical points of a function.� Use the second derivative test to classify the critical points as max/min/saddle.� Find the global max/min of a function f(x, y) on a region R:

(1) Find critical points inside R(2) Find relative max/min on the boundary(3) Make a list of all the “interesting points” and compute values of f there to find

the true max and min.

12.9: Lagrange multipliers.

� Use Lagrange multipliers to find the maxima and minima of f(x, y) subject to theconstraint g(x, y) = 0

� Key formula: ∇f(x, y) = λ∇g(x, y).Translate this into three equations in the three variables x, y, and λ and solve

� Translate a word problem into a constrained optimization problem.


Chapter 13

13.1: Double integrals.

� Set up and compute the double integral of a function f(x, y) on a rectangular regiona ≤ x ≤ b, c ≤ y ≤ d.

� Find the average value of f(x, y) on such a region.� Fubini’s theorem: you can change the order of the integral.

13.2: Double integrals over other regions.

� Evaluate a double integral where the inner bounds depend on the outer variable.� Sketch the region of integration for a double integral, given the bounds.� Give the bounds on a double integral, given a description of the region.� Change the order of integration in a double integral over a non-rectangular region.

13.3: Double integrals in polar coordinates.

� Sketch the region of integration for a double integral in polar� Evaluate a double integral in polar.� Convert a double integral in rectangular coordinates into polar:

(1) Write the bounds in polar(2) Write the function in polar (substitute r cos θ for x, r sin θ for Y(3) Write r dr dθ instead of dx dy.

13.4: Triple integrals.

� Evaluate a triple integral.� Set up the bounds for a triple integral in rectangular coordinates (important shapes:

rectangular regions, tetrahedral regions)� Change the order of integration in a triple integral.

13.5: Cylindrical spherical coordinates.

� Find the rectangular coordinates for a point given (r, θ, z)

x = r cos θ, y = r sin θ, z = z.

� Find the cylindrical coordinates for a point given (x, y, z):� Set up the bounds for a triple integral in cylindrical coordinates (Important shapes:

cylinder, paraboloid.)� Convert a rectangular integral into cylindrical coordinates and evaluate (usual three

steps: convert the bounds, convert the function, use r dr dθ dz).� Find the rectangular coordinates for a point given (ρ, θ, φ):

x = ρ sinφ cos θ, y = ρ sinφ sin θ, z = ρ cosφ.

� Set up the bounds for a triple integral into spherical coordinates (Important shapes:spheres, hemispheres, ice cream cones.)

� Convert a rectangular integral into spherical coordinates and evaluate (usual threesteps: convert the bounds, convert the function, use ρ2 sinφ dρ dφ dθ.


13.7: Change of variables in multiple integrals.

� Given a change of coordinates x = g(u, v), y = h(u, v), compute the Jacobian J(u, v):

J(u, v) =

∣∣∣∣∂x∂u ∂x∂v

∂y∂u

∂y∂v

∣∣∣∣ =∂x

∂u

∂y

∂v− ∂x

∂v

∂y

∂u=∂(x, y)

∂(u, v).

� Change an integral in terms of x and y to an integral in terms of u and v:¨R

f(x, y) dA =

¨S

f(g(u, v), h(u, v)) |J(u, v)| dA.

� Find the uv bounds for an integral from the xy bounds (often helpful; convert theequation for each edge of the region into an equation involving uv, and sketch thecorresponding region in the uv-plane).

Chapter 14

14.1: Vector fields.

� Sketch a vector field from the equation.� Write a formula for a vector field given a description.� Compute the gradient field F = ∇f(x, y) for a function f(x, y)

14.2: Line integrals.

� Compute the line integral

ˆC

f(x, y) ds of a scalar function along a path.

• Method: parametrize the path C by r(t) = 〈x(t), y(t)〉. Use bounds as parametriza-tion as bounds on integral, plug in x and y to f , and use |r′(T )| dt for the ds:

ˆC

f ds =

ˆ b

a

f(x(t), y(t)) |r′(t)| dt.

� Compute the line integral of a

ˆC

F · dr of a vector function along a path (i.e. a

circulation integral).• Parametrize C by r(t), then plug in x and y to F, and dot that with r′(t) to get

a function of t.� Compute the flux of a vector field across a path C

• Same method, but use n = 〈y(t),−x(t)〉 instead of r′(t) in the above.

14.3: Conservative vector fields.

� Check whether a 2D vector field F = 〈f, g〉 is conservative.� Check whether a 3D vector field F = 〈f, g, h〉 is conservative.� Find a potential function φ(x, y) for a conservative vector field.� Use the fundamental theorem for line integrals to compute the line integral of a

conservative field along a path C:ˆC

F · dr = φ(end)− φ(begin).

Note: integral is indepedent of path.


14.4: Green’s theorem.

� Compute the divergence and curl of a 2D vector field (both are scalar functions, notvetor fields).

� Be able to apply both versions of Green’s theorem to double integrals over a regionR:• Circulation form: ˛

C

F · dr =

¨R

curl F dA

• Flux form: ˛C

F · n ds =

¨R

div F dA

� Use Green’s theorem to turn a line integral over a complicated path from A to Binto a line integral over a simpler path from A to B and a double integral over theregion between the two paths.

14.5: Divergence and curl.

� Compute the divergence of a 3D vector field ∇ · F (this is a scalar function).� Compute the curl of a 3D vector field:

∇× F =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

f g h

∣∣∣∣∣∣=

(∂h

∂y− ∂g

∂z

)i +

(∂f

∂z− ∂h

∂x

)j +

(∂g

∂x− ∂f

∂y

)k

NB: This is another vector field.

14.6: Surface integrals.

� Parametrize a surface in 3D(1) Cylinder of radius a; x = a cosu, y = a sinu, z = v(2) Sphere of radius a: x = a sinu cos v, y = a sinu sin v, z = a cosu.(3) Graph z = f(x, y) (for example, z = 2(x2 + y2)): x = u, y = v, z = f(u, v).

� Next, know how to compute the tangent vectors tu, tv, and the normal vector tu×tv.

� Compute surface integrals of scalar functions:

¨S

f(x, y, z) dS. Steps to convert to

a double integral in u, v:(1) The bounds are the bounds for your parametrization(2) Rewrite the function in terms of u and v (plug in your parametrization for x, y,

and z)(3) Use |tu × tv| du dv for dS.

� Compute flux integrals of vector fields across a surface:

¨S

F ·n dS. Steps to convert

to a double integral in u, v:(1) The bounds are the bounds for your parametrization(2) Rewrite the function in terms of u and v (plug in your parametrization for x, y,

and z)(3) Plug in x, y, z from parametrization to F(4) Dot that with tu × tv and integrate the result du dv.


14.7: Stokes’ theorem.

� Be able to apply Stokes’ theorem, either to turn a computation of the flux of a curlinto a circulation integral, or to turn a circulation integral into the flux of a curl.¨

S

(∇× F) · n dS =

˛C

F · dr.

� Know how to compute both sides!� Know which way to go around the curve C to make this work (right-hand rule).� Apply Stokes’ theorem on regions with multiple boundaries.

14.8: Divergence theorem.

� Be able to apply divergence theorem, in either direction.‹S

F · n dS =

˚D

∇ · F dV.

� Know how to compute both sides!

Math 210 (Lesieutre)11.1: Vectors in the planeJanuary 9, 2017

Problem 1. Let u = 〈1, 3〉 and v = 〈0,−2〉.

a) Compute u + 2v, both geometrically and algebraically. Do your answers match?

We haveu + v = 〈1, 3〉+ 2 〈0,−2〉 = 〈1, 3〉+ 〈0,−4〉 = 〈1,−1〉 .

b) Compute u− v, both geometrically and algebraically.

For this one,u− v = 〈1, 3〉 − 〈0,−2〉 = 〈1, 5〉 .

c) Compute 3u, both geometrically and algebraically.

Multiplying both components by 3, we obtain

3u = 〈3, 9〉 .

d) What is the magnitude of v (i.e. |v|?) Does the formula match the picture?

For this one, the formula gives |v| =√

02 + (−2)2 =√

4 = 2. This obviously matches thelength we get if we draw the vector, which points straight towards the bottom of the pageand has length 2.

Problem 2. a) What is the vector pointing from (1, 1) to (4,−3)?

Again, try to do this one by drawing the picture. We want the vector

v = 〈4,−3〉 − 〈1, 1〉 = 〈3,−4〉 .

b) Find a unit vector parallel to the vector in your answer from (a).

We need

u =v

|v|=

〈3,−4〉√32 + (−4)2

=〈3,−4〉

5=

⟨3

5,−4

5

⟩.

c) Find a vector with length 7 parallel to the vector in your answer from (a).

We want 7 times our answer from (b), which is

w =

⟨21

5,−28

5

⟩.

Problem 3. Relative to the air, an airplane is flying 30 degrees west of north, with speed500 MPH. The wind is traveling due north at 100 MPH. What is the velocity vector of theairplane relative to the ground?

1

First we need to translate to find the components of the airplane vector. In polar coordinates,our vector has angle 90 + 30 = 120. The x-component is 500 cos 120 = 500(−1/2) = −250.The y-component is 500 sin 120 = 500(

√3/2) = 250

√3. So the velocity relative to the air is⟨

−250, 250√

3⟩. The velocity of the air relative to the ground is 〈0, 100〉.

Let me write vX/Y for the speed of X relative to Y . As we saw in class,

vplane/ground = vplane/air + vair/ground =⟨−250, 250

√3⟩

+ 〈0, 100〉 =⟨−250, 250

√3− 100

⟩.

Problem 4. A 10-pound weight is suspended from two strings, each making a 45 degreeangle with the ceiling. How much force is exerted on the mass by each of the strings?

This is a classic sort of problem. Since the situation is symmetric, the two forces from thestrings are equal in magnitude, but in different directions. Let’s say M is the answer. Thetwo force vectors from the strings are

F1 =

⟨−M√

2

2,M

√2

2

⟩

F2 =

⟨M

√2

2,M

√2

2

⟩

(the first of those is M cos 45◦ for the left string, etc; similar to the previous problem)

The gravitational force is g = 〈0,−10〉: a force of 10 directly downward.

Adding these up we should get 0: if the object isn’t moving, the forces have to balance out:

F1 + F2 + g = 0.

So⟨0,√

2M⟩

+ 〈0,−10〉 = 〈0, 0〉, which means√

2M = 10, so M = 10/√

2 = 5√

2. (Thistells use how strong each string needs to be to keep the mass from falling: each individuallyis holding the same force as if it were supporting a single 5

√2 ≈ 7.07.

2

Math 210 (Lesieutre)11.2: Vectors in three dimensionsJanuary 11, 2017

Problem 1. Consider the two points in three dimensions with coordinates P = (1, 2, 3) andQ = (−1, 2, 1).

a) What is−→PQ? What is

−→QP? How do these differ?

−→PQ is given by (−1, 2, 1) − (1, 2, 3) = 〈−2, 0,−2〉.

−→QP is given by (1, 2, 3) − (−1, 2, 1) =

〈2, 0, 2〉. These differ by a factor of −1: they have the same length, but point in oppositedirections.

b) What is the length of−→PQ?

It’s given by√

(−2)2 + 02 + (−2)2 =√

8 = 2√

2.

c) Find a vector of length 3 parallel to−→PQ.

We want 3−→PQ/

∣∣∣−→PQ∣∣∣ =

⟨−62√2, 0,− −6

2√2

⟩=⟨−3√2

2, 0,−3

√2

2

⟩.

d) What is the midpoint of the line segment between P and Q?

The midpoint formula says it’s ((1 + (−1))/2, (2 + 2)/2, (3 + 1)/2) = (0, 2, 2).

Problem 2. Describe the sphere with equation (x− 1)2 + (y + 2)2 + (z − 3)2 = 25.

The sphere has center (1,−2, 3) and radius 5.

Problem 3. Relative to the air, an airplane is flying 30 degrees west of north, with speed500 MPH. The wind is traveling due north at 100 MPH. What is the velocity vector of theairplane relative to the ground?

First we need to translate to find the components of the airplane vector. In polar coordinates,our vector has angle 90 + 30 = 120. The x-component is 500 cos 120 = 500(−1/2) = −250.The y-component is 500 sin 120 = 500(

√3/2) = 250

√3. So the velocity relative to the air is⟨

−250, 250√

3⟩. The velocity of the air relative to the ground is 〈0, 100〉.

Let me write vX/Y for the speed of X relative to Y . As we saw in class,

vplane/ground = vplane/air + vair/ground =⟨−250, 250

√3⟩

+ 〈0, 100〉 =⟨−250, 250

√3− 100

⟩.

Problem 4. A 10-pound weight is suspended from two strings, each making a 45 degreeangle with the ceiling. How much force is exerted on the mass by each of the strings?

1

This is a classic sort of problem. Since the situation is symmetric, the two forces from thestrings are equal in magnitude, but in different directions. Let’s say M is the answer. Thetwo force vectors from the strings are

F1 =

⟨−M√

2

2,M

√2

2

⟩

F2 =

⟨M

√2

2,M

√2

2

⟩

(the first of those is M cos 45◦ for the left string, etc; similar to the previous problem)

The gravitational force is g = 〈0,−10〉: a force of 10 directly downward.

Adding these up we should get 0: if the object isn’t moving, the forces have to balance out:

F1 + F2 + g = 0.

So⟨0,√

2M⟩

+ 〈0,−10〉 = 〈0, 0〉, which means√

2M = 10, so M = 10/√

2 = 5√

2. (Thistells use how strong each string needs to be to keep the mass from falling: each individuallyis holding the same force as if it were supporting a single 5

√2 ≈ 7.07.

Problem 5. Let R be a parallelogram with legs given by the vectors u and v. Prove thatboth diagonals of R have the same midpoint.

Place u and v with their tails at the origin. The diagonal from the origin to the far cornerof the parallelogram is u + v, and so the position vector for the midpoint is (u + v)/2.

The second diagonal (oriented to start at u and end at v) is the vector v − u. To get theposition vector, we need to add half of this to the position of its tail: u+(v−u)/2 = (u+v)/2.So the two answers match.

2

Math 210 (Lesieutre)11.3: Dot productsJanuary 13, 2017

Problem 1. For each of the following, try to estimate the dot product without making anycalculations. Then check your answer using the formula.

a) 〈3, 4〉 · 〈4, 3〉These both have length 5, and point in roughly the same direction. That means cos θ isgoing to be positive and more or less close to 1. So we expect the dot product to be a littleless than 25.

The correct value is (3)(4) + (4)(3) = 24.

b) 〈3, 1〉 · 〈−1, 4〉If you sketch these, you’ll see that they’re almost perpendicular, but not quite. The dotproduct will be small, at least compared to the lengths of the vectors. Positive or negativemight be hard to tell unless you draw a good picture, but it should be slightly positive.

The actual value is 1.

c) 〈1, 0, 0〉 · 〈0, 1,−1〉The first vector points along the x-axis, while the second lies in the yz-plane . The twovectors are orthogonal, and the dot product will be exactly 0.

The formula confirms this.

d) 〈1, 2, 3〉 · 〈−2,−2,−2〉Both vectors have length somewhere between 3 and 4. They point in basically oppositedirections, so cos θ is going to be fairly close to −1. So the answer should be a negativenumber in the ballpark of −10.

Multiplying it all out, you’ll find the actual value is −12.

Problem 2. Compute the angle between the two vectors in 1(a). (Your answer might be interms of an cos−1).

We know that u · v = |u| |v| cos θ, so

cos θ =u · v|u| |v|

=24

25.

That means θ = cos−1(24/25), which is about 16◦.

Problem 3. Suppose that v = 〈a, b〉 is a vector. What is v · v?It’s 〈a, b〉 · 〈a, b〉 = a2 + b2 = |v|2. This works for 3d vectors too!

1

Problem 4. Sketch the vectors u = 〈2, 1〉 and v = 〈1, 1〉, and draw the vector projv u. Whatis scalv u?

(Sorry, I’m omitting the drawing.)

The formula for projection gives us

projv u =(u · vv · v

)v =

(3

2

)〈1, 1〉 =

⟨3

2,3

2

⟩.

This is a vector in the same direction as v, and represents the part of u that’s in the directionof v.

For the scalar component, we use

scalv u =u · v|v|

=3√2

=3√

2

2.

This is the length of projv u.

Problem 5. A force F = 2, 1, 1 (Newtons) pushes an object from (1, 0, 0) to (3, 0, 0) (me-ters). Calculate the work done (Joules).

It’s just W = F · d = 〈2, 1, 1〉 · 〈2, 0, 0〉 = 4.

Problem 6. A 20 pound block sits on a plane with slope 30 degrees. Compute the componentsof the gravitation force that are parallel and perpendicular to the plane.

The force is F = 〈0,−20〉. The vector “down” the slope is v =⟨√

32,−12

⟩. The vector into

the plane is u =⟨

12,−√32

⟩.

We find

projv F =

(F · vv · v

)v =

⟨5√

3,−5⟩

and

projuF =

(F · uu · u

)u =

⟨−5√

3,−15⟩.

The one into the plane is usually called the normal component and written N.

Does the object slide down the plane? It’s a question of whether the frictional force is strongenough to overcome the force projv F. This depends on the coefficient of friction, whichtakes us a little further into physics than we’re really going to go. . .

Problem 7. a) Describe the set of all vectors v for which v · 〈1, 1, 1〉 = 0.

This is all vectors that are perpendicular to 〈1, 1, 1〉. These form the plane whose normalvector is 〈1, 1, 1〉.

2

b) For which vectors is proj〈1,0,0〉 = 5?

This works for any vector 〈5, a, b〉; the set of such vectors is the plane x = 5.

3

Math 210 (Lesieutre)11.3: Dot productsJanuary 13, 2017

Matt taught this day.

1

Math 210 (Lesieutre)11.5: Lines and curves in spaceJanuary 20, 2017

Problem 1. Consider the line from (1, 1, 1) to (1, 2, 3).

a) Express a parametrization of the line in vector form.

The direction vector is v = (1, 2, 3)− (1, 1, 1) = 〈0, 1, 2〉. Then we want the line

r(t) = 〈1, 1, 1〉+ t 〈0, 1, 2〉 .

This seems to work: when we plug in t = 0, we get 〈1, 1, 1〉, and when we plug in t = 1, weget 〈1, 2, 3〉.

b) Express your parametrization as three functions x(t), y(t), and z(t).

Our parametrization was r(t) = 〈1 + 0t, 1 + 1t, 1 + 2t〉, so we want

x(t) = 1,

y(t) = 1 + t,

z(t) = 1 + 2t.

c) At what time t does the line cross the plane z = 10?

We have z(t) = 1 + 2t, so we just want to know when 1 + 2t = 10, i.e. t = 9/2.

d) What is the projection of the line onto the xy-plane?

It’s x(t) = 1 and y(t) = 1 + t (we’re in the xy-plane, so it’s implicit that z = 0 for all t).

Problem 2. Sketch the parametrized curves indicated below.

a) r(t) = 〈cos t, sin t, 1〉 for t ≤ 0 ≤ 4π.

This one is a circle contained in the horizontal plane z = 1. For t in the range in question,it goes around the circle twice, in a counterclockwise direction when viewed from above.

1

b) r(t) = 〈0, 0, t〉 for 0 ≤ t ≤ 5

This is just a straight line from (0, 0, 0) moving upwards. It ends at (0, 0, 5).

c) r(t) = 〈cos t, sin t, t〉 for 0 ≤ t ≤ 4π.

This one is a helix. It moves around in a circle, while simultaneously going upwards. Sincewe go to 4π, it makes two full cycles around.

Problem 3. Do the two lines

r1(s) = 〈s, 1 + s, 2s− 3〉 ,r2(t) = 〈2− t, 2t, t〉

intersect?

2

To intersect, there must be values of s and t for which s = 2− t, 1 + s = 2t, and 2s− 3 = t.This is an overdetermined system: there may or may not be any solutions. The first two aresolved by s = 1, t = 1, but then the third equation isn’t satisfied. So there is no s and t thatmeets both requirements.

Problem 4. For what value of a do the two lines

r1(s) = 〈3 + s, 1 + s, a+ s〉 ,r2(t) = 〈2t+ a, t+ a, 0〉

intersect?

What does it mean for them to intersect? It means that there is a value s and a value t forwhich r1(s) = r2(t) (they might pass through the same point, but at two different values ofthe parameter).

So we need to solve the equations 3 + s = 2t, 1 + s = t, and a+ s = 0 for s and t, in termsof a. For most values of a there are going to be no solutions, but for some special a therewill be one.

First we use the first two to solve for s and t (in principle our answer could involve a, butit’s not going to this time). Subtracting the second equation from the first (or doing rowreduction, or matrices,. . . ), we get s = 1 + a and t = 2.

Now we need to know whether the third coordinates are also equal for these values of s andt. Is a + s = 0? Well, s = 1 + a, so we are asking whether 2a + 1 = 0. This is the case ifa = −1/2.

Again: if a has some other value, there is no s and t that make all the components equal.But if a = −1/2, the lines do intersect.

Problem 5. What is limt→0+ v(t), where v(t) =⟨t, sin t

t, e−1/t

⟩?

We just take the limit of each one of the components, whcih gives us 〈0, 1, 0〉.

3

Math 210 (Lesieutre)11.6: Calculus of vector valued functionsJanuary 23, 2016

Problem 1. The position of a planet is given by r(t) = 〈3 cos t, 2 sin t〉. Find a unit tangentvector to the path of the planet.

First we find thatr′(t) = 〈−3 sin t, 2 cos t〉 .

This gives

|r′(t)| =√

9 sin2 t+ 4 cos2 t.

Nothing to be done with that, really, but it gives

r′(t)

|r′(t)|=

⟨3 cos t√

9 sin2 t+ 4 cos2 t,

2 sin t√9 sin2 t+ 4 cos2 t

⟩.

Problem 2. Consider the function r(t) = 〈cos t, sin t, t〉

a) Sketch r(t) for 0 ≤ t ≤ 4π.

This one we did last time. It’s the helix sketched below.

b) Compute r′(t) and r′′(t). Plug in t = 0 and t = π/2 to your answer. Do these makesense? Can you think of any other “sanity checks”?

The derivatives are given by

r′(t) = 〈− sin t, cos t, 1〉r′′(t) = 〈− cos t,− sin t, 0〉

1

When t = 0, we have r′(0) = 〈0, 1, 1〉. This is a vector whose projection to the xy-planepoints in the y-direction, but also has z part equal to 1. This seems to match up with thepicture. Similarly, we find that r′′(0) = 〈−1, 0, 1〉, which points in the negative x-directionand upwards. Seems plausible.

You might also notice that r′′(t) = 0 for all t. This also makes sense: the rate at which theparticle is moving upwards is not changing.

c) Find a unit tangent vector for the parametrized curve, in terms of t.

We know that the tangent vector is given by

r′(t) = 〈sin t, cos t, 1〉 ,

for any t. The problem is that this isn’t a unit vector, since

|r′(t)| =√

sin2 t+ cos2 t+ 12 =√

2.

To get a unit vector, we want

r′(t)

|r′(t)|=

⟨sin t√

2,cos t√

2,

1√2

⟩.

Problem 3. Consider the path given by r(t) = 〈t, t2, t3〉.

a) Write down an equation for the tangent line to r(t) at t = 1.

The direction of the line is given by r′(t) = 〈1, 2t, 3t2〉. At t = 1 this is 〈1, 2, 3〉. It goesthrough the point r(1) = 〈1, 1, 1〉, so the equation is given by

x(t) = 〈1, 1, 1〉+ 〈1, 2, 3〉 t.

b) What is ddt

(r(t) · 〈1, 2, 3〉)?Let s(t) = 〈1, 1, 3〉. Then d

dt(r(t) · s(t)) = r′(t) · s(t) + r(t) + s′(t) = 〈1, 2t, 3t2〉 · 〈1, 2, 3〉 +

〈t, t2, t3〉 · 〈0, 0, 0〉 = 1 + 4t+ 9t2.

Problem 4. a) What isd

dt|x(t)|2 ,

in terms of x(t) and its derivatives?

This isd

dtx(t) · x(t) = x(t) · x′(t) + x′(t) · x(t) = 2x(t) · x′(t).

b) Check your answer to (a) for x(t) = 〈cos t, sin t〉. Does your answer make sense?

According to the product rule, it’s x(t) · x′(t) + x′(t) · x(t) = 〈cos t, sin t〉 · 〈− sin t, cos t〉 +〈− sin t, cos t〉 · 〈cos t, sin t〉 = 0.

2

c) Suppose that x(t) has constant length. What does this tell you about the relationshipbetween x(t) and x′(t)?

It tells you that they are orthogonal for all values of t!

Problem 5. Let s(t) = 〈− sin t, cos t, 1〉. Compute

∫s(t) dt.

We just integrate each component:∫〈− sin t, cos t, 1〉 = 〈cos t, sin t, t〉+ 〈c1, c2, c3〉 .

This undoes what we did in the second problem. We’ll see next time that if s(t) is thevelocity of a particle, r(t) is its position.

3

Math 210 (Lesieutre)11.7: Motion in spaceJanuary 25, 2017

Problem 1. Consider the path r(t) = 〈cos t, sin t, t2〉. Compute the velocity, speed, andacceleration, all as functions of t.

First let’s find the velocity and acceleration, which are just a matter of taking derivatives.We get

r(t) =⟨cos t, sin t, t2

⟩,

v(t) = 〈− sin t, cos t, 2t〉 ,a(t) = 〈− cos t,− sin t, 2〉 .

The speed should be a scalar function, and to get it we take the length of velocity, as afunction of time:

|v(t)| =√

(− sin t)2 + (cos t)2 + (2t)2 =√

1 + 4t2.

This seems plausible, I think. When t is big, it’s moving upwards faster and faster, which isreflected in the speed increasing as t increases.

Problem 2. A particle moves in a circular pattern, given by r(t) = 〈2 cos t, 2 sin t〉.

a) Compute the velocity, acceleration, and speed, as functions of t.

Taking derivatives, we get

r(t) = 〈2 cos t, 2 sin t〉 ,v(t) = 〈−2 sin t, 2 cos t〉 ,a(t) = 〈−2 cos t,−2 sin t〉 .

The speed is|v(t)| =

√(−2 sin t)2 + (2 cos t)2 =

√4 = 2,

which doesn’t depend on t. This makes sense, since it’s moving at a constant rate. Noticethat the velocity vector v(t) does depend on t: this reflects the fact that the direction ischanging.

b) Sketch the path of the particle. For t = π/4, draw the vectors r(t), v(t), and a(t). Dothese seem to make physical sense?

The vectors are r(t) =⟨√

2,√

2⟩, v(t) =

⟨−√

2,√

2⟩, and a(t) =

⟨−√

2,√

2⟩.

The velocity vector is tangent to the circle at the point, which makes sense: velocity issupposed to be the tangent vector. The acceleration points in to the circle, which may seema little weird at first, but this is how things work for circular motion. This is consistent withNewton’s law F = ma: because gravitation force pulls the particle inward, the accelerationshould also be inward.

1

Problem 3. Suppose that a particle moves in a straight-line path r(t) = r0 + vt. What arethe velocity and acceleration?

The velocity will just be the vector v, while the acceleration is the zero vector 0. (Becausethe particle is moving at constant velocity, there is no acceleration.)

Problem 4. A particle has acceleration a(t) = 〈2t,−1〉. Suppose that at time 0, it hasposition r(0) = 〈2, 3〉 and velocity v(t) = 〈0, 1〉. Find r(t).

We know that

v(t) =

∫a(t) dt =

∫〈2t,−1〉 dt =

⟨t2,−t

⟩+ C1

Since v(0) = 〈0, 1〉, we get 〈0, 0〉+ C1 = 〈0, 1〉, so C1 = 〈0, 1〉. That means

v(t) =⟨t2,−t

⟩+ 〈0, 1〉 =

⟨t2, 1− t

⟩.

Then

r(t) =

∫v(t) dt =

∫ ⟨t2, 1− t

⟩dt =

⟨t3

3, t− t2

2

⟩+ C2

Plugging in t = 0, we get〈0, 0〉+ C2 = r(0) = 〈2, 3〉 ,

which means that C2 = 〈2, 3〉. So at last we have

r(t) =

⟨t3

3, t− t2

2

⟩+ 〈2, 3〉 =

⟨t3

3+ 2, t− t2

2+ 3

⟩.

Problem 5. A batter hits a baseball with initial velocity 〈100, 100, 100〉 (a pop-up down theright field line; let’s say the units are ft/sec).

a) What is a(t)? Assume the only force acting on the ball is gravity.

It’s a constant downward force, 〈0, 0,−32〉. The 32 here is gravitation acceleration.

b) Solve for v(t) and r(t).

Well,

v(t) =

∫a(t) dt = 〈0, 0,−32t〉+ C1.

Since v(0) = 〈100, 100, 100〉, plugging in t = 0 we find that C1 = 〈100, 100, 100〉. This meansthat

v(t) = 〈0, 0,−32t〉+ 〈100, 100, 100〉 = 〈100, 100, 100− 32t〉 .Then

r(t) =

∫v(t) dt =

⟨100t, 100t, 100t− 16t2

⟩+ C2.

Since r(0) = 0, the constant vector C2 is 0 too. Thus

r(t) =⟨100t, 100t, 100t− 16t2

⟩.

2

c) At what time t does the ball hit the ground?

This will happen when 100t− 16t2 = 0, so that t = 100/16 = 6.25 seconds.

Problem 6. a) What isd

dt|x(t)|2 ,

in terms of x(t) and its derivatives?

This isd

dtx(t) · x(t) = x(t) · x′(t) + x′(t) · x(t) = 2x(t) · x′(t).

b) Suppose that x(t) has constant length. What does this tell you about the relationshipbetween x(t) and x′(t)?

It tells you that they are orthogonal for all values of t!

3

Math 210 (Lesieutre)11.8: Length of curvesJanuary 27, 2017

Problem 1. Let R be a circle centered at 0 with radius 5. Set up a parametrization r(t) forthe circle, and use it to compute the circumference.

This one is pretty painless. Use

r(t) = 〈5 cos t, 5 sin t〉

with 0 ≤ t ≤ 2π. Then r′(t) = 〈−5 sin t, 5 cos t〉, which gives us

|r′(t)| =√

(−5 sin t)2 + (5 cos t)2 =√

25 = 5.

This doesn’t depend on t! So the arc length is∫ 2π

0

5 dt = 10π.

This matches up with the 2πR you probably learned before you took calculus.

Problem 2. Consider the curve r(t) =⟨cos t, sin t, 2

3t3/2⟩. Compute the length of this curve

between t = 0 and t = 10.

We haver′(t) =

⟨− sin t, cos t, t1/2

⟩,

which gives|r′(t)| =

√(− sin t)2 + (cos t)2 + t =

√1 + t.

The formula for arc length then tells us that

L =

∫ 10

0

√1 + t dt =

(2

3(1 + t)3/2

)∣∣∣∣100

=2

3113/2 − 2

3=

2

3(11√

11− 1).

Sanity check? That’s about 23.65, which seems in the right ballpark at least.

Problem 3. Consider a cardioid r = 1 + cos θ. Set up the integral for the arc length, andevaluate it if you can.

1

We have f(θ) = 1 + cos θ and so f ′(θ) = − sin θ. The formula we just saw gives

L =

∫ β

α

√f(θ)2 + f ′(θ)2 dθ

=

∫ 2π

0

√(1 + cos θ)2 + (− sin θ)2 dθ

=

∫ 2π

0

√1 + 2 cos θ + cos2 θ + sin2 θ dθ

=

∫ 2π

0

√2 + 2 cos θ dθ = 2

∫ π

0

√2 + 2 cos θ dθ

=

∫ 2π

0

√4 cos2(θ/2) dθ.

There is a subtle point here. It’s pretty tempting to just write√

4 cos2(θ/2) = 2 cos(θ/2).But here’s the catch: it’s really |2 cos(θ/2)|, which isn’t the same thing if cos(θ/2) is negative.So we can’t make this simplification. However, notice that the cardioid is symmetrical: thelength from 0 to π is equal to the length from π to 2π, so it is true that∫ 2π

0

√4 cos2(θ/2) dθ = 2

∫ π

0

√4 cos2(θ/2) dθ.

Now, when 0 ≤ θ ≤ π, we have 0 ≤ θ/2 ≤ π/2, and so cos(θ/2) is guaranteed to be positive.That means for θ in this range, it’s actually true that

√4 cos2(θ/2) = 2 cos(θ/2). So our

integral becomes

4

∫ π

0

cos(θ/2)dθ = 8 sin(θ/2)|π0 = 8.

That was a little painful, but I think the result is interesting: the length of the cardioid is8. No square roots, no π, just 8.

Problem 4. The circle in the first problem was not parametrized by arc length. Give anotherparametrization in which it is.

The issue is that the particle is moving too fast: it’s going at a constant speed of 5. If weuse r(t) = 〈5 cos(t/5), 5 sin(t/5)〉 instead, it will slow down, and now is parametrized by arclength. Then r′(t) = 〈− sin(t/5), cos(t/5)〉, which gives us

|r′(t)| =√

(− sin(t/5))2 + (cos(t/5))2 =√

1 = 1,

which is what it means to be parametrized by arc length.

Problem 5. A batter hits a baseball with initial velocity 〈100, 100, 100〉 (a pop-up down theright field line; let’s say the units are ft/sec).

2

a) What is a(t)? Assume the only force acting on the ball is gravity.

It’s a constant downward force, 〈0, 0,−32〉. The 32 here is gravitational acceleration.

b) Solve for v(t) and r(t).

Well,

v(t) =

∫a(t) dt = 〈0, 0,−32t〉+ C1.

Since v(0) = 〈100, 100, 100〉, plugging in t = 0 we find that C1 = 〈100, 100, 100〉. This meansthat

v(t) = 〈0, 0,−32t〉+ 〈100, 100, 100〉 = 〈100, 100, 100− 32t〉 .

Then

r(t) =

∫v(t) dt =

⟨100t, 100t, 100t− 16t2

⟩+ C2.

Since r(0) = 0, the constant vector C2 is 0 too. Thus

r(t) =⟨100t, 100t, 100t− 16t2

⟩.

c) At what time t does the ball hit the ground?

This will happen when 100t− 16t2 = 0, so that t = 100/16 = 6.25 seconds.

d) Set up the integral for the length of the path of the ball between the moment it left the batand the moment it hit the ground.

We have r′(t) = 〈100, 100, 100− 32t〉, and so

L =

∫ 6.25

0

|r′(t)| dt =

∫ 6.25

0

√1002 + 1002 + (100− 32t)2 dt =

∫ 6.25

0

√20000 + (100− 32t)2 dt

=625

2

(√

3 + 2 arcsinh

(√2

2

))≈ 952.815.

(I’d say you don’t need to worry about how to actually do that integral.)

3

Math 210 (Lesieutre)12.1: Planes and surfaces, part 1January 30, 2017

Problem 1. a) What is the equation for a plane passing through (1, 2, 3) and with normalvector 〈2,−1, 3〉?The equation is 2(x− 1)− (y − 2) + 3(z − 3) = 0, which simplifies to 2x− y + 3z = 9.

b) At what point does this plane meet the x-axis?

A point on the x-axis has y = 0 and z = 0. So we need 2(x − 1) − (0 − 2) + 3(0 − 3) = 0,which means 2x = 9. So the point is (9/2, 0, 0).

c) At what point does this plane meet the line r(t) = 〈−t− 1, 2t, t + 1〉?We need to know for what value of t the equation of the plane is satisfied, which means that2(−1− t)− 2t + 3(1 + t) = 9. This gives 1− t = 9, so t = −8.

We are supposed to find the point where they intersect, rather than the time t. So we shouldplug this back in to our original equation: r(−8) = 〈7,−16,−7〉. This does indeed satisfy2x− y + 3z = 9

Problem 2. a) Find the equation of the plane containing the three points P = (1, 3, 1),Q = (1, 0, 3), and R = (5, 4,−1).

We need to find the normal vector. It will be perpendicular to both ~PQ and ~PR, so we wantto take the cross product to find n. We obtain

~PQ = (1, 0, 3)− (1, 3, 1) = (0,−3, 2)

~PR = (5, 4,−1)− (1, 3, 1) = (4, 1,−2)

~pq × ~pr =

∣∣∣∣∣∣i j k0 −3 21 1 −2

∣∣∣∣∣∣ = 4i + 8j + 12k = 〈4, 8, 12〉 .

That’s our normal vector. To get the equation, we just need to plug in a point it passesthrough. We might as well use P . You could also use Q or R; you’ll get the same answerfor the plane at the end of the day.

So the equation is

a(x− x0) + b(y − y0) + c(z − z0) = 0

4(x− 1) + 8(y − 3) + 12(z − 1) = 0

4x + 8y + 12z = 40

x + 2y + 3z = 10.

b) Draw a sketch of the plane by computing its traces in the three coordinate planes.

In the xy-plane, we plug in z = 0 and get 4x + 2y = 13. In the xz-plane, we plug in y = 0and get 4x + 3z = 13. In the yz-plane, we plug in x = 0 and get 2y + 3z = 13.

These three planes are plotted below.

1

xy-plane:

-10 -5 0 5 10

-10

-5

0

5

10

xz-plane:

-10 -5 0 5 10

-10

-5

0

5

10

yz-plane:

-10 -5 0 5 10

-10

-5

0

5

10

Putting it all together, we get the following plane. I’m drawing a “back view” that makesit easier to see the axes.

Problem 3. Graph the cylinder x2 + 2x + 4z2 = 15.

This doesn’t involve y, so we should try to graph the x and z part and then “draw the walls”by including all possible y-values.

2

First, complete the square to obtain (x − 1)2 + 4z2 = 16, or perhaps easier, ((x − 1)/4)2 +(z/z)2 = 1 This you should recognize: it’s an ellipse. It looks like this:

-5 0 5

-5

0

5

Problem 4. Consider the three planes given by the following equations:

x + 2y + 3z = 0

x− y + z = 2

x + 2y + 3z = 6.

a) Two of these planes are parallel: which two? Describe the intersection of these planes.

It’s the first and the third: the way you can tell is that they both have the same normalvector n = 〈1, 2, 3〉.

b) The first and second planes intersect in a line. Give a parametrization of this line, andcheck that your line is actually contained in the first plane.

First we need to find a point on the line. There are many points, and you can solve for onehowever you want. One way is just plug in z = 0 and then figure out what x and y have tobe. You can also just guess one, which is what I am about to do: (3, 0,−1).

Now we need to find the direction vector. How to do that? Well, it’s perpendicular to bothof the normal vectors, which are 〈1, 2, 3〉 and 〈1,−1, 1〉. So we can use the cross productv = 〈1, 2, 3〉 × 〈1,−1, 1〉 as our normal vector! This guy is v = 〈5, 2,−3〉.So our line is

r(t) = r0 + tv = 〈3, 0,−1〉+ t 〈5, 2,−3〉 = 〈3 + 5t, 2t,−1− 3t〉 .

To see whether it indeed line in the plane, just plug this in to the equation for the plane.You get x+ 2y + 3z = (3 + 5t) + 2(2t) + 3(−1− 3t) = 0, so it works as it should. You couldcheck the same thing with the second plane if you were so inclined.

3

Math 210 (Lesieutre)12.1: Planes and surfaces, part 2February 1, 2017

Problem 1. Consider the two planes given by the following equations:

x + 2y + 3z = 0

x− y + z = 2

These planes intersect in a line. Give a parametrization of this line, and check that yourline is actually contained in the first plane.

First we need to find a point on the line. There are many points, and you can solve for onehowever you want. One way is just plug in z = 0 and then figure out what x and y have tobe. You can also just guess one, which is what I am about to do: (3, 0,−1).

Now we need to find the direction vector. How to do that? Well, it’s perpendicular to bothof the normal vectors, which are 〈1, 2, 3〉 and 〈1,−1, 1〉. So we can use the cross productv = 〈1, 2, 3〉 × 〈1,−1, 1〉 as our normal vector! This guy is v = 〈5, 2,−3〉.So our line is

r(t) = r0 + tv = 〈3, 0,−1〉+ t 〈5, 2,−3〉 = 〈3 + 5t, 2t,−1− 3t〉 .

To see whether it indeed line in the plane, just plug this in to the equation for the plane.You get x+ 2y + 3z = (3 + 5t) + 2(2t) + 3(−1− 3t) = 0, so it works as it should. You couldcheck the same thing with the second plane if you were so inclined.

Problem 2. a) Graph the cylinder x2 + 2x + 4z2 = 15.

This doesn’t involve y, so we should try to graph the x and z part and then “draw the walls”by including all possible y-values.

First, complete the square to obtain (x − 1)2 + 4z2 = 16, or perhaps easier, ((x − 1)/4)2 +(z/z)2 = 1 This you should recognize: it’s an ellipse. It looks like this:

Problem 3. Consider the surface given by

x2

25+

y2

25+ z2 = 1.

a) Sketch the xy-, xz-, and yz- traces of this figure. Use these to guide a drawing of theentire surface.

xy-plane:

1

-6 -4 -2 0 2 4 6

-6

-4

-2

0

2

4

6

xz-plane:

-6 -4 -2 0 2 4 6

-6

-4

-2

0

2

4

6

yz-plane:

-6 -4 -2 0 2 4 6

-6

-4

-2

0

2

4

6

Putting it all together, we get the following figure. I’m drawing a “back view” that makesit easier to see the axes.

b) Now consider the line r(t) = 〈3t, 4t, 1− t〉. At what points does the line intersect thesurface?

We do the same procedure we used on Monday: plug in x(t) = 3 − 3t, y(t) = 4 − 4t, and

2

z(t) = 5t to our equation and solve for t. This gives

x2

25+

y2

25+ z2 = 1

x2 + y2 + 25z2 = 25

(3− 3t)2 + (4− 4t)2 + 25(1− t)2 = 25

25− 50t + 50t2 = 25

−50t + 50t2 = 0,

which means either t = 0 or t = 1. The points are then r(0) = 〈0, 0, 1〉 and r(1) = 〈3, 4, 0〉.

Problem 4. Consider the figure x2 + 2x + y2 − 4y − z2 + 4 = 0.

a) First, complete the square to simplify the equation for the surface.

x2 + 2x + y2 − 4y − z2 = −4

x2 + 2x+1 + y2 − 4y+4− z2 = 5− 4

(x + 1)2 + (y − 2)2 − z2 = 1

b) Next, sketch some traces of the figure. You can use the coordinate planes, but it might bebetter to use planes parallel to coordinate planes (for example, y = 2 might be a good one)

The plane z = 0. The trace is (x + 1)2 + (y − 2)2 = 1, a circle of radius 1.

-4 -2 0 2 4

-4

-2

0

2

4

The plane y = 2. The trace is (x + 1)2 − z2 = 1, a hyperbola.

-4 -2 0 2 4

-4

-2

0

2

4

the plane x = −1:

3

-4 -2 0 2 4

-4

-2

0

2

4

c) Use your traces to sketch the 3D surface.

Putting it all together, we get the following figure. I’m drawing a “back view” that makesit easier to see the axes.

4

Math 210 (Lesieutre)12.3: Limits and continuityFebruary 6, 2016

Problem 1. a) Evaluate the limit:

lim(x,y)→(1,2)

2x2 +√xy

This one we can do using the rules for limits.

lim(x,y)→(1,2)

2x2 +√xy = lim

(x,y)→(1,2)2x2 + lim

(x,y)→(1,2)

√xy

= 2

(lim

(x,y)→(1,2)x2)2

+√

lim(x,y)→(1,2)

xy = 2 +√

2.

b) Evaluate the limit:

lim(x,y)→(3,0)

3x2 + y

x+ y

This is another one where you can just “plug in” and things will work OK. The importantthing to remember is that to take a limit of a quotient, you can just take quotient of thelimits, unless the limit of the denominator is 0!

Since we’re looking at the point (3, 0), the limit is just 9 (what you get when you plug inx = 3 and y = 0).

Problem 2. Evaluate the following limits along the path y = mx. Does the limit exist?

a)

lim(x,y)→(0,0)

x4 + y4

x2 + y2

This gives us

lim(x,y)→(0,0)along y = mx

x4 + y4

x2 + y2= lim

x→0

x4 + (mx)4

x2 + (mx)2

= limx→0

(1 +m4)x4

(1 +m2)x2= lim

x→0

1 +m4

1 +m2x2 = 0,

no matter what m is. So at least for these paths, the limit doesn’t depend on the path. Wecan’t really be sure from this that the limit actually exists, but in fact it does.

b)

lim(x,y)→(0,0)

(x+ y)2

x2 + y2

1

Let’s again use y = mx as our path.


(x+ y)2

x2 + y2= lim

x→0

(x+mx)2

x2 + (mx)2

= limx→0

(1 +m)2

1 +m2=

(1 +m)2

1 +m2.

This answer depends on m, which means that the limit doesn’t exist.

Problem 3. a) Give an example of a function f(x, y) that is not continuous at (1, 2). Canyou come up with an example that is continuous for all (x, y) other than (1, 2)?

One function you could use is

f(x, y) =1

(x− 1)2 + (y − 2)2.

As (x, y) gets closer and closer to (1, 2), this gets closer and closer to infinity, and the functionis not defined there.

b) Give an example of a function g(x, y) that is not continuous for any (x, y) satisfyingy = x2.

You can use a similar idea here as on the last one: think about the function

g(x, y) =1

y − x2.

If y = x2, you can’t plug it in to this equation.

Problem 4. Is the function

p(x, y) =

{x2y2

x4+y4if (x, y) 6= (0, 0),

0 if (x, y) = (0, 0).

continuous at (0, 0)?

Being continuous means three things: the function is defined at (0, 0), the limit exists, andthe limit is equal to the function. It is defined there (p(0, 0) = 0 by fiat). So we need tothink about the limit. Let’s try the limit along the line y = mx.


x2y2

x4 + y4= lim

x→0

x2(mx)2

x4 + (mx)4

= limx→0

(m2)

(1 +m4)=

m2

1 +m4,

so the direction limit depends on the path, which means that the limit doesn’t exist, andthe function isn’t continuous.

2

Problem 5. Set up the integral for the arc length of r(t) = 〈cos t, sin t〉 for 0 ≤ t ≤ π.

We have

r′(t) = 〈− sin t, cos t〉|r′(t)| =

√(− sin t)2 + (cos t)2 = 1,

and so

L =

∫ π

0

√|r′(t)| dt =

∫ π

0

1 dt = π.

3

Math 210 (Lesieutre)12.4/12.5: Partial derivatives and the chain ruleApril 28, 2017

Problem 1. Compute the indicated partial derivatives:

a) ∂∂x

(x2y)

It’s 2xy: just think of this as being a constant times x2, and the derivative is the constant(i.e. y) times 2x.

b) ∂∂y

(x2y)

This one is x2, just as the derivative of 7y is 7 – you have to get used to thinking of the x’sas constants.

c) fy, where f(x, y, z) = exyz

(xz)exyz: treat the x and z in the exponent as a constant, and use the chain rule.

d) gy, where g(x, y) = y cosx

It’s just cosx.

Problem 2. Consider the function f(x, y) = x cos(xy). Compute the four second partialderivatives.

The first partials are:

fx = cos(xy) − xy sin(xy)

fy = −x2 sin(xy)

That gives the second partials as:

fxx = −y sin(xy) − y(xy cos(xy) + sin(xy))

= −xy2 cos(xy) − 2y sin(xy)

fxy = −x sin(xy) − x(xy cos(xy) + sin(xy))

= −x2y cos(xy) − 2x sin(xy)

fyx = −(x2(y cos(xy)) + 2x sin(xy))

= −x2y cos(xy) − 2x sin(xy)

fyy = −x3 cos(xy)

The thing I want you to notice here is that fxy = fyx. This is always happens, and it’s called“Clairaut’s theorem”.

1

Problem 3. Suppose that f(x, y) = x2 + y2, and that x(t) = 2t, y(t) = sin t. What is dfdt

?

Here we have to use the chain rule. The chain rule for one independent variable tells us that

df

dt=

∂f

∂x

dx

dt+

∂f

∂y

dy

dt

= (2x)(2) + (2y)(cos t) = (2 · 2t)(2) + (2 sin t)(cos t)

= 8t + 4 sin t cos t = 8t + 2 sin(2t).

Problem 4. Suppose that f(x, y) = xy2, and x(s, t) = 2s + t and y(s, t) = s cos t. Computethe partial derivative ∂f

∂s.

This is a little more painful. Again we have to use the chain rule.

∂f

∂s=

∂f

∂x

∂x

∂s+

∂f

∂y

∂y

∂s

= (y2)(2) + (2xy)(cos t) = 2(s cos t)2 + (2)(2s + t)(s cos t)(cos t)

= 2s2 cos2 t + (4s2 + 2st) cos2 t = 6s2 cos2 t + 2st cos2 t.

Problem 5. Consider the ellipse defined by F (x, y) = 0, where F (x, y) = x2 + xy + y2 − 1.Compute dy

dx.

The formula for implicit differentiation gives

dy

dx= −Fx

Fy

= −2x + y

2y + x.

Let’s make sure we actually understand what this is saying. Here’s the ellipse:

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

(It’s tilted because of the xy-term; you probably haven’t plotted one like that, and it’s toughuntil you’ve taken Math 310.)

2

One point on the ellipse is (x, y) = (1, 1). Our formula is telling us that dydx

, which is the

slope of the tanegnt line, is given by −2x+y2y+x

= −33

= −1 at this point. That appears tomatch up with the figure. So we were able to find the slope of the tangent line, even thoughwe don’t actually have a formula for y as a function of x.

3

Math 210 (Lesieutre)12.6: Directional derivatives and the gradientApril 28, 2017

Problem 1. Suppose that f(x, y) = xy2, and x(s, t) = 2s+ t and y(s, t) = s cos t. Computethe partial derivative ∂f

∂s.

This is a little more painful. Again we have to use the chain rule.

∂f

∂s=∂f

∂x

∂x

∂s+∂f

∂y

∂y

∂s

= (y2)(2) + (2xy)(cos t) = 2(s cos t)2 + (2)(2s+ t)(s cos t)(cos t)

= 2s2 cos2 t+ (4s2 + 2st) cos2 t = 6s2 cos2 t+ 2st cos2 t.

Problem 2. Consider the ellipse defined by F (x, y) = 0, where F (x, y) = x2 + xy + y2 − 1.Compute dy

dx.

The formula for implicit differentiation gives

dy

dx= −Fx

Fy

= −2x+ y

2y + x.

Let’s make sure we actually understand what this is saying. Here’s the ellipse:

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

(It’s tilted because of the xy-term; you probably haven’t plotted one like that, and it’s toughuntil you’ve taken Math 310.)

One point on the ellipse is (x, y) = (1, 1). Our formula is telling us that dydx

, which is the

slope of the tanegnt line, is given by −2x+y2y+x

= −33

= −1 at this point. That appears tomatch up with the figure. So we were able to find the slope of the tangent line, even thoughwe don’t actually have a formula for y as a function of x.

1

Problem 3. Consider the function f(x, y) = 10− x2 − 4y2.

a) Compute the gradient ∇f(x, y).

The gradient is given by 〈fx, fy〉 = 〈−2x,−8y〉.

b) Find the derivative in the direction of the vector v = 〈1, 1〉 at the point (1, 1). (Watchout! This isn’t a unit vector.)

First we need to know a unit vector in the direction of v. That’s given by u =⟨√

22,√22

⟩.

Now the directional derivative is

Duf(a, b) = ∇f(a, b) · u = 〈−2,−8〉 ·

⟨√2

2,

√2

2

⟩= −√

2− 4√

2 = −5√

2.

c) Find the directional derivative in the direction of the vector u = 〈0,−1〉 at (1, 1).

Same as strategy as above:

Duf(a, b) = ∇f(a, b) · u = 〈−2,−8〉 · 〈0,−1〉 = 8.

d) Sketch some level curves of f(x, y), including the level curve with z = 5.

The level curves look like x2 + 4y2 = C, which are ellipses. Here are a few.

-30

-30

-30

-30

-25

-25

-20

-20

-15

-15

-10

-10

-5

-5

0

0

5

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

z = 5 is the innermost one that’s marked. Notice that the point (1, 1) is contained in thislevel curve.

Here’s a graph of the whole surface, for what it’s worth:

2

e) Find the unit vectors in the directions of steepest ascent and descent at the point (1, 1).Do your answers make sense?

The gradient is given by 〈−2,−8〉, which means steepest ascent is in the direction of 〈−2,−8〉.A unit vector in this direction is

⟨−2/√

68,−8/√

68⟩

(a mess, sorry).

Looking at this on the plot, this vector points “inwards” on the ellipse. Makes sense: thewhole surface is shaped like a “hill”, and the inwards direction is uphill.

Steepest descent is in the opposite direction, which is⟨2/√

68, 8/√

68⟩. This is “downhill”,

the direction a ball would roll if placed on the graph.

f) Find the directional derivative in the direction of steepest ascent. Is this steeper than theanswers you got for directional derivatives earlier in the problem?

The directional derivative is given by

Duf(a, b) = ∇f(a, b) · u = 〈−2,−8〉 ·⟨−2/√

68,−8/√

68⟩

= 68/√

68 =√

68.

This is slightly more than the 8 that we got in part (c), so it’s plausible that this is indeedthe direction of steepest ascent.

g) Find a direction that is tangent to the level curve z = 5 at the point (1, 1). What is thedirectional derivative in this direction?

We want a direction that’s perpendicular to the gradient. One option is⟨8/√

68,−2/√

68⟩,

which will work because the dot product is 0. This points to the right and slightly down,which looks plausible for a tangent direction to the level curve at (1, 1), based on the picture.

The directional derivative is 0, because it’s tangent to a level curve: the function doesn’tchange in this direction. Taking the dot product with the gradient confirms this.

Problem 4. Suppose that a function has gradient ∇f(0, 0) = (1, 1).

a) What is the directional derivative of this function in a direction with angle θ?

The unit vector we want is 〈cos θ, sin θ〉, and so the directional derivative is

∇f(0, 0) · u = cos θ + sin θ.

3

b) Plot the directional derivative for 0 ≤ θ ≤ 2π. For what θ is it maximized? Zero?

Here’s a plot:

1 2 3 4 5 6

-1.5

-1.0

-0.5

0.5

1.0

1.5

It’s maximized at θ = π/4, which is the direction parallel to ∇f . It’s 0 at 3π/4 and 7π/4,which is orthogonal to the gradient.

4

Math 210 (Lesieutre)12.6 Directional derivatives, and some reviewApril 28, 2017

Problem 1. Suppose that a function has gradient ∇f(0, 0) = (1, 1).

a) What is the directional derivative of this function in a direction with angle θ?

The unit vector we want is 〈cos θ, sin θ〉, and so the directional derivative is

∇f(0, 0) · u = cos θ + sin θ.

b) Plot the directional derivative for 0 ≤ θ ≤ 2π. For what θ is it maximized? Zero?

Here’s a plot:

1 2 3 4 5 6

-1.5

-1.0

-0.5

0.5

1.0

1.5

It’s maximized at θ = π/4, which is the direction parallel to ∇f . It’s 0 at 3π/4 and 7π/4,which is orthogonal to the gradient.

Problem 2. Consider the function f(x, y) = 10− x2 − 4y2.

a) Sketch some level curves of f(x, y), including the level curve with z = 5.

The level curves look like x2 + 4y2 = C, which are ellipses. Here are a few.

-30

-30

-30

-30

-25

-25

-20

-20

-15

-15

-10

-10

-5

-5

0

0

5

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

1

z = 5 is the innermost one that’s marked. Notice that the point (1, 1) is contained in thislevel curve.

Here’s a graph of the whole surface, for what it’s worth:

b) Find the unit vectors in the directions of steepest ascent and descent at the point (1, 1).Do your answers make sense?

The gradient is given by 〈−2,−8〉, which means steepest ascent is in the direction of 〈−2,−8〉.A unit vector in this direction is

⟨−2/√

68,−8/√

68⟩

(a mess, sorry).

Looking at this on the plot, this vector points “inwards” on the ellipse. Makes sense: thewhole surface is shaped like a “hill”, and the inwards direction is uphill.

Steepest descent is in the opposite direction, which is⟨2/√

68, 8/√

68⟩. This is “downhill”,

the direction a ball would roll if placed on the graph.

c) Find the directional derivative in the direction of steepest ascent. Is this steeper than theanswers you got for directional derivatives earlier in the problem?

The directional derivative is given by

Duf(a, b) = ∇f(a, b) · u = 〈−2,−8〉 ·⟨−2/√

68,−8/√

68⟩

= 68/√

68 =√

68.

This is slightly more than the 8 that we got in part (c), so it’s plausible that this is indeedthe direction of steepest ascent.

d) Find a direction that is tangent to the level curve z = 5 at the point (1, 1). What is thedirectional derivative in this direction?

We want a direction that’s perpendicular to the gradient. One option is⟨8/√

68,−2/√

68⟩,

which will work because the dot product is 0. This points to the right and slightly down,which looks plausible for a tangent direction to the level curve at (1, 1), based on the picture.

The directional derivative is 0, because it’s tangent to a level curve: the function doesn’tchange in this direction. Taking the dot product with the gradient confirms this.

Problem 3. Find the parametrization for a line which. . .

2

a) Has r(0) = (1, 2, 3) and r(1) = (1,−2, 1).

For the point r0, use 1, 2, 3). For the direction, use (1,−2, 1)− (1, 2, 3) = (0,−4,−2). So theequation is

r(t) = r0 + vt = 〈1, 2, 3〉+ t 〈0,−4, 2〉 = 〈1, 2− 4t, 3 + 2t〉 .

b) Normal to the plane 3x− 2y + z = 0 and passes through the origin.

We want the line to be in the direction of the normal vector, so v = 〈3,−2, 1〉.The point is r0 = 〈0, 0, 0〉, and so:

r(t) = 〈0, 0, 0〉+ 〈3,−2, 1〉 t = 〈3t,−2t, t〉 .

c) The intersection of the planes x+ y + z = 3 and x− y + 2z = 1.

This time the direction should be the cross product of the normal vectors, which is 〈1, 1, 1〉×〈1,−1, 2〉. This comes out to

〈1, 1, 1〉 × 〈1,−1, 2〉 =

∣∣∣∣∣∣i j k1 1 11 −1 2

∣∣∣∣∣∣ = 〈3,−1,−2〉 .

We also need to find a point on the line. There are many, so to narrow down our search andmake it so there’s only one answer (which is then easy to find), let’s try to find a point withz = 0. Then we need x + y = 3 and x − y = 1. The solution is x = 2, y = 1, and so ourpoint is 〈2, 1, 0〉. That means the line in question is given by

r(t) = 〈2, 4, 8〉+ 〈1, 4, 12〉 t = 〈2 + t, 4 + 4t, 8 + 12t〉 .

d) Tangent to the curve r(t) = 〈t, t2, t3〉 at t = 2.

The direction is given by the derivative, which is r′(t) = 〈1, 2t, 3t2〉. We want to plug int = 2 to get the direction vector at that time, and so the direction of the tangent line isv = 〈1, 4, 12〉.What point does the tangent line need to go through? It’s r(2) itself, which is 〈2, 4, 8〉. Soour equation for the line is

`(t) = 〈2, 4, 8〉+ 〈1, 4, 12〉 t = 〈2 + t, 4 + 4t, 8 + 12t〉 .

(Note: I’m calling the line `(t) since r(t) was already in use for the original curve. It doesn’tmatter what you call it, as long as you’re clear about what you’re doing.)

Problem 4. Consider the two vectors u = 〈1, 2, 3〉 and v = 〈−1,−1,−1〉.

a) What is the angle between u and v?

We know u · v = |u| |v| cos θ, and so the angle is given by

θ = cos−1(

u · v|u| |v|

)= cos−1

(−6√14√

3

).

3

b) If a triangle has (0, 0, 0) as a vertex, with u and v the two edges from this vertex, what isthe vector for the third edge?

It’s u−v = 〈2, 3, 4〉 (or the other way, depending on which direction we want the edge vectorto point).

c) What is the area of this triangle?

Remember that magnitude of the cross product gives the area of the parallelogram spannedby the vectors, and we want half of that.

u× v =

∣∣∣∣∣∣i j k1 2 3−1 −1 −1

∣∣∣∣∣∣ = 〈1,−2, 1〉 .

So

area =1

2|〈1,−2, 1〉| = 1

2

√6.

4

Math 210 (Lesieutre)Exam 1 reviewApril 28, 2017

Problem 1. Find the parametrization for a line which. . .

a) . . . has r(0) = (1, 2, 3) and r(1) = (1,−2, 1).

For the point r0, use 1, 2, 3). For the direction, use (1,−2, 1)− (1, 2, 3) = (0,−4,−2). So theequation is

r(t) = r0 + vt = 〈1, 2, 3〉+ t 〈0,−4, 2〉 = 〈1, 2− 4t, 3 + 2t〉 .

b) . . . is normal to the plane 3x− 2y + z = 0 and passes through the origin.

We want the line to be in the direction of the normal vector, so v = 〈3,−2, 1〉.The point is r0 = 〈0, 0, 0〉, and so:

r(t) = 〈0, 0, 0〉+ 〈3,−2, 1〉 t = 〈3t,−2t, t〉 .

c) . . . is the intersection of the planes x+ y + z = 3 and x− y + 2z = 1.

This time the direction should be the cross product of the normal vectors, which is 〈1, 1, 1〉×〈1,−1, 2〉. This comes out to

〈1, 1, 1〉 × 〈1,−1, 2〉 =

∣∣∣∣∣∣i j k1 1 11 −1 2

∣∣∣∣∣∣ = 〈3,−1,−2〉 .

We also need to find a point on the line. There are many, so to narrow down our search andmake it so there’s only one answer (which is then easy to find), let’s try to find a point withz = 0. Then we need x + y = 3 and x − y = 1. The solution is x = 2, y = 1, and so ourpoint is 〈2, 1, 0〉. That means the line in question is given by

r(t) = 〈2, 4, 8〉+ 〈1, 4, 12〉 t = 〈2 + t, 4 + 4t, 8 + 12t〉 .

d) . . . is tangent to the curve r(t) = 〈t, t2, t3〉 at t = 2.

The direction is given by the derivative, which is r′(t) = 〈1, 2t, 3t2〉. We want to plug int = 2 to get the direction vector at that time, and so the direction of the tangent line isv = 〈1, 4, 12〉.What point does the tangent line need to go through? It’s r(2) itself, which is 〈2, 4, 8〉. Soour equation for the line is

`(t) = 〈2, 4, 8〉+ 〈1, 4, 12〉 t = 〈2 + t, 4 + 4t, 8 + 12t〉 .

(Note: I’m calling the line `(t) since r(t) was already in use for the original curve. It doesn’tmatter what you call it, as long as you’re clear about what you’re doing.)

Problem 2. Consider the two vectors u = 〈1, 2, 3〉 and v = 〈−1,−1,−1〉.

1

a) What is the angle between u and v?

We know u · v = |u| |v| cos θ, and so the angle is given by

θ = cos−1(

u · v|u| |v|

)= cos−1

(−6√14√

3

).

b) If a triangle has (0, 0, 0) as a vertex, with u and v the two edges from this vertex, what isthe vector for the third edge?

It’s u−v = 〈2, 3, 4〉 (or the other way, depending on which direction we want the edge vectorto point).

c) What is the area of this triangle?

Remember that magnitude of the cross product gives the area of the parallelogram spannedby the vectors, and we want half of that.

u× v =

∣∣∣∣∣∣i j k1 2 3−1 −1 −1

∣∣∣∣∣∣ = 〈1,−2, 1〉 .

So

area =1

2|〈1,−2, 1〉| = 1

2

√6.

Problem 3. Let u = 〈−13, 2, 1〉 and v = 〈0, 21,−2〉. In what general direction does u × vpoint? (Use the right-hand rule.)

It’s more or less straight down: when we use the right hand rule, the first vector points tothe left in the xy-plane (it is almost parallel to −i), while the second points straight ahead(roughly j). The right hand rule says the cross product points down.

Problem 4. Use the two-path test to explain why the limit

lim(x,y)→(0,0)

y2 − 3x

y2 − xdoes not exist.

I’ve trained you to always try the path y = mx first. In this case we get

limx→0

(mx)2 − 3x

(mx)2 − x= lim

x→0

m2x− 3

x2x− 1= 3,

which doesn’t depend on m. So it seems like the test isn’t going to help us. But there areother paths we might want to try too. In this case, we can use the path x = 0, and we get

limy→0

y2

y2= 1,

which isn’t the same. So the answer actually does depend on the path, even though y = mxall give the same answer. The two-path test tells us the limit doesn’t exist.

2

Problem 5. A moving particle has v(t) = 〈− sin t, cos t, 1〉.

a) What is the tangent vector to its path at t = 0?

That’s just v(0) = 〈0, 1, 1〉 (if I’d given you r(t) instead, you’d first calculate v(t) = r′(t),and then plug in 0 to that).

b) Find the length of the curve from t = 0 to t = 2π. What is the tangent vector at t = 0?

The length is

` =

∫ 2π

0

|v(t)| dt =

∫ 2π

0

√(− sin t)2 + (cos t)2 + (1)2 dt =

∫ 2π

0

√2 = 4π.

c) Suppose that r(0) = 〈0, 0, 0〉. Find a formula for r(t).

We know r(t) =∫v(t) + C, and so

r(t) = 〈cos t+ c1, sin t+ c2, t+ c3〉 .

Plugging in t = 0, we get 〈0, 0, 0〉〈1 + c1, c2, c3〉, and so c1 = −1, c2 = 0, and c3 = 0. Thus

r(t) = 〈cos t− 1, sin t, t〉 .

Problem 6. An object sits at (1, 1) on a surface sloped 45◦. The gravitational force isg = 〈0,−10〉.

a) What is the component of the gravitational force parallel to the surface?

We want the component in the direction u = 〈−1,−1〉. This is given by

proju g =(g · uu · u

)u =

10

2〈−1,−1〉 = 〈−5,−5〉 .

b) The object slides down the slope to the point (0, 0). Find the work done by gravity.

It’s just W = F · d, where d is the displacement. This is 〈0,−10〉 · 〈−1,−1〉 = 10 (Joules,perhaps).

Problem 7. Let f(x, y) = x2y + 3x− 2.

a) Find the gradient ∇f .I got

∇f = 〈fx, fy〉 =⟨2xy + 3, x2

⟩.

b) Find the direction of fastest increase at the point (x, y) = (0, 1). What is the rate ofincrease?

Fastest increase occurs in the direction of the gradient, which is ∇f(1, 1) = 〈5, 1〉. Therate of increase (i.e. the directional derivative in this direction) is given by the length of thegradient, which is

√26.

3

Problem 8. What is the domain of the function f(x, y) = ln(x2 + y2 − 4)? Sketch somelevel curves.

The domain is everywhere that the formula is going to make sense. The only thing that cango wrong is that we try to take the logarithm of a negative number, which will happen ifx2 + y2 − 4 < 0. This means that x2 + y2 < 4, which is the inside of a circle of radius 2.

That’s where the function isn’t defined – the domain is everywhere that it is, which meansit’s the area outside of a circle of radius 2 centered at the origin.

The level curve at z = 2 is all (x, y) with ln(x2 + y2 − 4) = 2, so x2 + y2 − 4 = e2. This is acircle of radius

√4 + e2; all other level curves are also circles too.

4

Math 210 (Lesieutre)12.7: Tangent planes and linear approximationApril 28, 2017

Problem 1. Let’s start with a tangent plane to a sphere.

a) Write down the equation f(x, y, z) for a sphere with center (1, 2, 3) and radius 3.

We want the sphere to be

(x− 1)2 + (y − 2)2 + (z − 3)2 = 25.

b) One point on the sphere is (a, b, c) = (2, 0, 5). Compute the gradient ∇f , and evaluate∇f(a, b, c).

The gradient is just∇f = 〈2(x− 1), 2(y − 2), 2(z − 3)〉 .

Plugging in the points in question, we get

∇f(2, 0, 5) = 〈2,−4, 4〉 .

c) Use your answer to write down the equation for the tangent plane to the sphere at (a, b, c).

The normal direction is 〈2,−4, 4〉, and it goes through the point (2, 0, 5). Using the regularold formula for the equation of a sphere, we get

2(x− 2)− 4(y − 0) + 4(z − 5) = 0.

d) Try to plot the sphere and the plane and convince yourself that this answer is reasonable.

I cheated and used a computer. Things look good, however:

(Note: I’ve rotated the picture somewhat to give us a better view, so yours might lookdifferent.)

1

Problem 2. Let f(x, y) = 1− x2 − y2.

a) Find the tangent plane to the graph at (x, y) = (0, 0). Does your answer make sense?

The formula isz = fx(a, b)(x− a) + fy(a, b)(y − b) + f(a, b)

In our situation,

f(x, y) = 1− x2 − y2

f(0, 0) = 0

fx(x, y) = −2x

fy(x, y) = −2y

fx(0, 0) = 0

fy(0, 0) = 0

So the plane is z = 0(x− 0) + 0(y − 0) + 1 i.e. a horizontal plane z = 1. This makes sense,based on the graph: it’s the tangent plane at the very top of the graph.

b) Find the tangent plane to the graph at (x, y) = (1,−1). Does your answer make sense?

2

This time we get

f(1,−1) = −1

fx(1,−1) = −2

fy(1,−1) = 2

The plane isz = −1− 2(x− 1) + 2(y + 1)

Seems plausible. Here’s a graph:

Problem 3. Consider the function f(x, y) = 1x2+y2

. Use a linear approximation to approxi-

mate the value of f(1.1, 1.9).

(1.1, 1.9) is close to the much cleaner value (1, 2), so we’re going to use (a, b) = (1, 2) in theapproximation formula. First, let’s figure out what the formula actually gives is in this case.The derivative needs a little thought. To get fx, we treat y as a constant C, and then ourfunction is g(h(x)) where g(x) = 1/x and h(x) = x2 + C. Notice that g′(x) = −1/x2 andh′(x) = 2x, and so according to the chain rule the derivative is g′(h(x))h′(x) = −1/(x2 +C)2 · 2x = − 2x

(x2+y2)2.

f(x, y) =1

x2 + y2

fx = − 2x

(x2 + y2)2

fy = − 2y

(x2 + y2)2

3

Now plug in the values (1, 2):

f(1, 2) =1

5

fx(1, 2) = − 2

25

fy(1, 2) = − 4

25.

The linear approximation is then

L(x, y) ≈ 1

5− 2

25(x− 1)− 4

25(y − 2).

We want x = 1.1 and y = 1.9, in which case this gives

L(x, y) ≈ 1

5− 2

25(0.1)− 4

25(−0.1) =

50

250− 2

250+

4

250=

52

250= 0.208.

How did we do? The true value is about 0.207469, so we’re off by 0.00053.

Problem 4. A cylinder has radius 2 and height 3. Suppose that the radius and height eachincrease by 0.1. Approximately how much will the volume change?

The volume is given by V (r, h) = πr2h. We have dV = Vr(a, b) dr + Vh(a, b) dh. In this caseVr = 2πrh so Vr(2, 3) = 12π and Vh = πr2 so Vh(2, 3) = 4π. Then

dV = Vr(, b) dr + Vh(a, b) dh = (12π)(0.1) + (4π)(0.1) = 1.6π.

Let’s check it. In fact the cylinder has volume 12π, and the new cylinder has volume 13.671π.The increase is 1.671π, which is just about what we expected. (This “true increase” is whatthe book calls ∆z.)

4

Math 210 (Lesieutre)12.8: Maxima and minimaApril 28, 2017

Problem 1. A cylinder has radius 2 and height 3.

a) Suppose that the radius and height each increase by 0.1. Approximately how much will thevolume change?

The volume is given by V (r, h) = πr2h. We have dV = Vr(a, b) dr + Vh(a, b) dh. In this caseVr = 2πrh so Vr(2, 3) = 12π and Vh = πr2 so Vh(2, 3) = 4π. Then

dV = Vr(a, b) dr + Vh(a, b) dh = (12π)(0.1) + (4π)(0.1) = 1.6π.

Let’s check it. In fact the cylinder has volume 12π, and the new cylinder has volume 13.671π.The increase is 1.671π, which is just about what we expected. (This “true increase” is whatthe book calls ∆z.)

b) Give a formula for the linear approximation to V (r, h) near (r, h) = (2, 3).

This one’s supposed to be a review of last time. The formula for the linear approximation is

L(r, h) = V (2, 3) + Vr(2, 3)(r − 2) + Vh(2, 3)(h− 3) = (12π)(r − 2) + 4π(h− 3).

Problem 2. Here is a graph of the function.

f(x, y) = e−(x−1)2−(y/3)2 + e−(x+1)2−(y/3)2 .

How many critical points can you identify on the graph? Are they maxima, minima, orsaddle points?

I see three: there are two maxima, which are the two “mountain peaks”, and there is asaddle point, which is the point directly between them. (One way to think about this: acritical point is a point where you could balance a marble on the graph and it wouldn’t rollin any direction.) There are no local minima.

1

Problem 3. For each of the following functions, compute all four second derivatives. Checkthat each function has a critical point at (0, 0), and classify it as a maximum, a minimum,or a saddle point.

Let me preface these solutions with a warning. I made all the functions in this examplevery simple (just quadratics), which has the side effect that the second derivatives are allconstant. This is not how things work in general! For other functions, you will need to plugin the (a, b) for your critical points into fxx(a, b) etc. before using the test.

a) f(x, y) = x2 + y2

Here we get

fx = 2x

fy = 2y

fxx = 2

fxy = fyx = 0

fyy = 2

Plugging in (x, y) = 0, we get fx(0, 0) = fy(0, 0) = 0, as needed, so it is a critical point, asclaimed by the problem statement.

To figure out whether it’s a max or a min, we need to use the 2d version of the secondderivative test. We have

D(x, y) =

∣∣∣∣fxx fxyfyx fyy

∣∣∣∣ =

∣∣∣∣2 00 2

∣∣∣∣ = 4,

which is positive. So it’s either a maximum or a minimum, and since fxx > 0, it’s a minimum.

b) f(x, y) = x2 − y2

This time,

fx = 2x

fy = −2y

fxx = 2

fxy = fyx = 0

fyy = −2

As before, plugging in (x, y) = 0, we get fx(0, 0) = fy(0, 0) = 0, so this is also a criticalpoint.

Now we compute the determinant of the second partials. This time,

D(x, y) =


∣∣∣∣ =

∣∣∣∣2 00 −2

∣∣∣∣ = −4.

This is negative, so it’s a saddle point. Roughly, what’s going on is: if you fix y = 0 andjust think of it as a function of x, it’s a minimum (the function is just x2, after all). If youfix x = 0 and think of it as a function of y, it’s a maximum (since the function is −y2). Sowhether the function increases or decreases depends on which direction you move.

2

c) f(x, y) = −x2 − 2y2

This time,

fx = −2x

fy = −4y

fxx = −2

fxy = fyx = 0

fyy = −4



D(x, y) =


∣∣∣∣ =

∣∣∣∣−2 00 −4

∣∣∣∣ = 8.

positive, so it’s either a max or a min. since fxx < 0, in this case we know it has to be amaximum.

d) f(x, y) = x2 + xy + y2

This one’s a little more interesting.

fx = 2x+ y

fy = x+ 2y

fxx = 2

fxy = fyx = 1

fyy = 2



D(x, y) =


∣∣∣∣ =

∣∣∣∣2 11 2

∣∣∣∣ = 3.

Positive, so max or min. To tell which, we check fxx. Since it’s positive, the point is aminimum.

Problem 4. For each of the following functions, find the critical points. Pick one, anddetermine whether it is a maximum, minimum, or saddle point.

a) x4 + y4 − 16xy

We have fx = 4x3 − 16x and fy = 4y3 − 16y. To get both 0, we need x3 − 4x = 0 andy3 − 4y = 0, so x = 0,±2 and y = 0,±2. Any combination of these works, so the functionhas nine critical points.

3

The problem only asks us to look at one of these points, so let’s use (2, 2). We have

fxx = 12x− 16

fxy = 0

fyy = 12y − 16.

At (2, 2), this gives fxx(2, 2) = 8 and fyy(2, 2) = 8

D(2, 2) = fxxfyy − f 2xy = 64, which is positive. That means it’s a max or a min. Since

fxx > 0, it’s a minimum.

b) f(x, y) = x2 − 2x+ y2 − 4y + xy + 5

We have

fx = 2x− 2 + y,

fy = 2y − 4 + x.

We need both of these to be 0, so 2x + y = 2 and x + 2y = 4. Solving the linear systemusing whatever your favorite method is (probably just eliminating one of the variables), weget x = 0 and y = 2. This is the only critical point.

Is it a maximum or a minimum? Well,

fxx = 2

fxy = fyx = 1

fyy = 2

look at the determinant: ∣∣∣∣2 11 2

∣∣∣∣ = 3.

It’s either or maximum or a minimum. To know which, we have fxx = 2, so it’s a minimum.

c) g(x, y) = x2ye−x2−y2 .

This is going to be fun! Let’s get all of the derivatives out of the way right off the bat.

gx = x2ye−x2−y2(−2x) + 2xye−x

2−y2 = 2(2xy − 2x3y)e−x2−y2 ,

gy = x2e−x2−y2 + (x2y)e−x

2−y2(−2y) = (x2 − 2x2y2)e−x2−y2 .

So critical points happen whenever gx = gy = 0. Since e−x2−y2 is never 0, we need to solve

2(2xy − 2x3y) = 0 and x2 − 2x2y2 = 0. So xy(1 − x2) = 0 and x2(1 − 2y2) = 0.

There are two possibilities: either x = 0 and y is another, or y = ± 1√2

and x = ±1.

I’m not actually going to figure out what are the max and mins and saddle points. The pointof this problem is simply to scare you: sometimes there are infinitely many critical points!

4

Math 210 (Lesieutre)12.8: Max/min, continuedApril 28, 2017

Problem 1. For each of the following functions, find the critical points. Pick one, anddetermine whether it is a maximum, minimum, or saddle point.

a) x4 + y4 − 16xy

We have fx = 4x3 − 16y and fy = 4y3 − 16x. The first of these then gives y = x3/4, andplugging this into the second, we obtain

4

(x3

4

)3

− 16x = 0

4x9

64− 16x = 0

x9

16− 16x = 0

x9 − 256x = 0

x(x8 − 256) = 0

The second factor x8 − 256 is 0 if x = 2 or x = −2 (there are six other solutions overthe complex numbers, but we’re only worried about the real solutions). If x = 0, theny = x3/4 = 0; if x = 2, then y = 2; if x = −2, then y = −2. So there are three criticalpoints: (2, 2), (0, 0), and (−2,−2).

We need to figure out which of these are maxes and which are mins. To do that we’re goingto need the second derivatives, so we might as well get it over with and calculate them now.

fxx = 12x2,

fxy = −16,

fyy = 12y2.

At (2, 2), this gives fxx(2, 2) = 48, fyy(2, 2) = 48, and fxy(2, 2) = −16. Thus D(2, 2) =(48)(48)− (−16)2 = 2048. The test tells us that this is a minimum.

At (−2,−2), this gives fxx(−2,−2) = 48, fyy(−2,−2) = 48, and fxy(−2,−2) = −16. ThusD(−2,−2) = (48)(48)− (−16)2 = 2048. The test tells us that this is a minimum.

At (0, 0), this gives fxx(0, 0) = 0, fyy(0, 0) = 0, and fxy(0, 0) = −16. Thus D(0, 0) =(0)(0)− (−16)2 = −256. So that’s a saddle point.

b) f(x, y) = x2 − 2x+ y2 − 4y + xy + 5

We have

fx = 2x− 2 + y,

fy = 2y − 4 + x.

1

We need both of these to be 0, so 2x + y = 2 and x + 2y = 4. Solving the linear systemusing whatever your favorite method is (probably just eliminating one of the variables), weget x = 0 and y = 2. This is the only critical point.

Is it a maximum or a minimum? Well,

fxx = 2

fxy = fyx = 1

fyy = 2

look at the determinant: ∣∣∣∣2 11 2

∣∣∣∣ = 3.

It’s either or maximum or a minimum. To know which, we have fxx = 2, so it’s a minimum.

c) g(x, y) = x2ye−x2−y2 .

This is going to be fun (not really)! Let’s get all of the derivatives out of the way right offthe bat.

gx = x2ye−x2−y2(−2x) + 2xye−x

2−y2 = 2(2xy − 2x3y)e−x2−y2 ,

gy = x2e−x2−y2 + (x2y)e−x

2−y2(−2y) = (x2 − 2x2y2)e−x2−y2 .

So critical points happen whenever gx = gy = 0. Since e−x2−y2 is never 0, we need to solve

2(2xy − 2x3y) = 0 and x2 − 2x2y2 = 0. So xy(1− x2) = 0 and x2(1− 2y2) = 0.

There are two possibilities: either x = 0 and y is another, or y = ± 1√2

and x = ±1.

I’m not actually going to figure out what are the max and mins and saddle points. The pointof this problem is mostly to warn you: sometimes there are infinitely many critical points!

Problem 2. Find the point on the surface x2− yz = 1 which is closest to the origin. (Hint:minimize the square of the distance, instead of the distance itself).

The square of the distance is x2 + y2 + z2. Since x2 = 1 + yz on the surface, we want tominimize 1 + yz+ y2 + z2 (a function of the independent variables y and z). So we’ll use thesecond-derivative test:

∂f

∂y= 2y + z,

∂f

∂z= y + 2z.

These vanish simultaneously at (0, 0), where x = ±1. So the two critical points of thedistance are (1, 0, 0) and (−1, 0, 0). The second derivatives are

D(y, z) =

∣∣∣∣fyy fyzfzy fzz

∣∣∣∣ =

∣∣∣∣2 11 2

∣∣∣∣ = 3

As this is positive, and fyy > 0, this is a minimum as expected.

2

Problem 3. Find the maximum value of the function f(x, y) = (x − 1)2 + y2 on or insidea circle of radius 2 centered at (0, 0).

First we find the critical points on the interior, i.e. on the inside of the circle. fx = 2(x− 1)and fy = 2y, so the unique critical point is (1, 0). By inspection it is a minimum, and so themaximum of the circle is going to be on the boundary.

To find the maximum value on the boundary, use a parametrization r(t) = (2 cos t, 2 sin t).This parametrizes the circle, and so we just need to find the t for which f(r(t)) is as largeas possible. Plugging in, we want to maximize

(2 cos t− 1)2 + (2 sin t)2 = 4 cos2 t− 4 cos t+ 1 + 4 sin2 t = 5− 4 cos t.

This has a maximum at t = π, when cos t = −1. The corresponding point is γ(π) = (−2, 0)and the value of f there is 3.

Problem 4. Find the maximum and minimum values of the function f(x, y) = 5 − (x −1)2 − (y − 1)2 on a triangle with vertices (0, 0), (8, 0), and (0, 4).

First we need to find the critical points. That’s the easy part. We have

fx = −2(x− 1),

fy = −2(y − 1).

The critical points are where both of these vanish. There is only one: the point (x, y) = (1, 1).This is inside the triangle in question, as needed.

Is that point a maximum or a minimum? Maybe you can guess from the formula (it surelooks like a max. . . ), but the way to check is to use the second derivative test with theHessian.

The second derivatives are

fxx = −2,

fxy = 0,

fyy = −2.

At (1, 1), we have fxx(1, 1) = 2, fxy(1, 1) = 0, fyy(1, 1) = 2. Thus D(1, 1) = (−2)(−2) −(0)(0) = 4, which is positive. Since fxx(1, 1) > 0, the point is a maximum. For futurecomparison, notice that f(1, 1) = 5.

However, we still don’t actually know that it’s a global maximum: maybe the functionachieves an even bigger value somewhere, but it’s on the edge, and so it isn’t a critical point.So we need to find the maxima and minima going around the outside of the triangle.

We have to check each edge. To do that, we’ll parametrize it, and then solve for the value(s)of t that maximize and minimize the function in question. Let’s start with the edge from(0, 4) to (8, 0). It’s parametrized by r1(t) = 〈8t, 4− 4t〉, where 0 ≤ t ≤ 1 (here I used ourusual strategy for parametrizing a line. Then we get:

3

f(r1(t)) = 5− (x− 1)2 − (y − 1)2 = 5− (3− 4t)2 − (−1 + 8t)2 = −80t2 + 40t− 5.

We now do max/min, math 180-style. The derivative is −160t+40, which is 0 when t = 1/4.This corresponds to the point (2, 3), for which f(r1(1/4)) = 0. Since f ′′(r1(t)) < 0, it’s alocal max, and we find f(r1(5/2)) = 1

2. We also need to check the two endpoints: t = 0 and

t = 4. At t = 0, we’re at the point r1(0) = (0, 4), and f(r1(0)) = −5. At t = 4, it’s (4, 0),and f(r1(4)) = −5 too.

That’s only one edge. There are two more. Let’s go from (0, 0) to (8, 0) now. This one hasr2(t) = 〈8t, 0〉 with 0 ≤ t ≤ 1. Then f(r2(t)) = 5 − (8t − 1)2 − (−1)2 = 4 − (8t − 1)2 =−64t2 + 16t + 3. What are the extrema? Well, d

dtf(r2(t)) = 16 − 128t, which vanishes for

t = 1/8, which is the point (1, 0). Here the function has the value 4. There are also theendpoints r2(0) = (0, 0) and r2(1) = (8, 0), where the values are respectively 3 and −45.

The last edge goes from (0, 0) to (0, 4), and is parametrized by r3(t) = 〈0, 4t〉. Thenf(r3(t)) = 5 − (4t − 1)2 − (−1)2 = −16t2 + 8t + 3. Then d

dtf(r3(t)) = 8 − 32t, which

vanishes for t = 1/4, corresponding to the point r3(1/4) = (0, 1), for which f(r3(1/4)) = 4.The two endpoints are (0, 0) and (0, 4), which are already on the list.

All told, the possible extrema come from the edges and the critical points of the 2D function.They’re all listed below.

Part of R Possible extremum Value TypeInterior (1, 1) 5 MaxEdge #1 (2, 3) 0 Min

(8, 0) −45 ?(0, 4) −5 ?

Edge #2 (1, 0) 4 Min(0, 0) 3 ?(8, 0) −45 ?

Edge #3 (0, 1) 4 Min(0, 0) 3 ?(0, 4) −5 ?

(The ?’s indicate that at the corners of the region, we don’t really know if the points aregoing to be maxima or minima, because we can’t use the second derivative test. We justhave to compare the values against the other values.

Those are the candidates. To see the actual global max and min on the region, look for thebiggest and smallest numbers. The global max occurs at the critical point (1, 1), which thevalue is 5. The global min is at (8, 0), where the value is −45.

4

Math 210 (Lesieutre)12.8: Max/min, continuedApril 28, 2017

Problem 1. An airline will let you carry on any rectangular bag for which the sum of thedimensions x, y, and z is less than 60. Suppose that you want to bring a bag with the largestpossible volume. To find the appropriate x, y, and z, what function should you maximize,and on what region?

The volume is xyz. We can assume that z = 60 − x − y, since otherwise we could make zbigger without breaking the rules. Then the function we should maximize is xy(60− x− y).The region is where x ≥ 0, y ≥ 0, and x+ y ≤ 60, which is a triangle.

Problem 2. Find the maximum and minimum values of the function f(x, y) = 5 − (x −1)2 − (y − 1)2 on a triangle with vertices (0, 0), (8, 0), and (0, 4).

First we need to find the critical points. That’s the easy part. We have

fx = −2(x− 1),

fy = −2(y − 1).

The critical points are where both of these vanish. There is only one: the point (x, y) = (1, 1).This is inside the triangle in question, as needed.

Is that point a maximum or a minimum? Maybe you can guess from the formula (it surelooks like a max. . . ), but the way to check is to use the second derivative test with theHessian.

The second derivatives are

fxx = −2,

fxy = 0,

fyy = −2.

At (1, 1), we have fxx(1, 1) = 2, fxy(1, 1) = 0, fyy(1, 1) = 2. Thus D(1, 1) = (−2)(−2) −(0)(0) = 4, which is positive. Since fxx(1, 1) > 0, the point is a maximum. For futurecomparison, notice that f(1, 1) = 5.

However, we still don’t actually know that it’s a global maximum: maybe the functionachieves an even bigger value somewhere, but it’s on the edge, and so it isn’t a critical point.So we need to find the maxima and minima going around the outside of the triangle.

We have to check each edge. To do that, we’ll parametrize it, and then solve for the value(s)of t that maximize and minimize the function in question. Let’s start with the edge from(0, 4) to (8, 0). It’s parametrized by r1(t) = 〈8t, 4− 4t〉, where 0 ≤ t ≤ 1 (here I used ourusual strategy for parametrizing a line. Then we get:

1

f(r1(t)) = 5− (x− 1)2 − (y − 1)2 = 5− (3− 4t)2 − (−1 + 8t)2 = −80t2 + 40t− 5.

We now do max/min, math 180-style. The derivative is −160t+40, which is 0 when t = 1/4.This corresponds to the point (2, 3), for which f(r1(1/4)) = 0. Since f ′′(r1(t)) < 0, it’s alocal max, and we find f(r1(5/2)) = 1

2. We also need to check the two endpoints: t = 0 and

t = 4. At t = 0, we’re at the point r1(0) = (0, 4), and f(r1(0)) = −5. At t = 4, it’s (4, 0),and f(r1(4)) = −5 too.

That’s only one edge. There are two more. Let’s go from (0, 0) to (8, 0) now. This one hasr2(t) = 〈8t, 0〉 with 0 ≤ t ≤ 1. Then f(r2(t)) = 5 − (8t − 1)2 − (−1)2 = 4 − (8t − 1)2 =−64t2 + 16t + 3. What are the extrema? Well, d

dtf(r2(t)) = 16 − 128t, which vanishes for

t = 1/8, which is the point (1, 0). Here the function has the value 4. There are also theendpoints r2(0) = (0, 0) and r2(1) = (8, 0), where the values are respectively 3 and −45.

The last edge goes from (0, 0) to (0, 4), and is parametrized by r3(t) = 〈0, 4t〉. Thenf(r3(t)) = 5 − (4t − 1)2 − (−1)2 = −16t2 + 8t + 3. Then d

dtf(r3(t)) = 8 − 32t, which

vanishes for t = 1/4, corresponding to the point r3(1/4) = (0, 1), for which f(r3(1/4)) = 4.The two endpoints are (0, 0) and (0, 4), which are already on the list.

All told, the possible extrema come from the edges and the critical points of the 2D function.They’re all listed below.

Part of R Possible extremum Value TypeInterior (1, 1) 5 MaxEdge #1 (2, 3) 0 Min

(8, 0) −45 ?(0, 4) −5 ?

Edge #2 (1, 0) 4 Min(0, 0) 3 ?(8, 0) −45 ?

Edge #3 (0, 1) 4 Min(0, 0) 3 ?(0, 4) −5 ?

(The ?’s indicate that at the corners of the region, we don’t really know if the points aregoing to be maxima or minima, because we can’t use the second derivative test. We justhave to compare the values against the other values.

Those are the candidates. To see the actual global max and min on the region, look for thebiggest and smallest numbers. The global max occurs at the critical point (1, 1), which thevalue is 5. The global min is at (8, 0), where the value is −45.

Problem 3. Consider the function f(x, y) = e−x2−2y2.

2

a) Find the maximum and minimum on or inside a square with vertices (±2,±2).

Step 1: Find the critical points.

We have

fx(x, y) = −e−x2−2y2(−2x) = 2xe−x2−2y2 ,

fy(x, y) = −e−x2−2y2(−4y) = 4ye−x2−2y2 .

How can the first of these be 0? It’s only if 2x = 0, since e−x2−2y2 can’t possibly be 0. How

about the second? Only if 4y = 0, for the same reason. So the only critical point is (0, 0).

We’ll need to know the second partials to determine the types of the critical point. This isa little more messy, since we have to use the product rule.

fxx(x, y) = −2e−x2−2y2 + 4x2e−x

2−2y2x2

fxy(x, y) = 8xye−x2−2y2

fyy(x, y) = −4e−x2−2y2 + 16y2e−x

2−2y2

At (0, 0), the values of these things are respectively −2, 0, and −4. Then D(0, 0) =(−2)(−4)− 02 = 8, which is positive, and since fxx(0, 0) < 0, we conclude that the point isa maximum.

The annoying part is that the maximum or minimum could be on an edge, and there arefour of them. Here are the parametrizations:

r1(t) = 〈2,−2 + 4t〉 (top)

r2(t) = 〈−2,−2 + 4t〉 (bottom)

r3(t) = 〈−2 + 4t,−2〉 (left)

r4(t) = 〈−2 + 4t, 2〉 (right)

Let’s do r1(t) in detail. We need to plug this in to f(t) and think of it as a function of thesingle variable t, which we then find max/min for using the usual single-variable calculusstrategies. Plugging in, we get

f(r1(t)) = e−4−2(−2+4t)2 = e−12+32t−32t2 .

The derivative is ddtf(r1(t)) = (−64t + 32)e−12+32t−32t2 , which is 0 when t = 1/2. This

corresponds to the point (2, 0), and we find f(2, 0) = 1/e4. This is a local maximum (on theedge). We also need to check the endpoints, which are (2, 2) and (2,−2). Both of these give1/e12.

Now we need to do the second edge. We find that f(r2(t)) = e−12+32t−32t2 , which is the samething that we got with r1(t). There is again a maximum at t = 1/2, which is the point(−2, 0), and the value is again 1/e4. The two endpoints (−2,−2) and (−2, 2) give values of1/e12.

3

Now the third edge r3(t) = 〈−2 + 4t,−2〉 gives us f(r3(t)) = e−12+16t−16t2 . The derivative isddtf(r3(t)) = e−12+16t−16t2(16 − 32t), which again has a max at t = 1/2, which this time is

the point (0,−2), where we get 1/e8.

It’s getting very late, so I’m not going to write down the fourth edge. Sorry! Take my wordfor it that it works the same as the third one, or email me if you want me to add some moredetails.

Part of R Possible extremum Value TypeInterior (0, 0) 1 MaxEdge #1 (2, 0) 1/e4 Max

(2,−2) 1/e12 ?(2, 2) 1/e12 ?

Edge #2 (−2, 0) 1/e4 Max(−2,−2) 1/e12 ?

(−2, 2) 1/e12 ?Edge #3 (0, 2) 1/e8 Max

(−2, 2) 1/e12 ?(2, 2) 1/e12 ?

Edge #4 (0,−2) 1/e8 Max(−2,−2) 1/e12 ?

(2,−2) 1/e12 ?

The global max is (0, 0), where the function is f(0, 0) = 1. The global min is at each of thefour corners (±2,±2), where f takes the value 1/e12.

b) Find the maximum of the same function on the unit circle.

The strategy is the same, and this one is actually a little easier, since a circle only has oneedge. Parametrize it by r(t) = 〈cos t, sin t〉. Plugging that in, we get

f(r(t)) = e− cos2 t−2 sin2 t = e−32+ 1

2cos(2t)

(I used a trig identity here, but you’d get the same answer even without it.) We have

d

dtf(r(t)) = e−

32+ 1

2cos(2t)(−2 sin(2t)).

The critical points are t = 0, π/2, π, and 3π/2. The values of the function are

t f(r(t))0 e−1

π/2 e−2

π e−1

3π/2 e−2

The critical points are all either on the edge, or in the interior of the unit disk, where wealready found that the only critical point is (0, 0). Hence the max is at (0, 0), where f(0, 0) =1, and the minimum occurs on the boundary at r(π/2) = (0, 1) and r(3π/2) = (0,−1). Atthese points, the value of the function is e−2.

4

Math 210 (Lesieutre)12.9: Lagrange multipliersApril 28, 2017

Problem 1. You want to find a rectangle with perimeter 16 and area as large as possible.

a) Convert this into a constrained optimization problem: let x and y be the two sides ofthe rectangle. What function f(x, y) are you trying to maximize? What constraint must besatisfied by x and y?

We are trying to maximize f(x, y) = xy subject to g(x, y) = 2x+ 2y = 16.

b) Use the method of Lagrange multipliers to find the maximum value of the function.

We have

∇f = 〈y, x〉∇g = 〈2, 2〉

The equations are ∇f(x, y) = λ∇g(x, y) and g(x, y) = 0, which gives three equations inthree variables:

y = 2λ

x = 2λ

2x+ 2y = 16

The second equation becomes 2(2λ)+2(2λ) = 16, so that λ = 2. Then x = 4 and y = 4 mustbe the maximum. The value of the function is given by f(4, 4) = 16. Hence the maximumpossible area is achieved when the function is a square.

Problem 2. Find the maximum value of the function f(x, y) = y2 − 4x2 subject to theconstraint g(x, y) = x2 + 2y2 = 4.

In this case, we have

∇f(x, y) = 〈−8x, 2y〉∇g(x, y) = 〈2x, 4y〉

Our three equations come from ∇f(x, y) = λ∇g(x, y), so that

−8x = λ2x

2y = λ4y

x2 + 2y2 = 4

This is a fairly typical set-up for one of these problems. The equations aren’t so hard tosolve, but you need to be extremely careful not to forget that the variables can be 0.

1

If x is not 0, then dividing the first equation through by x gives λ = −4. Then the secondequation gives 2y = −16y, which means that y = 0. The third then says that x2 = 4, sox = −2 or x = 2. This gives two points: (x, y) = (−2, 0) and (x, y) = (2, 0). We havef(−2, 0) = −16 and f(2, 0) = −16.

If x is 0, then the first equation is true no matter what λ is. The third equation reads2y2 = 4, so y = ±

√2. Since y is nonzero, the second equation just turns into λ = 1/2 (but

again, we don’t really care what λ is). So we have two more solutions: (x, y) = (0,−√

2)and (x, y) = (0,

√2). Plugging in we get f(0,−

√2) = 2 and f(0,

√2) = 2. So the function

is maximized at (0,−√

2) and (0,√

2), where the value is 2, and minimized at (2, 0) and(−2, 0), where the value is −16.

Problem 3. Use Lagrange multipliers to find the point on the parabola y = x2 − 1 which isclosest to the origin.

We want to minimize the distance function h(x, y) =√x2 + y2. Nobody likes square roots,

so notice that we can just as well minimize the distance squared, which is f(x, y) = x2 + y2.This is subject to the constraint y = x2−1. This is better written as g(x, y) = y−x2+1 = 0.

We have

∇f(x, y) = 〈2x, 2y〉∇g(x, y) = 〈−2x, 1〉

Our equations are thus 2x = −2xλ, 2y = λ, and y − x2 + 1 = 0. Assume first that x 6= 0.Then λ = −1 from the first equation, so y = −1/2. Then (−1/2) − x2 + 1 = 0 so that

x = ±√

1/2. So we get two points:(√

22,−1

2

). The square of the distance at either one is

f(√

22,−1

2

)= 3

4.

If x = 0, then the third equation gives y = −1. The distance here is 1. So the closest pointswere the first two that we found.

Problem 4. You are making a open-top drawer out of wood. The material for the sides andback costs $2 per square foot, and material for the bottom costs $1, and the material for thefront costs $4.

a) Suppose that the dimensions of the drawer are x (side to side), y (top to bottom), and z(front to back). What is the total cost of the materials?

It should be

cost = costfront + costback + costsides + costbottom

= 4xy + 2xy + 2(2yz) + xz = 6xy + 4yz + xz.

2

b) What are the dimensions of the cheapest drawer with volume 24 cubic feet?

We want to minimize f(x, y, z) = 6xy + 4yz + xz subject to the constraint xyz = 24. Thegradients are

∇f(x, y, z) = 〈6y + z, 6x+ 4z, 4y + x〉∇g(x, y, z) = 〈yz, xz, xy〉

Our four equations are

(6y + z) = λ(yz)

(6x+ 4z) = λ(xz)

(4y + x) = λ(xy)

xyz = 24

After some painful algebra, you’ll get that x = 4, y = 1, z = 6, and λ = 2. So the minimumis a drawer that’s 4× 1× 6.

3

Math 210 (Lesieutre)12.9, Lagrange multipliers, and a little 13.1April 28, 2017

Problem 1. Find the maximum value of the function f(x, y) = x+y subject to the constraintx2 + y2 = 1.

First let’s come up with the equations. The functions are f(x, y) = x + y and g(x, y) =x2 + y2 − 1.

∇f(x, y) = 〈1, 1〉∇g(x, y) = 〈2x, 2y〉 .

The equations are then 〈1, 1〉 = λ 〈2x, 2y〉, so

2xλ = 1,

2yλ = 1,

x2 + y2 = 1.

Notice that λ can’t possibly be 0: then the first equation wouldn’t have any solution. As-suming λ 6= 0, the first two equations give x = y, from which the third gives 2x2 = 1, i.e.

x = ±√22

. Thus there are two possibilities: (x, y) =(√

22,√22

)and (x, y) = −

(−

√22,−

√22

).

(x, y) f(x, y)(√22,√22

) √2(

−√22,−

√22

)−√

2

The first point is the max, and the second point is the min.

Problem 2. You are making a open-top drawer out of wood. The material for the sides andback costs $2 per square foot, and material for the bottom costs $1, and the material for thefront costs $4.

a) Suppose that the dimensions of the drawer are x (side to side), y (top to bottom), and z(front to back). What is the total cost of the materials?

It should be

cost = costfront + costback + costsides + costbottom

= 4xy + 2xy + 2(2yz) + xz = 6xy + 4yz + xz.

b) What are the dimensions of the cheapest drawer with volume 24 cubic feet?

1

We want to minimize f(x, y, z) = 6xy + 4yz + xz subject to the constraint xyz = 24. Thegradients are

∇f(x, y, z) = 〈6y + z, 6x+ 4z, 4y + x〉∇g(x, y, z) = 〈yz, xz, xy〉

Our four equations are

(6y + z) = λ(yz)

(6x+ 4z) = λ(xz)

(4y + x) = λ(xy)

xyz = 24

After some painful algebra, you’ll get that x = 4, y = 1, z = 6, and λ = 2. So the minimumis a drawer that’s 4× 1× 6.

Problem 3. Compute the following double integrals.

a) ∫ 1

0

∫ 2

1

xy dy dx

The inner integral is everything from the second integral sign to the dwhatever.∫ 2

1

xy dy =y2

2x∣∣∣2y=1

=22

2x− 12

2x =

3

2x.

The outer integral is then ∫ 1

0

3

2x dx =

3

2

x2

2

∣∣∣1x=0

=3

4− 0 =

3

4.

That’s our final answer.

b) ∫ 1

0

∫ 2

0

yexy dx dy

The inner integral is given by∫ 2

0

exy dx = exy∣∣∣2x=0

= e2y − e0y = e2y − 1.

The final answer is∫ 1

0

e2y − 1 dy =

(1

2e2y − y

)∣∣∣∣1y=0

=

(1

2e2 − 1

)−(

1

2− 0

)=

1

2e2 − 3

2.

Notice that if we tried to do it in the other order, it would be tough going:∫yexy dy would

require an integration by parts. In some cases, it’s even worse: the integral simply can’t bedone unless the order is right.

2

Math 210 (Lesieutre)13.1: Double integrals over rectangular regionsApril 28, 2017

Problem 1. Compute the following double integrals. Sketch the region R over which theintegral is being taken.

a)

∫ 1

0

∫ 2

1

xy dy dx

The inner integral is everything from the second integral sign to the dwhatever.∫ 2

1

xy dy =y2

2x∣∣∣2y=1

=22

2x− 12

2x =

3

2x.

The outer integral is then ∫ 1

0

3

2x dx =

3

2

x2

2

∣∣∣1x=0

=3

4− 0 =

3

4.


b)

∫ 2

1

∫ 1

0

xy dx dy

Inner: ∫ 1

0

xy dx = yx2

2

∣∣∣1x=0

=y

2.

Outer: ∫ 2

1

y

2dy =

y2

4

∣∣∣21

=3

4.

Same answer as before. Coincidence? No – Fubini’s theorem. Changing the order of thevariables doesn’t change the value of the integral.

Problem 2. a)

∫ 2

0

∫ 1

0

yexy dy dx

The inner integral is

∫ 1

0

yexy dy, which is moderately unpleasant, since we have to integrate

by parts. If we change the order of integration, it’s a little better:∫ 1

0

∫ 2

0

yexy dx dy

The inner integral is given by∫ 2

0

yexy dx = exy∣∣∣2x=0

= e2y − e0y = e2y − 1.

1

The final answer is∫ 1

0

e2y − 1 dy =

(1

2e2y − y

)∣∣∣∣1y=0

=

(1

2e2 − 1

)−(

1

2− 0

)=

1

2e2 − 3

2.

Notice that if we tried to do it in the other order, it would be tough going:∫yexy dy would

require an integration by parts. In some cases, it’s even worse: the integral simply can’t bedone unless the order is right.

b) What is the average value of f(x, y) = yexy on the region R?

It’s given by1

area(R)

∫ 2

0

∫ 1

0

yexy dy dx =1

2

(1

2e2 − 3

2

)=

1

4e2 − 3

4.

Problem 3. a) Sketch the region of integration for∫ 2

0

∫ x2

0

y dy dx.

It’s the region between y = 0 and y = x2 with 0 ≤ x ≤ 2.

b) Evaluate the integral.

Inner: ∫ x2

0

y dy =y2

2

∣∣∣x2

0=

x4

2.

Outer: ∫ 2

0

x4

2dx =

x5

10

∣∣∣20

=32

10=

16

5.

c) Rewrite the integral with the variables in the opposite order.

This time y is going to go on the outside. Based on the picture, we need to go from y = 0to y = 4. For a given value of y, what x’s do we want? It’s frmo x = 0 to

√y to 2.∫ 4

0

∫ 2

√y

y dx dy.

d) Evaluate the integral.

Inner: ∫ 2

√y

y dx = y(2−√y) = 2y − y3/2.

Outer: ∫ 4

0

2y − y3/2 dy =

(y2 − 2

5y5/2

)∣∣∣∣40

= 16− 2

532 =

16

5.

2

Math 210 (Lesieutre)13.3: Double integrals in other coordinate systemsApril 28, 2017

Problem 1. Evaluate the integral ∫ 1

0

∫ 1

x2

xy dy dx

Inner: ∫ 1

x2

xy dy = xy2

2

∣∣∣1x2

=x

2− x5

2.

Outer: ∫ 1

0

(x

2− x5

2

)dx =

(x2

4− x6

12

) ∣∣∣10=

(1

4− 1

12

)− (0, 0) =

1

6.

Problem 2. Sketch the region of integration for each of the following double integrals.

a)

∫ 1

0

∫ x

0

f(x, y) dy dx

Here x goes from 0 to 1, and for a given value of x, y goes from 0 to x. This makes a triangle.

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

b)

∫ 1

0

∫ 1

y2f(x, y) dx dy

Here y goes from 0 to 1, and for a given value of y, x goes from y2 to 1. Remembering thatx = y2 is a sideways parabola, here’s the picture we get:

1

0.2 0.4 0.6 0.8 1.0

0.2

0.4

0.6

0.8

1.0

c)

∫ 2

0

∫ 4x

x3

f(x, y) dy dx

Here x goes from 0 to 2, and for a given value of x, y goes from x3 to 4x. Here it is:

0.5 1.0 1.5 2.0

2

4

6

8

Problem 3. For each of the double integrals in the previous problem, write bounds in newcoordinates to reverse the order of integration.

a)

∫ 1

0

∫ x

0

f(x, y) dy dx

Based on the picture, the new bounds are∫ 1

0

∫ 1

y

f(x, y) dx dy

b)

∫ 1

0

∫ 1

y2dx dy

2

Again, go to the picture. x is going to run from 0 to 1, and for a given x, y goes from 0 to√x. ∫ 1

0

∫ √x

0

f(x, y) dy dx.

c)

∫ 2

0

∫ 4x

x3

dy dx

Now we need to go in the other direction. We can see that y goes from 0 to 8. What are thebounds for a given value of y? The lower bound is the orange curve, which is y = 4x, akax = y/4. The upper bound is the blue curve, y = x3, so x = 3

√y. Hence the integral is∫ 8

0

∫ 3√y

y/4

f(x, y) dx dy.

Problem 4. Find the area of the region in (b) by integrating the function f(x, y) = 1. Checkthat you get the same answer for either order of integration.

First, we want to evaluate ∫ 1

0

∫ 1

y21 dx dy.

Inner: ∫ 1

y21 dx = x

∣∣∣1y2

= 1− y2

Outer: ∫ 1

0

1− y2 dy =

(y − y3

3

)∣∣∣∣10

=2

3.

Let’s try to do the same thing the other way:∫ 1

0

∫ √x

0

1 dy dx.

Inner: ∫ √x

0

1 dy = y∣∣∣√x

0=√x.

Outer: ∫ 1

0

√x dx =

2

3x3/2

∣∣∣10=

2

3.

That’s the same answer we got before, as it should be.

3

Math 210 (Lesieutre)13.3: Double integrals in polar coordinatesApril 28, 2017

Problem 1. Sketch the regions corresponding to the following double integrals.

a)

∫ 2

1

∫ π

π/2

f(r, θ) r dr dθ

It’s the region between orange and blue curves in the following illustration, which is a sortof quarter-ring.

-2.0 -1.5 -1.0 -0.5

0.5

1.0

1.5

2.0

b)

∫ 1+cos θ

0

∫ 2π

0

f(r, θ) r dr dθ

This is the inside of the cardioid pictured inthe following picture.

0.5 1.0 1.5 2.0

-1.0

-0.5

0.5

1.0

Problem 2. Find the area inside the cardioid r(θ) = 1 + cos θ.

1

This is a classic one. To find the area, we have to integrate 1 dA = 1 r drdθ on the region inquestion. Don’t forget the “r”!. This is∫ 2π

0

∫ 1+cos θ

0

r dr dθ.

The inner integral is:∫ 1+cos θ

0

r dr =r2

2

∣∣∣1+cos θ

0=

(1 + cos θ)2

2=

1 + 2 cos θ + cos2 θ

2

=1

2+ cos θ +

1

2

(1 + cos 2θ

2

)=

3

4+ cos θ +

1

4cos 2θ.

The full integral is now given by∫ 2π

0

3

4+ cos θ +

1

4cos 2θ dθ =

3

4(2π) + 0 + 0 =

3π

2.

Seems plausible.

Problem 3. Find the volume of the region under the graph f(x, y) = 1 − x2 − y2 above theunit circle R.

The volume is going to be given by

∫∫R

f(r, θ) dA. There are three things we need to do: (i)

express the bounds on the integral in polar (ii) express the function in question in polar (iii)express the area element dA in polar. Then we need (iv): actually compute the integral.

The function f(x, y) = 1−x2−y2 is given in terms of θ by f(r, θ) = 1−r2. The area elementis dA = r dr dθ (that’s what you always want to use here). The volume is then given by∫ 2π

0

∫ 1

0

(1 − r2) r dr dθ.

The inner integral is∫ 1

0

(1 − r2) r dr =

∫ 1

0

(r − r3) dr =

(r2

2− r4

4

)∣∣∣∣10

=1

4.

The outer integral is ∫ 2π

0

1

4dθ = (2π)

1

4=π

2.

That’s the volume. Seems plausible.

2

Problem 4. a) Rewrite the following integral in Cartesian coordinates:∫∫

R2xy dA, where

R is part of a circle of radius 4 that lies in the first quadrant.

Again, we need to get the bounds in polar, the function in polar, and dA is polar. Thehardest part is the function. Remember that x = r cos θ and y = r sin θ, and so f(x, y) =2xy = 4r2 cos θ sin θ = 2r2 sin 2θ. This gives us∫ π/2

0

∫ 4

0

2r2 sin(2θ) r dr dθ =

∫ π/2

0

∫ 4

0

2r3 sin(2θ) dr dθ

b) Evaluate the integral.

This one is∫ π/2

0

∫ 4

0

2r3 sin(2θ) dr dθ =

(∫ π/2

0

sin 2θ dθ

)(∫ 4

0

2r3 dr

)=

(− cos 2θ

2

∣∣∣π/20

)(r4

2

∣∣∣40

)=

(1

2− −1

2

)(256

2− 0

)= 128.

3

Math 210 (Lesieutre)13.4: Triple integralsApril 28, 2017

Problem 1. Compute the following triple integrals over rectangular regions.

a)

∫ 1

0

∫ 2

1

∫ 2

0

xyz dz dy dx

Inner: ∫ 2

0

xyz dz = xyz2

2

∣∣∣20

= 2xy.

Middle: ∫ 2

1

2xy dy = xy2∣∣∣2y=1

= (4x) − (1x) = 3x.

Outer: ∫ 1

0

3x dx =3x2

2

∣∣∣10

=3

2.


b)

∫ 2

0

∫ 2

1

∫ 1

0

xyz dx dy dz

Inner: ∫ 1

0

xyz dx = yzx2

2

∣∣∣1x=0

=yz

2.

Middle: ∫ 2

1

yz

2dy =

zy2

4

∣∣∣2y=1

=z22

4− z12

4=

3z

4.

Outer: ∫ 2

0

3z

4dz =

3z2

8

∣∣∣20

=3

2.

Of course, you’re supposed to notice that this is the same integral as the one from part (a),but with the order changed. We got the same answer, which is reassuring. In fact, there arefour other ways to rearrange it, and they’ll all give you the same thing.

Problem 2. Set up bounds for integrating a function f(x, y, z) on a cylinder of height 3 andradius 2, with base centered at (0, 0, 0).

This one we want to use ∫ 3

z=0

∫ 2

x=−2

∫ √4−x2

y=−√4−x2

f(x, y, z) dx dy dz

(Note that sometimes people like to skip the “x =”, “y =”, “z =” at the bottom of theintegral, since you can surmise which is which from the order of the d’s. But I think thismakes it easier to keep things straight.)

1

Problem 3. Evaluate the triple integral∫ 1

0

∫ √1−x2

0

∫ 2−x

0

yz dz dy dx

Inner: ∫ 2−x

0

yz dz = yz2

2

∣∣∣2−xz=0

= y(2 − x)2

2.

Middle: ∫ √1−x2

0

y(2 − x)2

2dy =

y2

2

(2 − x)2

2

∣∣∣√1−x2

y=0=

(1 − x2)(2 − x)2

4.

Outer: ∫ 4

0

(1 − x2)(2 − x)2

4dx = −144

5.

The last integral is a hassle: I don’t see any better way to do it than to just expand it thewhole way.

Problem 4. Change the bounds on the following integral from dx dy dz to dy dx dz.∫ 4

0

∫ 1

0

∫ 2y

0

f(x, y, z) dx dy dz.

This is a triangular prism. The base is a triangle: y goes from 0 to 1, and x from 0 to 2y,which gives the triangle with vertices at (0, 0), (2, 1), and (0, 1). In the other order, we get∫ 4

0

∫ 2

0

∫ 1

x/2

f(x, y, z) dy dx dz.

2

Math 210 (Lesieutre)13.5: Triple integrals in cylindrical coordinatesApril 28, 2017

Problem 1. a) Find the cylindrical coordinates for the point (x, y, z) = (0, 1, 1).

We have r =√x2 + y2 =

√02 + 12 = 1, z = 1, and θ = π/2 (same as polar).

b) Find the rectangular coordinates for the point (1, π/4, 3).

We get

x = r cos θ = (1)(√

2/2) =√

2/2,

y = r sin θ = (1)(√

2/2) =√

2/2,

z = 3,

and so the point is (√

2/2,√

2/2, 3).

c) Sketch the cylindrical surface defined by z = 2r.

This is a cone with vertex at the origin:

Problem 2. Use a triple integral in cylindrical coordinates to compute the volume of acylinder with radius R and height h.

To find the area of a region, you integrate the function 1. Ditto to find the volume of aregion in 3d: we need to integrate the function 1 over this region. We are going to need0 ≤ r ≤ R, 0 ≤ θ ≤ 2π, and 0 ≤ z ≤ h. The integral then becomes

V =

∫ R

0

∫ 2π

0

∫ h

0

1 dV

=

∫ R

0

∫ 2π

0

∫ h

0

1 r dz dθ dr

Inner: ∫ h

0

r dz = rh

1

Middle: ∫ 2π

0

rh dθ = 2πrh

Outer: ∫ R

0

2πrh dr = πr2h∣∣R0

= πR2h.

That’s the right answer.

Problem 3. You want to integrate the function x2 +y2 +z2 over a cone with base a circle ofradius 3 in the xy-plane centered at the origin, and vertex at the rectangular point (0, 0, 6).Set up the corresponding integral in cylindrical coordinates.

This is a little more interesting, because one of the bounds depends on the other: the rangeof z’s that we integrate over depends on the value of r. We need 0 ≤ θ ≤ 2π and 0 ≤ r ≤ 3.For a given r, the maximum z is 6− 2r: this gives 6 at r = 0 and 0 at r = 3.

•What are the bounds? It’s 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 3, and 0 ≤ z ≤ 6− 2r.

•What’s the function? x2 + y2 + z2 = r2 + z2.

•What’s dv? In spherical, dv = r dr dθ dz

Thus the integral is ∫ 2π

0

∫ 3

0

∫ 6−2r

0

(r2 + z2) r dz dr dθ.

Problem 4. Find the volume of the solid bounded by the paraboloid z = 4− x2− y2 and thexy-plane.

The paraboloid intersects the xy-plane where 4 − x2 − y2 = 2, which is a circle of radius 2We have 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 2, and 0 ≤ z ≤ 4− r2. The volume then works out to be∫ 2π

0

∫ 2

0

∫ 4−r2

0

1 r dz dr dθ.

Inner: ∫ 4−r2

0

r dz = r(4− r2) = 4r − r3,

(since we’re integrating a constant dz) Middle:∫ 2

0

4r − r3 dr =

(2r2 − r4

4

)∣∣∣∣20

= 4− 0 = 4.

Outer: ∫ 2π

0

4 dθ = 8π.

2

Problem 5. Consider the integral∫ 2

0

∫ π/4

0

∫ 3

0

r3z dz dθ dr.

a) Sketch the region of integration.

Take a cylinder with radius 2 and height 3, and then take the part that’s above the 1stquadrant in the xy-plane. That’s your region.

b) Convert this to an integral in rectangular coordinates.

The bounds are going to be ∫ 3

z=0

∫ √4−x2y=0

∫ 2

x=0

dx dy dz

We also need to convert the volume element: r dz dθ dr = dx dy dz. So when we switch over,we are going to lose a factor of r from the integrand. The function is then r2z, which is(x2 + y2)z. The integral is therefore∫ 3

z=0

∫ √4−x2y=0

∫ 2

x=0

(x2 + y2)z dx dy dz

3

Math 210 (Lesieutre)Exam 2 reviewApril 28, 2017

Problem 1. Find the point on the plane 4x+ 3y + z = 10 nearest to (2, 0, 1).

The first thing to realize is that this is really a constrained optimization problem. Thedistance from (x, y, z) to (2, 0, 1) is just

√(x− 2)2 + y2 + (z − 1)2. We want to minimize

this, subject to the constraint 4x+ 3y+ z = 10. As usual, it will be simpler to minimize thesquare of the distance, which will find the correct closest point but save us a lot of troubleon the algebra. So we are going to use Lagrange multipliers with

f(x, y, z) = (x− 2)2 + y2 + (z − 1)2

g(x, y, z) = 4x+ 3y + z − 10.

The gradients are

∇f = 〈2x− 4, 2y, 2z − 2〉∇g = 〈4, 3, 1〉 .

We need ∇f = λ∇g, so〈2x− 4, 2y, 2z − 2〉 = λ 〈4, 3, 1〉 .

We get four equations in four variables:

2x− 4 = 4λ

2y = 3λ

2z − 2 = λ

4x+ 3y + z = 10.

As usual this isn’t very fun to solve, but we can do it. There are many approaches to thealgebra. I’m going to replace all three variables in the last equation with λ:

4x = 8 + 8λ

3y =9

2λ

z = 1 +1

2λ

4x+ 3y + z = 9 + 13λ = 10,

so λ = 113

. Plugging this back in to the original equations, we end up with

x =28

13, y =

3

26, z =

27

26.

So that’s our closest point.

1

Problem 2. Write down an iterated triple integral that expresses the volume of the tetrahe-dron bounded by the xy-plane, the yz-plane, the xz-plane, and the plane 2x + 4y + 6z = 8.Do not evaluate the integral.

This one is lifted from an old exam. First step is to draw the thing. You can use the traceslike we did in class, but with planes often the best strategy is to plot the points where itintersects the axes. It meets the x-axis where y = 0 and z = 0, so that 2x = 8, i.e x = 4. Thisis the point (4, 0, 0). Similarly it goes through (0, 2, 0) and (0, 0, 4/3). So it looks somethinglike the plane sketched below:

To set up the integral, we should first figure out the bounds on x and y that give the baseof the tetrahedron. The base is the triangle in the xy-plane sketched below:

What’s the equation for the line? Well, this is the line with z = 0, so 2x + 4y = 8, whichamounts to y = 2 − x

2. From the picture, we can see that x goes from 0 to 4. For a given

value of x, y goes from 0 to 2− x2.

At last, we need the bounds on z. For given (x, y), the lower bound is 0 (since the tetrahedronhas base on the xy-plane), and the upper bound is the plane in question. The formula for zwe get for the plane is z = 1

6(8− 2x− 4y) = 1

3(4− x− 2y).

We want to integral the function 1 to get volume, and dV = dx dy dz. So our volume is

V =

∫∫∫D

dV =

∫ 4

x=0

∫ 2−x2

y=0

∫ 13(4−x−2y)

z=0

dz dy dx.

Problem 3. Find the maximum value of the function x2y such that (x, y) lies on the unitcircle.

2

We are trying to maximize f(x, y) = x2y subject to the constraint x2 + y2 − 1 = 0, sog(x, y) = x2 + y2 − 1. Then ∇f = 〈2xy, x2〉 and ∇g = 〈2x, 2y〉. The Lagrange multiplierequation is ∇f = λ∇g, so 2xy = 2xλ and x2 = λ2y. In addition to these two equations, wehave the third equation x2 + y2 − 1 = 0.

Now, if x is not 0, the first equation just says y = λ, and the second then gives x2 = 2y2.Plugging this into the third equation, 2y2+y2−1 = 0, so y2 = 1/3, and we have y = ±1/

√3.

Then x2 = 2/3, so x = ±2/√

3.

It could be that x = 0 instead. Then the first equation will hold no matter what, and thethird says that y = ±1. The second equation is then satisfied if λ = 0. Altogether there aresix points we need to worry about:

(x, y) f(x, y)

(2/√

3, 1/√

3) 2/3√

3

(2/√

3,−1/√

3) −2/3√

3

(−2/√

3, 1/√

3) 2/3√

3

(−2/√

3,−1/√

3) −2/3√

3(0, 1) 0(1, 0) 0.

The maxima are (±2/√

3, 1/√

3), where the value of the function is 23√3.

Problem 4. a) What is the tangent plane to the surface x2+y2+z2 = 9 at the point (1, 2, 2)?

The tangent plane is normal to the gradient of the function defining the surface: f(x, y, z) =x2 + y2 + z2 − 9. The gradient is ∇f = 〈2x, 2y, 2z〉, which at the point in question is∇f(1, 2, 2) = 〈2, 4, 4〉. So the plane is normal to this vector. Since the tangent plane passesthrough the point (1, 2, 2), it must be

2(x− 1) + 4(y − 2) + 4(z − 2) = 0.

b) Consider the function f(x, y) = x√y. Use a linear approximation centered at (1, 1) to

approximate the value of f(1.1, 1.2).

We have

fx =√y

fy =x

2√y.

At the point (1, 1), these things evaluate to

f(1, 1) = 1

fx(1, 1) = 1

fy(1, 1) =1

2.

3

The formula for linear approximation says that

f(x, y) ≈ f(1, 1) + fx(1, 1)(x− 1) + fy(1, 1)(y − 1)

= 1 + 1(x− 1) +1

2(y − 1).

Plugging in (x, y) = (1.1, 1.2) as requested, we get

f(1.1, 1.2) ≈ 1 + 1(1.1− 1) +1

2(1.2− 1) = 1 + 0.1 + 0.1 = 1.2.

The true value is 1.20499. Not bad.

Problem 5. Let C be a cylinder between z = 1 and z = 3 with radius 2 and centered aroundthe z-axis, and let R be the portion of this cylinder that lies above the second quadrant in thexy-plane. Suppose you want to integrate the function 2xyz over R. Set up the correspondingintegral in cylindrical coordinates (you need not evaluate it).

We have 1 ≤ z ≤ 3, 0 ≤ r ≤ 2, and π/2 ≤ θ ≤ π. The volume element is dV = r dr dθ dz.The function is

2xyz = 2(r cos θ)(r sin θ)z = 2r2 cos θ sin θz = r2 sin(2θ)z.

So the integral is. . . ∫ 3

z=1

∫ 2

0

∫ π

π/2

r2z sin(2θ) r dθ dr dz

Problem 6. Consider the double integral∫ 1

x=0

∫ x

y=x22y dy dx.

a) Evaluate the integral directly.

Inner: ∫ x

y=x22y dy = y2

∣∣∣xy=x2

= (x)2 − (x2)2 = x2 − x4.

Outer: ∫ 1

x=0

x2 − x4 dx =

(x3

3− x5

5

)∣∣∣∣10

=1

3− 1

5=

2

15.

b) Sketch the region of integration, and switch the order of integration. (You do not need toevaluate the new integral.)

Here’s the region:

4

In the other order of integration, this works out to be∫ 1

y=0

∫ √yx=y

2y dx dy.

(You don’t have to evaluate it, but I did, and I got 2/15 this way as well, which is a goodsign. )

Problem 7. Consider the function f(x, y) = x2 − 2x+ y2.

a) Find the critical points of f(x, y) and classify the types of each.

We have

fx(x, y) = 2x− 2

fy(x, y) = 2y

fxx(x, y) = 2

fxy(x, y) = 0

fyy(x, y) = 2.

The critical point is where 2x− 2 = 0 and 4y3 = 0. This happens only for x = 1 and y = 0.At that point, we have

fxx(1, 0) = 2

fxy(1, 0) = 0

fyy(1, 0) = 2

Since fxxfyy − f 2xy = 4, which is positive, we conclude that the point is a minimum.

b) Find the absolute maximum and minimum of f(x, y) on or inside a circle of radius 2centered at the origin.

After having solved this, I realized it’s probably too much of a mess to appear on theexam. The method below is what you should do if faced with a problem like this, but don’tworry about being able to do one quite this messy. There are better examples in the olderworksheets.

We already found the critical points in part (a), and so we only need to worry about themax/min on the boundary of the region. To get at that, the best way is to parametrize it:


Here 0 ≤ t ≤ 2π. Plugging this in to the function, we get

g(t) = f(r(t)) = (2 cos t)2 − 2(2 cos t) + (2 sin t)2

= −4 cos t+ 4 cos2 t+ 4 sin2 t = 4− 4 cos t.

5

We want to know what t makes this have a max or a min. The way to do that is to set thederivative equal to 0 and see what the critical points are, Math 180-style.

g′(t) = 4 sin t

This is 0 when either t = 0 or t = π. We also need to consider the endpoints t = 0 andt = 2π. The corresponding (x, y) points are

t (x, y) = r(t)0 (2, 0)π (−2, 0)

2π (2, 0)

Now what we want to do is make a list of all the critical points on the inside of the regionand on the boundary, plug in values, and see what gives the actual max and min.

(x, y) f(x, y)(1, 0) −1(2, 0) 0

(−2, 0) 8

We see that the max is 8 at (−2, 0), and the min is −1 at (1, 0).

Problem 8. Consider the cardioid r = 1 + cos θ. Use an integral in polar coordinates tocompute the area of the portion of the cardioid that lies to the right of the y-axis.

The only trick here is to figure out the bounds we want on θ: in this case it isn’t from 0to 2π, but rather from −π/2 to π/2. The area is then given by the integral of the function1 dA, where dA = r dr dθ in polar. This gives us

A =

∫ π/2

−π/2

∫ 1+cos θ

0

1 r dr dθ.

Inner: ∫ 1+cos θ

0

r dr =r2

2

∣∣∣1+cos θ

0=

1

2(1 + cos θ)2

Outer: ∫ π/2

−π/2

1

2(1 + cos θ)2 dθ = 2 +

3π

4.

(Actually doing the outer integral requires some trig identities; I doubt we’ll have one likethat on the exam, though you might want to be ready for it.)

Problem 9. Consider the region bounded by the planes x = 0, x = 2, y = 1, y = 2, andz − 4x = 5. Use a triple integral to compute the valume of the region.

6

The volume is going to be given by∫ 2

y=1

∫ 2

x=0

∫ 5+4x

z=0

1 dz dx dy.

Inner: ∫ 5+4x

z=0

1 dz = 5 + 4x

Middle: ∫ 2

x=0

5 + 4x dx = 5x+ 2x2∣∣∣20

= 18

Outer: ∫ 2

y=1

18 dy = 18.

That’s our answer.

7

Math 210 (Lesieutre)13.5: Triple integrals in spherical coordinatesApril 28, 2017

Problem 1. a) Find the spherical coordinates for the point (x, y, z) = (0, 1, 1).

The angle with the z axis is π/4, so φ = π/4. The length is ρ =√

02 + 12 + 12 =√

2. Atlast, θ = π/2. So the point is (ρ, φ, θ) = (

√2, π/4, π/2).

b) Find the rectangular coordinates for the point (1, π/4, π/4).

We get

x = ρ sinφ cos θ = (1)(√

2/2)(√

2/2) = 1/2,

y = ρ sinφ sin θ = (1)(√

2/2)(√

2/2) = 1/2,

z = ρ cosφ = (1)(√

2/2) =√

2/2,

and so the point is (1/2, 1/2,√

2/2).

Problem 2. Use a triple integral to compute the volume of the unit sphere.

To find the area of a region, you integrate the function 1. Ditto to find the volume of aregion in 3D: we need to integrate the function 1 over this region. We are going to need0 ≤ ρ ≤ 1, 0 ≤ φ ≤ π, and 0 ≤ θ ≤ 2π. The integral then becomes

V =

∫ 1

0

∫ π

0

∫ 2π

0

1 dV

=

∫ 1

0

∫ π

0

∫ 2π

0

ρ2 sinφ dθ dφ dρ.

Inner: ∫ 2π

0

ρ2 sinφ dθ = 2πρ2 sinφ.

Middle: ∫ π

0

2πρ2 sinφ dφ = −2πρ2 cosφ∣∣∣π0

= 4πρ2.

Outer: ∫ 1

0

4πρ2 dρ = 4πρ3

3

∣∣∣10

=4π

3.

This matches up with that mysterious formula V = 43πR3 that you’ve probably heard before.

Problem 3. You want to integrate the function x2+y2+z2 over the portion of the earth withlatitude greater than 45◦ N. Convert this to an integral in spherical coordinates. (Assumethe radius of the earth is 4000.)

1

•What are the bounds? It’s 0 ≤ ρ ≤ 4000, 0 ≤ φ ≤ π/4, and 0 ≤ θ ≤ 2π.

•What’s the function? x2 + y2 + z2 = ρ2.

•What’s dV ? In spherical, dV = ρ2 sinφdρ dφ dθ

Thus the integral is∫ 4000

0

∫ π/4

0

∫ 2π

0

(ρ2)ρ2 sinφdθ dφ dρ =

∫ 4000

0

∫ π/4

0

∫ 2π

0


The problem doesn’t ask us to evaluate it, so I’m not going to.

Problem 4. a) Use a triple integral to compute the volume of the top half of the unit hemi-sphere.

This is practically identical to the first problem, with the exception that the bounds on φare now 0 ≤ φ ≤ π/2.

V =

∫ 1

0

∫ π=2

0

∫ 2π

0

1 dV

=

∫ 1

0

∫ π/2

0

∫ 2π

0


Now take a deep breath, and evaluate the integrals one at a time.

Inner: ∫ 2π

0

ρ2 sinφ dθ = 2πρ2 sinφ.

Middle: ∫ π/2

0

2πρ2 sinφ dφ = −2πρ2 cosφ∣∣∣π/20

= 2πρ2.

Outer: ∫ 1

0

2πρ2 dρ = 2πρ3

3

∣∣∣10

=2π

3.

b) Find the integral of the function z over the top half of the hemisphere.

The intgeral we want is∫ 1

0

∫ π/2

0

∫ 2π

0

z dV

=

∫ 1

0

∫ π/2

0

∫ 2π

0

(ρ cosφ)ρ2 sinφ dθ dφ dρ

=

∫ 1

0

∫ π/2

0

∫ 2π

0

1

2ρ3 sin(2φ) dθ dφ dρ.

Here I used the trig identity cosφ sinφ = 12

sin(2φ).

2

Ouch.

Inner: ∫ 2π

0

1

2ρ3 sin(2φ) dθ = πρ3 sin(2φ).

Middle: ∫ π/2

0

πρ3 sin(2φ) dφ = πρ3

Outer: ∫ 1

0

πρ3 dρ = πρ4

4

∣∣∣10

=π

4

c) To find the z-coordinate of the center of mass of a region in 3D, you can use the formula

zcm =

∫∫∫Rz dV∫∫∫

R1 dV

.

What is the center of mass of the northern hemisphere of the unit sphere?

Putting together our answers from (a) and (b), we get

zcm =π/4

2π/3=

3

8.

By symmetry the x and y coordinates are both 0. Hence the center of mass is (0, 0, 3/8).Seems plausible: a hemisphere is a little bit bottom-heavy, so the center of mass should bebelow the midpoint.

3

Math 210 (Lesieutre)13.7: Change of variables in multiple integralsApril 28, 2017

Problem 1. Consider the transformation T given by x = 2u, y = 4u+ v.

a) Let S be the region in uv-plane given by the unit square. To what region R in the xy-planedoes T send S?

We actually did some of these last semester: this particular change of coordinates is a lineartransformation, given by the matrix ( 2 0

4 1 ). In this case, perhaps the easiest way to see whatit does is to think about what happens to each of the edges of the rectangle.

The bottom edge is v = 0, 0 ≤ u ≤ 1, which turns into x = 2u, y = 4u, with 0 ≤ u ≤ 1. Thisis a line from (0, 0) to (2, 4). Making a similar check for the other variables, we concludethat the region is given by a paralellogram with vertices at (0, 0), (0, 1), (2, 4), and (2, 5) inthe xy-plane.

When you’re dealing with a linear transformation, it will always turn the unit square intoa parallelogram, and to figure out the vertices you can just check what the transformationdoes to the corners of the square. This is a pretty specific situation, but it’s one that occursfrequently and might be worth remembering.

b) What is the Jacobian for the transformation T?

It’s given by

J(u, v) =

∣∣∣∣∂x∂u ∂x∂v

∂y∂u

∂y∂v

∣∣∣∣ =

∣∣∣∣2 04 1

∣∣∣∣ = 2.

c) Compute∫∫

R

√2x(y − 2x) dA.

Using the substitution rule,∫∫R

√2x(y − 2x) dA =

∫∫S

√4uv |J(u, v)| dA

=

∫ 1

0

∫ 1

0

√4uv(2) du dv

=

∫ 1

0

8

3

√v dv =

16

9.

Problem 2. Consider the transformation T given by x = u cos v, y = u sin v.

a) Let R be the top half of a circle of radius 2, in the xy-plane. What region in the uv-planeis mapped to R by the transformation T?

This is easy: our u and v are just polar coordinates with different names! The region wewant is a rectangle with 0 ≤ u ≤ 2 (that’s the radius) and 0 ≤ v ≤ π (that’s the angle).

1

b) What is the Jacobian for the transformation T?

It’s given by

J(u, v) =

∣∣∣∣∂x∂u ∂x∂v

∂y∂u

∂y∂v

∣∣∣∣ =

∣∣∣∣cos v −u sin vsin v u cos v

∣∣∣∣ = u cos2 v + u sin2 v = u.

That means that when we integrate, the |J(u, v)| du dv replacing dx dy will be u du dv. Thisis the r dr dθ that we already know, but by another name!

c) Compute∫∫

Re−x

2−y2 dA

Using the substitution rule,∫∫R

e−x2−y2 dA =

∫∫S

e−u2 |J(u, v)| dA

=

∫ π

0

∫ 2

0

e−u2 · u du dv

= · · ·

=1

2

(1− e−4

)π.

Here’s the point of this problem: switching integrals into polar coordinates, like we’ve beendoing for awhile, actually just boils down to being a special case of this business of makinga change of variables and using the Jacobian.

Problem 3. Make a substitution to evaluate the integral of√

x+yx−y over a square R with

vertices at (2, 0), (3, 1), (3,−1), and (4, 0).

First order of business is to guess our change of variables. The natural guess would seem tobe u = x + y and v = x − y; that will make the integrand simpler. Remember that we’resupposed to give x and y in terms of u and v rather than the other way around, but we canget that from these equations. Adding together, we get u + v = 2x, so x = u+v

2. Similarly,

y = u−v2

.

The hard part is to figure out the region in uv-plane that gets sent to R under this transfor-mation. This is going to be the region that R is sent to under u = x+y and v = x−y, so wemight as well think about it that way. The corners of the square go to the (u, v) points (2, 2),(4, 2), (2, 4) and (4, 4). Since the transformation is linear, that’s our region S: 2 ≤ u ≤ 4and 2 ≤ v ≤ 4.

We also need to know the Jacobian. Using the formulas above, we get

J(u, v) =

∣∣∣∣∂x∂u ∂x∂v

∂y∂u

∂y∂v

∣∣∣∣ =

∣∣∣∣1/2 1/21/2 −1/2

∣∣∣∣ =1

2.

2

The integral is now ∫∫R

√x+ y

x− ydA =

∫ 4

2

∫ 4

2

√u

v

1

2dv du

=

∫ 4

2

(√

4−√

2)√u du

=40

3− 8√

2.

This integral could also be done without a change of variables, but it’s quite messy. Theset-up is ∫ 3

2

∫ x−2

−x+2

√x+ y

x− ydy dx+

∫ 4

3

∫ −x+4

x−4

√x+ y

x− ydy dx.

(I split it into the left half of the square and the right half of the square.) The computerconfirms that this gives the same result, but even it had to think about it for awhile.

Problem 4. Let R be the region bounded by x = −1, x = 1, y = x2, and y = x2 + 1.Compute the area by integrating the function 1 over the region, using a substitution.

It’s nice when we have a region where the sides are constants, right? The bound y = x2 + 1could be thought of as y − x2 = 1. So take u = y − x2 and v = x. Again, we need to knowx in terms of u and v to find the Jacobian, etc. So a little algebra is required: x = v andy = u+ v2

Our region in the uv-plane is now just 0 ≤ u ≤ 1 and −1 ≤ v ≤ 1. The annoying part isgoing to be the Jacobian, which is

J(u, v) =

∣∣∣∣∂x∂u ∂x∂v

∂y∂u

∂y∂v

∣∣∣∣ =

∣∣∣∣0 11 2v

∣∣∣∣ = 2v.

Rewriting the integral: ∫ 1

−1

∫ 1

0

1 |2v| du dv = 2.

3

Math 210 (Lesieutre)14.1: Vector fieldsApril 28, 2017

Problem 1. Sketch each of the following vector fields.

a) F1(x, y) = 〈x, y〉

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

b) F2(x, y) = 〈3, 0〉

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

c) F3(x, y) = yi− xj

1

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

d) F4(x, y) = student’s choice (make one up, not constant)

I made up the vector field F5(x, y) = 〈x, 0〉. Here’s a plot.

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

(You probably made up something different, so hopefully your plot doesn’t look the same.)

Problem 2. Write down formulas for the vector fields described.

a) A field F which always points clockwise, and has length 1 for any x and y.

To do this, we can start with the field from (c) of the previous problem. This already pointsin the right direction, so all we need to do to get the answer we want is adjust it so that thelength is always 1. To do that, just divide by the length:

F =

⟨y√

x2 + y2,

x√x2 + y2

⟩.

b) A 3D vector field which points directly in towards the origin, with length proportional to1/r2 (this could be a gravitational field).

2

The field 〈−x,−y,−z〉 points in the direction we want, but the length is r. We need todivide by r3 to get the length to be 1/r2, and so our field is going to be⟨

− x

(x2 + y2 + z2)3/2,− y

(x2 + y2 + z2)3/2,− z

(x2 + y2 + z2)3/2

⟩.

Problem 3. Suppose that a mass rests at (0, 0). The gravitational potential due to the massat a point (x, y) is given by f(x, y) = 1

r, where r is the distance from (x, y) to (0, 0).

a) Compute the gradient field associated to this potential function.

It’s just given by the gradient, which we compute in the usual way. The function is

f(x, y) =1√

x2 + y2

Notice that

∂

∂x

1√x2 + y2

=∂

∂x(x2 + y2)−1/2 = −1

2(x2 + y2)−3/2(2x)

= −x(x2 + y2)−3/2 = − x

(x2 + y2)3/2.

So the gradient is

∇f =

⟨− x

(x2 + y2)3/2,− y

(x2 + y2)3/2

⟩b) Try to sketch your vector field. Do the vectors get shorter or longer the further we getfrom the origin?

Here’s the sketch:

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

They get shorter. In fact, the length of the vector at position (x, y) is given by 1x2+y2

= 1r2

.This is what you know from Newton’s law of universal gravitation: the field is proportionalto 1/r2 (really it goes the other way: we found the formula for the potential assuming wealready knew what the field looks like).

3

Problem 4. Compute and sketch the gradient field associated with the function f(x, y) =tan−1(y/x).

First let’s compute the derivatives. Remember that the derivative of tan−1(x) is 11+x2 . We

have

∂

∂xtan−1(y/x) =

1

(y/x)2 + 1

(− y

x2

)= − x2

y2 + x2

y

x2= − y

x2 + y2,

∂

∂ytan−1(y/x) =

1

(y/x)2 + 1

(1

x

)=

x2

y2 + x2

1

x=

x

x2 + y2.

So the gradient field is

F = ∇f =

⟨− y

x2 + y2,

x

x2 + y2

⟩.

This is another radial field, with the vectors getting shorter as we get further from the origin.

-3 -2 -1 0 1 2 3

-3

-2

-1

0

1

2

3

Problem 5. Consider the two fields

F1(x, y) =⟨x2, y2

⟩, F2(x, y) =

⟨y2, x2

⟩.

One of these is the gradient field of some function, and the other one isn’t. Which is which?How can you tell?

Let’s try to find a potential function. If f(x, y) is a potential for the first function, thenfx = x2, and so f = x3

3+ g(y) (here g(y) could be any function, but it only depends on y,

which makes its x partial 0). Then we want fy = y2, and we know fy = gy, so gy = y2 and

g(y) = y3/3. So we found a potential: f(x, y) = x3

3+ y3

3.

How about the second one? Well, fx = y2, so f(x, y) = xy2 + g(y). This gives fy = 2xy+ gy,which we want to be y2. No function gy is going to make this work, so this one isn’t agradient.

(More on this later.)

4

Math 210 (Lesieutre)14.2: Line integralsApril 28, 2017

Problem 1. Consider the scalar function f(x, y) = xy. Compute

∫C

f ds for the listed

paths.

a) Let C be a straight line path from (2, 0) to (0, 2).

First we parametrize. Using our usual formula for a straight line parametrization, this comesout to

r(t) = 〈2, 0〉+ t 〈−2, 2〉 = 〈2− 2t, 2t〉 .

We’re also going to need to know

r′(t) = 〈−2, 2〉|r′(t)| =

√(−2)2 + (2)2 = 2

√2.

(In this case it doesn’t depend on t. Sometimes it will, when the speed of the particle isvarying.)

So x(t) = 2− 2t, y(t) = 2t. The integral is now∫C

f ds =

∫ 1

0

(2− 2t)(2t)(√

2) dt = 4√

2

∫ 1

0

t− t2 dt =2√

2

3.

b) Let C be the part of a circle of radius 2 centered at the origin that lies in the first quadrant,oriented counterclockwise.

We follow the same strategy, starting with a parametrization. In this case we want a circularpath, and so


with the range 0 ≤ t ≤ π/2. We’re also going to need to know

r′(t) = 〈−2 sin t, 2 cos t〉|r′(t)| =

√(−2 sin t)2 + (2cost)2 =

√4 = 2.

(Again, there’s no t dependence, because our particle is moving at constant speed.)

So x(t) = 2 cos t, y(t) = 2 sin t. The integral is now∫C

f ds =

∫ 1

0

(2 cos t)(2 sin t)(2) dt = 8

∫ π/2

0

cos t sin t dt = 81

2= 4.

c) What is the average value of f(x, y) along the path from (b)?

We just need to take our answer and divide it by the length of the path. It’s a quarter circle,so the length is 1

4(4π) = π. The average value is then going to be 4

π.

1

Problem 2. Now consider the vector field F(x, y) = 〈xy, 1 + x〉:

-2 -1 0 1 2

-2

-1

0

1

2

Let C1 be the quarter-circle path from problem 1b, and C2 the straight line path from (0, 0)to (2, 0).

a) Do you expect∫C1

F · dr to be positive, negative, or zero? How about∫C2

F · dr?

Along the path C1, we have y = 0, and so the vector field points directly upward. Thismeans that it’s perpendicular to our path, and so not doing any work: this wind is comingstraight from the side, and so neither helps nor hinders our walking. Thus we expect thatthe integral is 0.

For the second path, looking at the picture, it appears that the wind is more or less at ourbacks, at least until towards the end of the path. So it should help, and we expect that theintegral is positive.

b) Compute∫C1

F · dr.

The parametrization is r(t) = 〈2t, 0〉, so that x(t) = 2t and y(t) = 0. Plugging in, we get

F(t) = 〈(2t)(0), 1 + (2t)〉

We have r′(t) = 〈2, 0〉. So the integral is∫ 1

t=0

〈0, 1 + 2t〉 · 〈2, 0〉 dt =

∫ 1

0

0 dt = 0.

c) Compute∫C2

F · dr, or at least set up the corresponding single variable integral.

This is a little more work. We have r(t) = 〈2 cos t, 2 sin t〉. Then

F(t) = 〈xy, 1 + x〉 = 〈(2 cos t)(2 sin t), 2 cos t+ 1〉= 〈4 cos t sin t, 2 cos t+ 1〉

r′(t) = 〈−2 sin t, 2 cos 2〉

2

What we want is then∫ π/2

0

〈4 cos t sin t, 2 cos t+ 1〉 · 〈−2 sin t, 2 cos t〉 dt

=

∫ π/2

0

(−8 cos t sin2 t+ 4 cos2 t+ 2 cos t

)dt = π − 2

3.

It takes a bit of trig to integrate this mess. Here’s a plot of the function we’re integrating:

0.5 1.0 1.5

2

4

6

It’s mostly positive, which means that the wind is more or less at our backs, and so doingpositive work. So it should come as no surprise that the integral is positive.

Problem 3. Consider the change of variables T given by u = 2x + y and v = x + y. LetR be the parallelogram in the xy-plane with vertices at (0, 0), (−1, 2), (3,−3), and (2,−1).Compute ∫∫

R

xy dx dy.

First we need to express the region in uv-coordinates. What are these points? Using theformulas for u and v, we find that they are respectively (u, v) = (0, 0), (0, 1), (3, 0), and(3, 1). This is a coordinate rectangle!

We need x and y as a function of u and v, instead of the other way. Notice that u − v =(2x + y) − (x + y) = x. Then y = v − x = v − (u − v) = 2v − u. So x(u, v) = u − v andy(u, v) = 2v − u. (There are other ways to do the algebra to get this answer – it’s up toyou.)

Now, the function xy becomes (u− v)(2v − u).

At last, what to make of dx dy. We need the Jacobian, which is

J(u, v) =

∣∣∣∣∂x∂u ∂x∂v

∂y∂u

∂y∂v

∣∣∣∣ =

∣∣∣∣ 1 −1−1 2

∣∣∣∣ = 1.

That means we get off easy this time: dx dy = J(u, v) du dv = 1 du dv = du dv.

So our integral is ∫ 1

v=0

∫ 3

u=0

(u− v)(2v − u) du dv = · · · = −17

4.

3

Math 210 (Lesieutre)14.3: Conservative vector fields, IApril 28, 2017

Problem 1. Consider the two fields

F1(x, y) =⟨x2, y2

⟩, F2(x, y) =

⟨y2, x2

⟩.

One of these is a conservative field, and the other one isn’t. Which is which? Find thepotential function.

Let’s try to find a potential function. Our field is f(x, y) = x2 and g(x, y) = y2. If F = ∇φ,then φx = x2. This means that φ = x3

3+ c(y) (here c(y) could be any function, but it only

depends on y, which makes its x partial 0). Then we want φy = y2. Since φ = x3

3+ c(y), we

get y2 = cy, so that c = y3

3. This gives our potential as φ(x, y) = x3

3+ y3

3.

How about the second one? Well, fx = y2, so f(x, y) = xy2 + g(y). This gives fy = 2xy+ gy,which we want to be y2. No function gy is going to make this work, so this one isn’t agradient.

Problem 2. Check whether the field F = 〈sin y, x cos y + 1〉 is conservative. If it is, find apotential function.

We have fy = cos y and gx = cos y. Since there isn’t anything funny going on (both functionsare defined everywhere), this means that the field is going to be conservative. Now to actuallyfind the potential function.

We want our function φ to have φx = cos y, so φ = x cos y+ c(y). Then φy = x sin y+ cy(y).We want that to be x cos y + 1, so it had better be the case that cy = 1. That meansthat c(y) = y (plus a constant, if you want). All told, this means that our fnuction isφ(x, y) = x cos y + y.

Problem 3. Consider the vector field

F(x, y, z) = 〈yz, xz + zeyz, xy + yeyz + sin z〉 .

This field is conservative. Find a potential function.

We have φx = yz, so φ = xyz + c(y, z).

We then have φy = xz+ cy(y, z), which should be xz+ zeyz. So cy = zeyz, yielding c(y, z) =eyz + d(z). This gives φ = xyz + eyz.

Now, φz = xy+yeyz+dz, should should be xy+yeyz+sin z. So dz = sin z, and d(z) = − cos z.Our potential function is therefore

φ(x, y, z) = xyz + eyz − cos z.

1

Problem 4. Consider the field F1 = 〈x2, y2〉 from the first problem. Compute the lineintegral

∫CF1 ·dr for two paths from (0, 0) to (1, 1): first, a straight line path. Second, a path

that goes from (0, 0) to (1, 0) in a straight line, and then from (1, 0) to (1, 1) in a straightline.

Let’s do the first one first. We have r(t) = 〈t, t〉 with 0 ≤ t ≤ 1. Then r′(t) = 〈1, 1〉.Plugging in x(t) = t and y(t) = t, our integral becomes∫ 1

0

⟨t2, t2

⟩· 〈1, 1〉 dt =

∫ 1

0

2t2 dt =2

3.

Now comes the second path. We have to integrate over the two parts separately, then addthe result. For the first piece, r(t) = 〈t, 0〉 with 0 ≤ t ≤ 1. This gives us r′(t) = 〈1, 0〉, andwe have to integrate ∫ 1

0

⟨t2, 0

⟩· 〈1, 0〉 dt =

∫ 1

0

t2 dt =1

3.

For the second piece, r(t) = 〈0, t〉 and r′(t) = 〈0, 1〉. The corresponding integral is∫ 1

0

⟨0, t2

⟩· 0, 1 dt =

∫ 1

0

t2 dt =1

3.

The total integral along this bent path is 13

+ 13

= 23.

We got the same answer for both paths. Is it a coincidence? No – this always happens forconservative fields. We’ll talk about this next time.

2

Math 210 (Lesieutre)14.3: Conservative vector fields, IIApril 28, 2017

Problem 1. Consider the field F1 = 〈x2, y2〉. Last time we saw that this is a conservative

field, and that φ(x, y) = x3

3+ y3

3is a potential function.

a) Compute the line integral of F along a straight line path from (0, 0) to (1, 1).

Let’s do the first one first. We have r(t) = 〈t, t〉 with 0 ≤ t ≤ 1. Then r′(t) = 〈1, 1〉.Plugging in x(t) = t and y(t) = t, our integral becomes∫ 1

0

⟨t2, t2

⟩· 〈1, 1〉 dt =

∫ 1

0

2t2 dt =2

3.

b) Compute the line integral of F along a path that goes from (0, 0) to (1, 0) and then to(1, 1). What do you notice?

We have to integrate over the two parts separately, then add the results together. For thefirst piece, r(t) = 〈t, 0〉 with 0 ≤ t ≤ 1. This gives us r′(t) = 〈1, 0〉, and we have to integrate∫ 1

0

⟨t2, 0

⟩· 〈1, 0〉 dt =

∫ 1

0

t2 dt =1

3.

For the second piece, r(t) = 〈0, t〉 and r′(t) = 〈0, 1〉. The corresponding integral is∫ 1

0

⟨0, t2

⟩· 0, 1 dt =

∫ 1

0

t2 dt =1

3.

The total integral along this bent path is 13

+ 13

= 23.

What do I notice? Well, we got the same answer for both paths. This is suspicious.

c) Repeat the previous two calculations using the fundamental theorem for line integrals.

Both paths start at (0, 0) and end at (1, 1). We know that a potential function is

φ(x, y) =x3

3+y3

3,

so the integral is just ∫C

F · dr = φ(1, 1)− φ(0, 0)

Problem 2. Consider the vector field F(x, y) = 〈y + 1, x+ 1〉.

a) Verify that F is a conservative field and find a potential function φ(x, y).

We have f(x, y) = y + 1 and g(x, y) = x + 1. Then fy = 1 and gx = 1, and because thevector field is defined for all x and y that means it’s conservative.

We need φx = y + 1, so φ = xy + x + c(y). Then φy = x + cy, which is supposed to equalx+ 1. So cy(y) = 1, whence c(y) = y. Our potential function is then φ(x, y) = xy + x+ y.

1

b) Let C be a semicircular path from (1, 0) to (−1, 0). Use the fundamental theorem for line

integrals to compute

∫C

F · dr.

The path starts at (1, 0) and ends at (−1, 0). That means that∫C

F · dr = φ(−1, 0)− φ(1, 0) = −1− 1 = −2.

c) Let C be a path that goes the whole way around the unit circle. Check by a direct compu-tation that ∫

C

F · dr = 0.

We parametrize the path by r(t) = 〈cos t, sin t〉. Then r′(t) = 〈− sin t, cos t〉, with bounds0 ≤ t ≤ 2π. Now we plug this in to F = 〈y + 1, x+ 1〉 and integrate.

∫C

F · dr =

∫ 2π

0

〈sin t+ 1, cos t+ 1〉 · − sin t, cos t dt

=

∫ 2π

0

(− sin2 t− sin t+ cos2 t+ cos t

)dt

=

∫ 2π

0

cos(2t)− sin t+ cos t dt

=1

2sin(2t) + cos t+ sin t

∣∣∣∣2π0

= 0.

Problem 3. Let F be the vector field 〈x, y〉 and let C be a straight line from (1, 1) to (−1, 1).

a) Do you expect the flux of F across C to be positive, negative, or zero?

The field is flowing outward from the origin, across C. We expect the flux to be positive.

b) Compute the flux

∫C

F · n ds.

The path is parametrized by r(t) = 〈1− 2t, 1〉. The tangent vector is r′(t) = 〈−2, 0〉. Thenormal is n = 〈0, 2〉 (switch the two parts of r′ and then multiply the second by −1). Thisis an upward normal vector.

Our integral is then ∫ 1

0

〈1− 2t, 1〉 · 0, 2 dt =

∫ 1

0

2 dt = 2.

Positive, like we thought.

2

Math 210 (Lesieutre)14.4: Green’s theorem, IApril 28, 2017

Circulation form: Flux form:∮C

F · dr =

∫∫R

curlF dA

∮C

F · n ds =

∫∫R

divF dA

curlF = ∂g∂x− ∂f

∂ydivF = ∂f

∂x+ ∂g

∂y

Problem 1. Compute divF and curlF for the indicated vector fields.

a) F1(x, y) = 〈x, y〉Here f(x, y) = x and g(x, y) = y. The divergence is fx + gy = 1 + 1 = 2. This seemsphysically reasonable: the field flows outward everywhere.

The curl is gx − fy = 0− 0 = 0. Also sensible: from a sketch of the vector field, it looks likea small propellor wouldn’t turn.

b) F2(x, y) = 〈−y, x〉This time we get divF = 0 + 0 = 0 while curlF = 1 − (−1) = 2. This also makes sense:this field looks like a whirlpool around the original, and a propellor stuck in anywhere wouldspin counterclockwise.

Problem 2. Let F = 〈−y, x〉, and let C be a path around the unit circle starting at (1, 0)and going counterclockwise. Verify the circulation form of Green’s theorem by computingboth sides directly.

Let’s first do the line integral. We parametrize the path as r(t) = 〈cos t, sin t〉. Pluggingthis into the vector field we get F = 〈−y, x〉 = 〈− sin t, cos t〉. We also need to knowr′(t) = 〈− sin t, cos t〉 (yes, it’s the same thing as F; this is basically a coincidence since Iwanted an easy integral). We then compute∮

C

F · dr =

∫ 2π

0

〈− sin t, cos t〉 · 〈− sin t, cos t〉 dt =

∫ 2π

0

sin2 t+ cos2 t dt =

∫ 2π

0

1 dt = 2π.

Now let’s do the area integral. The curl is curlF = gx− fy = 1− (−1) = 2, and so we want:∫∫R

curlF dA =

∫∫R

2 dA = 2(area of R) = 2π.

Seems to work!

1

Problem 3. Let R be the region bounded by y = 1− x2 and y = 0, and let C be a path thatgoes around the region counterclockwise, starting at (1, 0). Verify the flux form of Green’stheorem for the vector field F = 〈x+ y, xy〉.Start with the double integral this time. The divergence is fx + gy = 1 + x, and so we wantto integrate the function 1 + x over the triangle. The top edge of the region y = 1− x2, sothe integral we want is∫ 1

x=−1

∫ 1−x2

y=0

1 + x dy dx =

∫ 1

x=−1(1− x2)(1 + x) dx =

∫ 1

x=−11 + x− x2 − x3

=

(x+

x2

2− x3

3− x4

4

)∣∣∣∣1−1

=11

12−(− 5

12

)=

16

12=

4

3.

Now we need to compute the flux across the edges, of which there are two. First, let’s dothe top, curved edge; call it C1

r(t) =⟨−t, (−t)2

⟩=⟨−t, 1− t2

⟩r′(t) = 〈−1,−2t〉n(t) = 〈−2t, 1〉 .

The range is −1 ≤ t ≤ 1. Plugging in, we have

F = 〈x+ y, xy〉 =⟨−t+ (1− t2),−t+ t3

⟩=⟨1− t− t2, t3 − t

⟩.

This normal vector points upward, which is out of the region; that’s as it should be, soeverything looks good.

The flux is ∫C1

F · n ds =

∫ 1

t=−1

⟨1− t− t2, t3 − t

⟩· 〈−2t, 1〉 dt

=

∫ 1

t=−1

(−2t+ 2t2 + 2t3

)+(t3 − t

)dt

=

∫ 1

t=−13t3 + 2t2 − 3t dt =

4

3.

Now we have to compute the flux across the bottom edge. This is a straight line from (−1, 0)to (1, 0), and so we get

r(t) = 〈2t, 0〉r′(t) = 〈2, 0〉n(t) = 〈0,−2〉 .

2

Our range this time is −1 ≤ t ≤ 1. Plugging in we have F = 〈x+ y, xy〉 = 〈2t, 0〉, and theintegral is ∫

C2

F · n ds =

∫ 1

t=0

〈2t, 0〉 · 〈0,−2〉 =

∫ 1

t=0

0 dt = 0.

The total flux is then∮C

F · n ds =

∫C1

F · n ds+

∫C2

F · n ds =4

3+ 0 =

4

3.

This confirms that ∮C

F · n ds =

∫∫R

divF dA.

Problem 4. What does the circulation form of Green’s theorem tell us when F is a conser-vative vector field?

If F is conservative, then fy = gx. This means that curlF = gx − fy = 0. The theorem saysthat ∮

C

F · dr =

∫∫R

curlF dA

The right-hand side is 0, and so the theorem is telling us that∮C

F · dr = 0.

Of course, we already knew that because of the fundamental theorem for line integrals!

3

Math 210 (Lesieutre)14.4: Green’s theorem, IIApril 28, 2017

Circulation form: Flux form:∮C

F · dr =

∫∫R

curlF dA

∮C

F · n ds =

∫∫R

divF dA

curlF = ∂g∂x− ∂f

∂ydivF = ∂f

∂x+ ∂g

∂y

Problem 1. Let F = 〈−y, x〉, and let C be a path around the unit circle starting at (1, 0)and going counterclockwise. Verify the circulation form of Green’s theorem by computingboth sides directly.

Let’s first do the line integral. We parametrize the path as r(t) = 〈cos t, sin t〉. Pluggingthis into the vector field we get F = 〈−y, x〉 = 〈− sin t, cos t〉. We also need to knowr′(t) = 〈− sin t, cos t〉 (yes, it’s the same thing as F; this is basically a coincidence since Iwanted an easy integral). We then compute∮

C

F · dr =

∫ 2π

0

〈− sin t, cos t〉 · 〈− sin t, cos t〉 dt =

∫ 2π

0

sin2 t+ cos2 t dt =

∫ 2π

0

1 dt = 2π.

Now let’s do the area integral. The curl is curlF = gx− fy = 1− (−1) = 2, and so we want:∫∫R

curlF dA =

∫∫R

2 dA = 2(area of R) = 2π.

Seems to work!

Problem 2. Suppose that you have a region R bounded by a curve C. What does the circu-lation form of Green’s theorem tell you when it’s applied to the field F = 〈0, x〉?Well, the curl of this field is

curlF = gx − fy = 1.

So Green’s theorem says ∮F · dr =

∫∫R

1 dA = area(R).

The line integral is the area!

There is actually a gadget called a planimeter that you can fix to a table and then track outa curve with a pencil. The device is geared so that as you trace it out, it adds up

∮F · dr.

Once you trace around a complete loop, it computes the area of the region. This is withoutusing any computers! You can find some description of how these things work if you googlearound a little bit.

1

Problem 3. Use Green’s theorem to compute the outward flux of the field F = 〈x2y, y〉across a counterclockwise semicircular path from (2, 0) to (0,−2).

One could probably do this by a direct integral, but there is a useful trick with Green’stheorem to avoid it.

Let C be a closed loop that starts at (2, 0), follows the path in question to (−2, 0) aroundthe top of the circle, and then goes in a straight line from (−2, 0) to (2, 0). Let’s give namesto the two parts of this path: C1 is the curved part around the top, and C2 is the straightpart around the bottom. Let R be the interior of C, which is a half disk.

Then Green’s theorem tells us that∫∫R

divF dA =

∮C

F · dr =

∫C1

F · dr +

∫C2

F · dr.

If we could compute

∫∫R

divF dA and

∫C2

F · dr, that would give us

∫C1

F · dr, which is a

much worse integral than either of the other two.

Let’s start with the double integral. The divergence is

divF = fx + gy = 2xy + 1.

So we want to take the double integral of that over the region R. Remember that we’reintegrating over a half-disk, so life is going to be a lot more pleasant if we do this integral inpolar coordinates. The function is

f(x, y) = 2xy + 1 = 2(r cos θ)(r sin θ) + 1

= 2r2 cos θ sin θ + 1

= r2 sin 2θ + 1.

Don’t forget to use r dr dθ for dA, since we’re working in polar.

Our integral is now: ∫∫R

divF dA =

∫ 2

r=0

∫ π

θ=0

(r2 sin(2θ) + 1

)r dθ dr

=

∫ 2

r=0

∫ π

θ=0

r3 sin(2θ) + r dθ dr

The inner one is ∫ π

θ=0

r3 sin 2θ + r dθ = r3∫ π

θ=0

sin 2θ dθ +

∫ π

θ=0

r dθ

= r3− cos 2θ

2

∣∣∣∣π0

+ πr = πr.

2

The outer is ∫ 2

r=0

πr dr =πr2

2

∣∣∣∣20

= 2π.

We also need to compute ∫C2

F · dr.

This is just an old-fashioned circulation integral. The parametrization is

r(t) = 〈−2, 0〉+ t 〈4, 0〉 = 〈−2 + 4t, 0〉dr(t)

dt= 〈4, 0〉 .

with 0 ≤ t ≤ 1.

Plugging that in,∫C

F · dr =

∫ 1

t=0

⟨(−2 + 4t)2(0), 0

⟩· 〈4, 0〉 dt =

∫ 1

t=0

0 dt = 0.

Going back to our original equation,∫∫R

divF dA =

∫C1

F · dr +

∫C2

F · dr

2π =

∫C1

F · dr + 0∫C1

F · dr = 2π,

which is what we were trying to compute. In this case the original integral wouldn’t havebeen all that bad, but there are situations where this trick is really very handy.

Problem 4. Evaluate the line integral

∮C

(2x+ y) dx+ (x+ y) dy, where C is a square with

opposite corners at (0, 0) and (1, 1).

First, what is this weird notation? I haven’t used it very heavily.

Think of it as ∮C

〈x+ y, 2x+ y〉 ·⟨dx

dt,dy

dt

⟩dt

So this is a circulation integral. (Sometimes you’ll see∫Cf dy− g dx; this is 〈f, g〉 · dy,−dx,

which is the flux.)

3

Anyway, nobody wants to do four separate line integrals, so we’ll use Green’s theorem instead.The field in question in F = 〈x+ y, 2x+ y〉, and so curlF = gx − fy = 2 − 1 = 1. ThenGreen’s theorem tells us that∮

C

(2x+ y) dx+ (x+ y) dy =

∫∫R

1 dA = 1 area(R) = 1,

since the square has area 1.

Problem 5. Let R be the unit circle. Convert the double integral∫∫

Rxy + sinx dA into a

line integral; you do not need to evaluate.

There’s more than one way to do this. If we can find a field F for which divF = 2xy+ sinx,we’d be done, because we could then use Green’s theorem. Here are a couple fields thatwould work as such an F:

F1(x, y) =⟨x2y − cosx, 0

⟩,

F2(x, y) =⟨x2y, y sinx

⟩then ∫∫

R

xy + sinx dA =

∮C

⟨x2y, y sinx

⟩· dr,

according to Green’s theorem. We could then evaluate this integral.

By finding a field F with curlF = 2xy+ sinx, we could accomplish the same thing using theother version of Green’s theorem.

4

Math 210 (Lesieutre)14.5: Divergence and curlApril 28, 2017

Problem 1. Let F be the vector field F = 〈xyz, yex, z〉.

a) Compute the divergence ∇ · F.

The divergence is a scalar function, given in this case by

∇ · F = fx + gy + hz = yz + ex + 1.

b) Compute the curl ∇× F.

The curl is given by

∇× F =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

f g h

∣∣∣∣∣∣=

(∂h

∂y− ∂g

∂z

)i +

(∂f

∂z− ∂h

∂x

)j +

(∂g

∂x− ∂f

∂y

)k

= (0− 0) i + (xy − 0) j + (yex − xz)k

= xyj + (yex − xz)k = 〈0, xy, yex − xz〉 .

c) Compute the divergence of the curl, ∇ · (∇× F).

We found that the curl was given by

G = ∇× F = 〈0, xy, yex − xz〉 .

The divergence is then

∇ ·G = fx + gy + hz = 0 + x+ (0− x) = 0.

In fact this is not a coincidence: the divergence of the curl is always 0.

Problem 2. Suppose that we have a 3D vector field of the form F = 〈f(x, y), g(x, y), 0〉 (i.e.f and g only depend on x and y, and h = 0). What is the (3D) curl ∇× F? What do younotice about this?

The curl is given by

∇× F =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

f g h

∣∣∣∣∣∣=

(∂h

∂y− ∂g

∂z

)i +

(∂f

∂z− ∂h

∂x

)j +

(∂g

∂x− ∂f

∂y

)k

= (0− 0)i + (0− 0)j + (gx − fy)k.

The gx − fy is the 2d curl of the corresponding 2d field F = 〈f, g〉.

1

Problem 3. Consider the function φ(x, y, z) = xy2z.

a) Compute the gradient F = ∇φ. (This F is a conservative field.)

The gradient is given by 〈φx, φy, φz〉 = 〈y2z, 2xyz, xy2〉.

b) Compute the curl of the gradient, ∇×∇φ.

Once again, the curl is

∇× F =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

f g h

∣∣∣∣∣∣=

(∂h

∂y− ∂g

∂z

)i +

(∂f

∂z− ∂h

∂x

)j +

(∂g

∂x− ∂f

∂y

)k

= (2xy − 2xy)i + (y2 − y2)j + (2yz − 2yz)k.

In fact the curl of a gradient is always 0, which gives us a way to check remember how tocheck whether a 3d field is conservative.

Problem 4. Consider the vector field F = 〈1, 0, 3〉 × r, where r = 〈x, y, z〉. This is anexample of a rotation vector field.

a) Multiply out the cross product to find a formula for F.

I get

F = 〈1, 0, 3〉 × 〈x, y, z〉 =

∣∣∣∣∣∣i j k1 0 3x y z

∣∣∣∣∣∣ = (0− 3y)i− (z − 3x)j + (y − 0)k = 〈−3y, 3x− z, y〉 .

b) Compute the curl of this field.

The curl is

∇× F =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

f g h

∣∣∣∣∣∣=

(∂h

∂y− ∂g

∂z

)i +

(∂f

∂z− ∂h

∂x

)j +

(∂g

∂x− ∂f

∂y

)k

= (1− (−1))i + (0− 0)j + (3− (−3))k = 2i + 6k = 〈2, 0, 6〉 .

It’s no coincidence that this is twice the vector 〈1, 0, 3〉 that we started with – this is howthings always work for rotation fields F = a× r.

2

Math 210 (Lesieutre)14.6: Surface integralsApril 28, 2017

Problem 1. Let S be the outer shell of a cylinder of height 3 and radius 4, with base centeredat the origin.

a) Give a parametrization for S, and compute tu × tv.

How do we describe a point on a cylinder? We use cylindrical coordinates (r, θ, z). In thiscase, r is always 3, so we’re going to use the other two. Take u to be θ and v to be z. Ourcoordinates are

r(u, v) = 〈4 cosu, 4 sinu, v〉 .Then

tu =∂r

∂u= 〈−4 sinu, 4 cosu, 0〉

tv =∂r

∂v= 〈0, 0, 1〉

tu × tv = 〈4 cosu, 4 sinu, 0〉|tu × tv| =

√(4 cosu)2 + (4 sinu)2 + 02 =

√16 = 4.

The bounds would be 0 ≤ u ≤ 2π and 0 ≤ v ≤ 3.

b) How would you compute the integral∫∫

Sxz dS?

The function xz becomes (4 cosu)(v) = 4v cosu in these coordinates. So we’d use the integral∫∫S

xz dS =

∫ 2π

u=0

∫ 3

v=0

4v cosu (4 du dv).

The extra 4 at the end comes from |tu × tv| du dv.

Problem 2. Give parametrizations for the following regions.

a) A sphere of radius 7.

We’re going to use the spherical coordinates θ and φ as our coordinates. Let’s take u to beφ and v to be θ. We’re working on the surface S, and so ρ is always going to be 7.

This givesr(u, v) = 〈7 sinu cos v, 7 sinu sin v, 7 cosu〉 .

In this case, we want 0 ≤ u ≤ π and 0 ≤ v ≤ 2π.

b) The part of the plane z = 5 + 2x+ y lying above the unit square in the xy-plane.

In this case our parameters are just going to be u = x and v = y, and we’ll use z = 5+2u+v.So we want

r(u, v) = 〈u, v, 5 + 2u+ v〉 .

1

Our bounds are 0 ≤ u ≤ 1, 0 ≤ v ≤ 1.

I guess the problem doesn’t ask us to compute the area element dS, but what’s the harm?

tu =∂r

∂u= 〈1, 0, 2〉

tv =∂r

∂v= 〈0, 1, 1〉

tu × tv = 〈−2,−1, 1〉|tu × tv| =

√6.

Problem 3. Consider the surface S comprising the portion of the paraboloid z = 1−x2−y2lying above the unit disk.

a) Give a parametrization of S; be sure to include bounds on u and v.

We’re just going to take x(u, v) = u and y(u, v) = v. Then z(u, v) = 1−u2− v2. We requireu and v to be in the unit disc R in the uv-plane.

It’s also possible to skip straight to parametrizing it using polar coordinates for the x andy coordinates, which is actually a little easier. I’m going to do it the long way, but on thehomework you might consider taking this approach the entire way through. In this case,we’d take u to be r and v to be θ. Then

x(u, v) = u cos v

y(u, v) = u sin v

z(u, v) = 1− x2 − y2 = 1− v2.

The bounds will just be 0 ≤ u ≤ 1 and 0 ≤ v ≤ 2π (remember that u and v are really justthe coordinates from polar).

b) Compute the tangent vectors tu = ∂r∂u, tv = ∂r

∂v, the cross product tu × tv, and the length

|tu × tv|.I get

tu = 〈1, 0,−2u〉tv = 〈0, 1,−2v〉

tu × tv =

∣∣∣∣∣∣i j k1 0 −2u0 1 −2v

∣∣∣∣∣∣ = 〈2u, 2v, 1〉

|tu × tv| =√

(2u)2 + (2v)2 + 1 =√

1 + 4u2 + 4v2.

If you were doing it the polar way, you’d instead take the derivatives of the alternate answergiven in the first part of problem 1.

2

c) Set up an integral for the surface area of this part of the paraboloid. If you have time,compute the integral.

We just want to integrate the function 1, so our integral is∫∫S

1 dS =

∫∫R

1√

1 + 4u2 + 4v2 du dv

Here R is the unit disk in the uv-plane.

To actually compute this, we have little choice but to convert the integral into polar. Theintegrand

√1 + 4u2 + 4v2 is

√1 + 4r2, and the bounds are 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π. We

also have to change the du dv into r dr dθ, and the integral then becomes∫∫S

1 dS =

∫∫R

1√

1 + 4u2 + 4v2 du dv

=

∫ 2π

θ=0

∫ 1

r=0

√1 + 4r2 r dr dθ

The inner one is∫ 1

r=0

r√

1 + 4r2 dr =1

12(1 + 4r2)3/2

∣∣∣10

=5√

5

12− 1

2=

5√

5− 1

12.

The outer is then ∫ 2π

θ=0

5√

5− 1

12=

(5√

5− 1)π

6.

That’s the surface area.

Problem 4. Let S be the top half of a sphere of radius 2 centered at the origin.

a) Give a parametrization of S; be sure to include bounds on u and v.

We’re going to use the spherical coordinates θ and φ as our coordinates. Let’s take u to beφ and v to be θ. We’re working on the surface S, and so ρ is always going to be 1.

This givesr(u, v) = 〈2 sinu cos v, 2 sinu sin v, 2 cosu〉 .

In this case, we want 0 ≤ u ≤ π/2 and 0 ≤ v ≤ 2π.

b) Compute the tangent vectors tu = ∂r∂u

and tv = ∂r∂v.

We have

tu =∂r

∂u= 〈2 cosu cos v, 2 cosu sin v,−2 sinu〉

tv =∂r

∂v= 〈−2 sinu sin v, 2 sinu cos v, 0〉

tu × tv = 4 sin2 u cos v, 4 sin2 u sin v, 4 sinu cosu

|tu × tv| = · · · = 4 sinu.

3

c) Compute the surface area of this portion of the sphere. Does your answer match the oneyou have learned before?

The surface area is given by∫∫S

1 dS =

∫ 2π

v=0

∫ π/2

u=0

1 (4 sinu) du dv

=

∫ 2π

v=0

4 dv = 8π.

The formula we know is that the surface area of a sphere is 4πR2. In this case the radius is2, so the surface area of a sphere of radius 2 is 16π. Since we only have half a sphere, ouranswer of 8π makes sense.

d) Compute the average value of the angle φ on this part of the sphere. What is the averagelatitude of a point in the northern hemisphere?

In this case, we would want to use1

area(S)

∫∫S

φ dS. The set-up is the same as above,

except we need to change the function we’re integrating to u (which is how to express φ inuv-coordinates). ∫∫

S

1 dS =

∫ 2π

v=0

∫ π/2

u=0

u (4 sinu) du dv.

The inner integral requires an integration by parts, which I’m going to omit. The value is4. The outer integral is then

∫ 2π

v=04 dv = 8π.

We know that the area is also 8π, so the average value of φ is just φ = 1. That’s one radian,which is about 57.29 degrees. Remember that φ is 90◦ minus the latitude, rather than thelatitude itself. So the corresponding average latitude is ≈ 90− 57.29 = 32.70.

This seems plausible: although there are points with every latitude between 0 and 90 degrees,there are a lot more points at small latitudes than large ones, so the average ought to beless than half of 90. This is roughly the latitude of Dallas.

There’s a little global perspective: most points on the earth are closer to the equator thanDallas, and almost the entire United States (except Hawaii and some of the most southerlyparts of the south) is further north than the global average.

4

Math 210 (Lesieutre)14.6: Surface integralsApril 28, 2017

Problem 1. Evaluate the indicated flux integrals

¨S

F ·n dS without making any computa-

tions.

a) S is the unit sphere, F = k is a constant upward field.

Imagine that this is the flow of fluid in a 3d region. The field is constant, so the amountgoing into the sphere is equal to the amount going out. The flux is 0.

b) S is the unit sphere, F = 〈x, y, z〉.The flow is always directly out of the sphere. Since on the sphere F is a unit vector, whenwe dot with the unit normal n we are going to get 1 for all points. So the integral is

˜S

1 dS,which is the surface area, which is 4π.

Problem 2. Compute the flux of the field 〈2z, 3, 2x〉 across the top face of a tetrahedron withvertices at (1, 0, 0), (0, 1, 0), and (0, 0, 1). (Hint: the equation for the top is z = 1− x− y.)

This one is given to us as a graph, so it should be easy to parametrize. We’ll use

r(u, v) = 〈u, v, 1− u− v〉 .

The bounds are a little tricky. We want u and v to vary over the base of this tetrahedron,just like if we were trying to do a triple integral over the region (there was one like this onthe last test). So our bounds are going to be

ˆ 1

u=0

ˆ 1−u

v=0

.

We also need the normal vector, which is

tu = 〈1, 0,−1〉tv = 〈0, 1,−1〉

tu × tv = 〈1, 1, 1〉 .

We’re now in position to actually compute the integral. Plug in our parametrization to theformula for the field: 〈2z, 3, 2x〉 = 〈2− 2u− 2v, 3, 2u〉.

¨S

F · n dS =

ˆ 1

u=0

ˆ 1−u

v=0

〈2− 2u− 2v, 3, 2u〉 · 〈1, 1, 1〉 dv du

=

ˆ 1

u=0

ˆ 1−u

v=0

5− 2v dv du.

1

Inner: ˆ 1−u

v=0

5− 2v dv = 5v − v2∣∣∣1−u0

= 5(1− u)− (1− u)2 = 4− 3u− u2.

Outer: ˆ 1

u=0

4− 3u− u2 =13

6.

Problem 3. Let V be a cylinder with base at the origin, radius 3, and height 4, and considerthe vector field F = 〈z2, 3y, 1〉.

a) Compute the flux of F across S1, the outer wall of the cylinder.

Looking at our table, we have r(u, v) = 〈3 cosu, 3 sinu, v〉, with 0 ≤ u ≤ 2π and 0 ≤ v ≤ 4.The normal vector is tu × tv = 〈3 cosu, 3 sinu, 0〉.When we plug in our parametrization to the vector field, we get F = 〈z2, 3y, 1〉 = 〈v2, 9 sinu, 1〉.That’s all we need to know to compute the integral.

¨S1

F · n dS =

ˆ 4

v=0

ˆ 2π

u=0

⟨v2, 9 sinu, 1

⟩· 〈3 cosu, 3 sinu, 0〉 du dv

=

ˆ 4

v=0

ˆ 2π

u=0

3v2 cosu+ 27 sin2 u du dv

=

ˆ 4

v=0

27π dv = 108π.

b) Compute the flux of F across S2, the top of the cylinder. (Use an upward-pointing normalvector.)

Here the surface is just r(u, v) = 〈u, v, 4〉, since the top is at height 4. We’re going to usethe upward normal vector 〈0, 0, 1〉. The region is R, a circle of radius 3.

¨S2

F · n dS =

¨R

〈16, 4v, 1〉 · 〈0, 0, 1〉 du dv

=

¨R

1 du dv = area(R) = 9π.

If that had come out to be a hard integral, we’d probably want to switch to polar. Luckily,there was no need.

c) Compute the flux of F across S3, the bottom of the cylinder. (Use a downward-pointingnormal vector.)

This is almost identical to the previous one. The surface is just r(u, v) = 〈u, v, 0〉, since thebottom is at height 0. We’re going to use the downward normal vector 〈0, 0,−1〉. The region

2

is R, a circle of radius 3.

¨S2

F · n dS =

¨R

〈0, 4v, 1〉 · 〈0, 0,−1〉 du dv

=

¨R

−1 du dv = −area(R) = −9π.

d) What is the total outward flux across the cylinder?

It’s

‹S

F · n dS = 27π + 9π − 9π = 27π.

Why did I make us go through all this? Soon we’ll learn the divergence theorem, that says

‹S

F · n dS =

˚V

∇ · F dV

We found that the left side is 27π. On the right, the divergence is ∇ · F = 0 + 3 + 0 = 3,and so ˚

V

∇ · F dV = 3 volume(V ) = 3(36π) = 108π.

So the divergence theorem checks out, at least in this example.

3

Math 210 (Lesieutre)14.7: Stokes’ theorem, 1April 28, 2017

Problem 1. Let S be the portion of the paraboloid z = 1 − x2 − y2 lying above the planez = 0. Check Stokes’ theorem for the field F = 〈x, y, z〉.Let’s do the line integral first. The boundary of S is the unit circle in the xy-plane, and soa parametrization is

r(t) = 〈cos t, sin t, 0〉 ,

with 0 ≤ t ≤ 2π. This hasr′(t) = 〈− sin t, cos t, 0〉 .

Plugging that in, we get∮C

F · dr =

∫ 2π

0

〈cos t, sin t, 0〉 · 〈− sin t, cos t, 0〉 dt =

∫ 2π

0

0 dt = 0.

How about the surface integral? The curl is

∇× F =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

x y z

∣∣∣∣∣∣ = (0− 0)i− (0− 0)j + (0− 0)k = 〈0, 0, 0〉 .

The other side of Stokes’ theorem is then∫∫S

(∇× F) · n dS =

∫∫S

0 · n dS =

∫∫S

0 dS = 0.

Luckily we don’t have to actually parametrize the paraboloid this time, since we’re integrat-ing 0. Make sure you know how to! There was one of these a couple worksheets ago.

Problem 2. Let S be the top half of a sphere of radius 2. Compute

∫∫S

k · n dS.

(Hint: for the field F = 〈0, x, 0〉, we get ∇× F = 〈0, 0, 1〉.)If there’s one thing that the last few days have taught us, it’s that computing flux integralsacross a sphere is a generally unpleasant undertaking: the parametrization has too manytrig functions to keep track of. Luckily for us, ∇×F = 〈0, 0, 1〉, and so Stokes’ theorem saysthat ∫∫

S

k · n dS =

∫∫S

(∇× F) · n dS =

∮C

F · dr,

where C is a path around the boundary of C. In this case, the boundary is a circle of radius2 in the xy-plane, which is easy to deal with.

1

This time things are oriented so that we want to go counterclockwise around a circle of radius2. A parametrization is

r(t) = 〈2 cos t, 2 sin t, 0〉r′(t) = 〈−2 sin t, 2 cos t, 0〉 .

Then ∮C

F · dr =

∫ 2π

t=0

〈0, 2 cos t, 0〉 · 〈−2 sin t, 2 cos t, 0〉 dt

=

∫ 2π

t=0

4 cos2 t dt =

∫ 2π

t=0

4

(1 + cos 2t

2

)dt

=

∫ 2π

t=0

2 + 2 cos 2t dt = (2t− sin 2t)∣∣2π0

= 4π.

Problem 3. What does Stokes’ theorem tell us in the case that ∇×F = 0? Does this seemplausible?

If ∇× F = 0 then certainly ∫∫S

(∇× F) · n dS = 0.

Stokes’ theorem tells us that this means∮C

F · dr = 0,

where C is a curve around the boundary of S. But of course we already knew that: ∇×F = 0means that F is conservative, and because C is a closed loop, the fundamental theorem forline integrals tells us that the integral around a closed loop is 0.

Problem 4. Let S be a cylinder of height 2 and radius 1 centered at the origin, not includingeither of the ends. This region has two boundary components. Find an orientation for each,and verify Stokes’ theorem for the field F = 〈yz,−xz, 0〉.Assume the normal vector to the field is pointing outward, we want we to go clockwise aroundthe top edge and counterclockwise around the bottom edge. The top edge is parametrizedby

r(t) = 〈cos t,− sin t, 1〉

(note: by “centered at” I mean the cylinder extends up to z = 1 and down to z = −1). Then

r′(t) = 〈− sin t,− cos t, 0〉 .

2

We get ∮C1

F · dr =

∫ 2π

0

〈(− sin t)(1), (cos t)(1), 0〉 · 〈− sin t,− cos t, 0〉 dt

=

∫ 2π

0

sin2 t+ cos2 t+ 0 dt =

∫ 2π

0

1 dt = 2π.

The bottom edge is parametrized by

r(t) = 〈cos t, sin t,−1〉 .

Thenr′(t) = 〈− sin t, cos t, 0〉 .

The integral is∮C2

F · dr =

∫ 2π

0

〈(sin t)(−1),−(cos t)(−1), 0〉 · 〈− sin t, cos t, 0〉 dt

=

∫ 2π

0


∫ 2π

0

1 dt = 2π.

So the total circulation is∮C

F · dr =

∮C1

F · dr +

∮C2

F · dr = 2π + 2π = 4π.

We need to check that this is equal to the other side of Stokes’ theorem, which is∫∫S

(∇× F) · n dS.

First step is to find the curl of F. For that, I get

∇× F =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

yz −xz 0

∣∣∣∣∣∣ = (0− (−x))i− (0− y)j + (−z − z)k = xi + yj = 〈x, y,−2z〉 .

We’ve seen before that for a cylinder of radius a we have a parametrization

r(u, v) = 〈a cosu, a sinu, v〉

which here is justr(u, v) = 〈cosu, sinu, v〉

since a = 1. The bounds are 0 ≤ u ≤ 2π and −1 ≤ v ≤ 1.

n = 〈a cosu, a sinu, 0〉 ,

3

which in this case givesn = 〈cosu, sinu, 0〉 .

Then plugging in x, y, and z to ∇× F, we get∫∫S

(∇× F) · n dS =

∫ 2π

u=0

∫ 1

v=−1〈cosu, sinu, 0〉 · 〈cosu, sinu, 0〉 dv du

=

∫ 2π

u=0

∫ 1

v=−1cos2 u+ sin2 u+ 0 dv du

=

∫ 2π

u=0

∫ 1

v=−11 dv du

= (2π)(2) = 4π.

This matches our first answer.

4

Math 210 (Lesieutre)14.6/7: Stokes’ and divergence theoremsApril 28, 2017

Problem 1. Let S be a cylinder of height 2 and radius 1 centered at the origin, not includingeither of the ends. This region has two boundary components. Find an orientation for each,and verify Stokes’ theorem for the field F = 〈yz,−xz, 0〉.Assume the normal vector to the field is pointing outward, we want we to go clockwise aroundthe top edge and counterclockwise around the bottom edge. The top edge is parametrizedby

r(t) = 〈cos t,− sin t, 1〉

(note: by “centered at” I mean the cylinder extends up to z = 1 and down to z = −1). Then

r′(t) = 〈− sin t,− cos t, 0〉 .

We get ∮C1

F · dr =

∫ 2π

0

〈(− sin t)(1), (cos t)(1), 0〉 · 〈− sin t,− cos t, 0〉 dt

=

∫ 2π

0


∫ 2π

0

1 dt = 2π.

The bottom edge is parametrized by

r(t) = 〈cos t, sin t,−1〉 .

Thenr′(t) = 〈− sin t, cos t, 0〉 .

The integral is∮C2

F · dr =

∫ 2π

0

〈(sin t)(−1),−(cos t)(−1), 0〉 · 〈− sin t, cos t, 0〉 dt

=

∫ 2π

0


∫ 2π

0

1 dt = 2π.

So the total circulation is∮C

F · dr =

∮C1

F · dr +

∮C2

F · dr = 2π + 2π = 4π.

We need to check that this is equal to the other side of Stokes’ theorem, which is∫∫S

(∇× F) · n dS.

1

First step is to find the curl of F. For that, I get

∇× F =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

yz −xz 0

∣∣∣∣∣∣ = (0− (−x))i− (0− y)j + (−z + z)k = xi + yj = 〈x, y, 0〉 .

We’ve seen before that for a cylinder of radius a we have a parametrization

r(u, v) = 〈a cosu, a sinu, v〉

which here is justr(u, v) = 〈cosu, sinu, v〉

since a = 1. The bounds are 0 ≤ u ≤ 2π and −1 ≤ v ≤ 1.

n = 〈a cosu, a sinu, 0〉 ,which in this case gives

n = 〈cosu, sinu, 0〉 .

Then plugging in x, y, and z to ∇× F, we get∫∫S

(∇× F) · n dS =

∫ 2π

u=0

∫ 1

v=−1〈cosu, sinu, 0〉 · 〈cosu, sinu, 0〉 dv du

=

∫ 2π

u=0

∫ 1

v=−1cos2 u+ sin2 u+ 0 dv du

=

∫ 2π

u=0

∫ 1

v=−11 dv du

= (2π)(2) = 4π.

This matches our first answer.

Problem 2. Find the flux of the field F = 〈x+ y sin z, xz, 4z〉 across a sphere of radius 2centered at the origin.

This would be a pretty bad integral to do directly. Instead, we’re going to use the divergencetheorem, which lets us convert it into a triple integral over the inside of the sphere.

The divergence theorem tells us that∫∫S

F · n dS =

∫∫∫D

∇ · F dV.

So we need to know the divergence, ∇ · F = 1 + 0 + 4 = 5. This reduces our problem toevaluating the triple integral∫∫∫

D

5 dV = 5 volume(D) = 5(4/3π(2)3) =160π

3.

2

Problem 3. Let S be portion of the paraboloid z = 1 − x2 − y2 lying above the xy-plane.Compute the flux of the vector field from Problem 2 across S. (Hint: the divergence theoremwill make your life easier.)

The surface S isn’t a closed surface, so it’s not so clear how to use the divergence theoremhere. Let S ′ be a “cap” on the bottom of the paraboloid, which makes it into a closedsurface. Now take D to be the 3d region bounded by the paraboloid and the xy-plane.

The divergence theorem tells us that∫∫∫D

∇ · F dV =

∫∫S

F · n dS +

∫∫S′F · n dS.

The first of these we’ll be able to compute without too much trouble. The second of theseis what we want to know, and it’s a big mess. The third one is also not so bad. So the planis to compute the first and third and use them to find the second.

Let’s do first things first and take the triple integral. We are trying to integrate the function 3.We’ve done some integrals on paraboloids before; it’ll be easiest to use cylindrical coordinatesand remember that z = 1− x2 − y2 = 1− r2.

∫∫∫D

∇ · F dV =

∫ 1

r=0

∫ 2π

θ=0

∫ 1−r2

z=0

5 r drdθ = · · · = 5π

2.

Next we want to calculate∫∫

S′ F · n dS. We’ll use polar: use u is θ and v is r. Then oursurface is

x(u, v) = v cosu

y(u, v) = v sinu

z(u, v) = 0.

Then

tu = 〈−v sinu, v cosu, 0〉tv = 〈cosu, sinu, 0〉

tu × tv = 〈0, 0,−v〉 .

We need to be a little careful here: is this the right normal vector? Since S ′ is the bottomof the 3d region, and we want an outward normal, we want it to point downward, which iswhat we got.

Plugging in x, y, and z to F = 〈x+ y sin z, xz, 4z〉, we just get 〈x, 0, 0〉. That’s easy!

So the flux across the bottom is now∫∫S′F · n dS =

∫ 1

v=0

∫ 2π

u=0

〈v cosu, 0, 0〉 · 〈0, 0,−v〉 du dv =

∫ 1

v=0

∫ 2π

u=0

0 du dv = 0.

3

So we’ve found ∫∫∫D

∇ · F dV =

∫∫S

F · n dS +

∫∫S′F · n dS

5π

2=

∫∫S

F · n dS + 0∫∫S

F · n dS =5π

2.

It was a pain, but definitely less work than doing the integral directly.

Problem 4. Check that the divergence theorem is true by computing both sides for the fieldF = 〈x, 2y, 3z〉 and the region D = {(x, y, z) : x2 + y2 + z2 ≤ 9}.Remember the theorem: ∫∫

S

F · n dS =

∫∫∫D

∇ · F dV.

This is a sphere of radius 3, and so (using the formula from Table 14.3)

r(u, v) = 〈3 sinu cos v, 3 sinu sin v, 3 cosu〉tu × tu =

⟨9 sin2 u cos v, 9 sin2 u sin v, 9 sinu cosu

⟩∫∫S

F · n dS =

∫ π

u=0

∫ 2π

v=0

〈3 sinu cos v, 6 sinu sin v, 9 cosu〉 ·⟨9 sin2 u cos v, 9 sin2 u sin v, 9 sinu cosu

⟩dv du

=

∫ π

u=0

∫ 2π

v=0

27 cos2(v) sin3(u) + 54 sin3(u) sin2(v) + 81 cos2(u) sin(u) dv du

= · · · = 36π + 72π + 108π = 216π.

(I’m leaving on the work on the integrals, but they’re actually pretty harmless.)

On the other hand, the divergence of the field is 1 + 2 + 3 = 6, and so∫∫∫D

∇ · F dV =

∫∫∫D

6 dV = 3(4/3π(3)3) = 216π.

So the divergence theorem checks out.

4

Math 210 (Lesieutre)Chapter 14 reviewApril 28, 2017

Problem 1. Compute the flux of the vector field F = 〈2x2, xz4, sin y〉 across a tetrahedronwith vertices at the points (1, 0, 0), (0, 2, 0), and (0, 0, 3).

We’d have to do flux integrals over four different surfaces to get this, so of course it’s a betteridea to just use the divergence and compute the triple integral of the divergence over thetetrahedron. The divergence is easy to calculate for this one: it’s

∇ · F = 4x.

So we need to compute ˚D

4x dV.

This is a trick review question: the first thing we need to do is come up with the equationfor the plane! Call the points A, B, and C. To get the normal vector, we’re going to useAB ×AC. We have AB = 〈−1, 2, 0〉 and AC = 〈−1, 0, 3〉. Then AB ×AC = 〈−6,−3,−2〉.A plane with this normal vector through (1, 0, 0) has equation

−6(x− 1)− 3(y − 0)− 2(z − 0) = 0.

This simplifies to 6x + 3y + 2z = 6. Solving for z, we get z = 6−6x−3y2

. The base in thexy-plane is obtained by taking z = 0, which gives y = 2− 2x.

ˆ 1

x=0

ˆ 2−2x

y=0

ˆ 6−6x−3y2

z=0

4x dz dy dx.

This comes out to

T =

ˆ 1

x=0

ˆ 2−2x

y=0

ˆ 6−6x−3y2

z=0

4x dz dy dx

=

ˆ 1

x=0

ˆ 2−2x

y=0

(6− 6x− 3y

24x

)dy dx

=

ˆ 1

x=0

ˆ 2−2x

y=0

12x− 12x2 − 6xy dy dx

=

ˆ 1

x=0

12x3 − 24x2 + 12x dx = 1.

1

Problem 2. What does each of the following types of integral represent? Can you rememberhow to compute one?

a)

ˆC

f ds

This is the integral of a scalar function along a path. To compute it, you need to parametrizethe path by r(t) = 〈x(t), y(t)〉, then plug in x and y to the function f , and use |r′(t)| dtfor ds. The bounds on the integral are the bounds on your parametrization, you’re justintegrating a function of t.

b)

ˆC

F · dr

This is a circulation integral, also known as a work integral, also known as the integral ofa vector field along a path. The procedure is similar: parametrize it by r(t) = 〈x(t), y(t)〉.Plug in x and y to the vector field F, to get a vector. Dot that vector with the vector r′(t),and integrate the result from t = a to t = b.

c)

ˆC

F · n ds

The procedure is mostly like the above, except you want to dot that vector with the vectorn(t) = 〈y(t),−x(t)〉, and integrate the result from t = a to t = b.

d)

¨S

f dS

This is the integral over a scalar function over a surface. This is the one that’s the most workof the whole bunch. You need to parametrize the surface S by r(u, v) = 〈x(u, v), y(u, v), z(u, v)〉.Then compute tu = ∂r

∂uand tv = ∂r

∂v, and use those to compute |tu × tv|. At last, take

|tu × tv|, a scalar function involving a square root unless you’re very fortunate.

To set up the double integral, use your bounds on the parametrization, then plug in the x,y, and z to f(x, y, z) to get a function of u and v. For dS, you want to use |tu × tv| du dv.

e)

¨S

F · n dS

Start off like the above, compute up to tu× tv. Plug in x, y, z to your field F to get a vectorfield in terms of u and v. Dot that with tu × tv to get a function of u and v. Integrate thatover the bounds of your parametrization.

Problem 3. Try to recall the main integration theorems from this chapter.

a) Fundamental theorem for line integrals.

This one says that if F is a conservative field, so that F = ∇φ for some scalar function φ,and if C is a curve, then ˆ

C

F · dr = φ(B)− φ(A),

where B is the endpoint of C and A is the beginning point of C.

2

b) Green’s theorem, circulation form.

This time we have a region R and a curve C going around R, counterclockwise. Then

˛C

F · dr =

¨R

curlF dA.

Here curlF is the curl (duh). If we have a 2D field F = 〈f(x, y), g(x, y)〉, then divF = gx−fy.

c) Green’s theorem, flux form.

This is similar to the previous one. It says that

˛C

F · n = ds =

¨R

div dA.

Here divF = ∇ · F is the divergence. If we have a 2D field F = 〈f(x, y), g(x, y)〉, thendivF = fx + gy.

d) Stokes’ theorem

This is a 3D thing. We have a surface S in 3D, and the boundary of the surface is a curveC (oriented appropriately – see the previous sheet).

˛C

F · dr =

¨S

(∇× F) · n dS.

e) Divergence theorem

The divergence theorem related a triple integral to a double integral. Let D be a 3D region,with boundary a surface S. Then

˚D

(∇ · F) dV =

‹S

F · n dS.

Problem 4. (We’ll vote on an integral to compute.)

3

Math 210 (Lesieutre)Exam review #1April 28, 2017

Problem 1. Try to recall the main integration theorems from this chapter.

a) Fundamental theorem for line integrals.

This one says that if F is a conservative field, so that F = ∇φ for some scalar function φ,and if C is a curve, then ˆ

C

F · dr = φ(B)− φ(A),

where B is the endpoint of C and A is the beginning point of C.

b) Green’s theorem, circulation form.

This time we have a region R and a curve C going around R, counterclockwise. Then˛C

F · dr =

¨R

curlF dA.

Here curlF is the curl (duh). If we have a 2D field F = 〈f(x, y), g(x, y)〉, then divF = gx−fy.

c) Green’s theorem, flux form.

This is similar to the previous one. It says that˛C

F · n = ds =

¨R

div dA.

Here divF = ∇ · F is the divergence. If we have a 2D field F = 〈f(x, y), g(x, y)〉, thendivF = fx + gy.

d) Stokes’ theorem

This is a 3D thing. We have a surface S in 3D, and the boundary of the surface is a curveC (oriented appropriately – see the previous sheet).

˛C

F · dr =

¨S

(∇× F) · n dS.

e) Divergence theorem

The divergence theorem related a triple integral to a double integral. Let D be a 3D region,with boundary a surface S. Then

˚D

(∇ · F) dV =

‹S

F · n dS.

Problem 2. Find the equation for the tangent line to the curve r(t) = 〈t, sin t, 3〉 at thepoint (π, 0, 3).

1

Any find-a-line question, the strategy is exactly the same. We need to know a point thatthe line goes through, and the direction vector for the line. In this case, the point is givento us: it’s (π, 0, 3). So we need the direction. To find the direction of the tangent, we needto take the derivative of r(t) (remember that that will give the tangent vector to a curve).Since our point is r(π), the tangent vector will be the vector r′(π). I get r′(t) = 〈1, cos t, 0〉,and so r′(π) = 〈1,−1, 0〉. So our line has equation

`(t) = r0 + vt = 〈π, 0, 3〉+ t 〈1,−1, 0〉 = 〈π + t,−t, 3〉 .

Problem 3. Let f(x, y) = x2−x+y2. Find the absolute maximum and minimum of f(x, y)on a disk of radius 3 centered at the origin.

This is a classic max/min with a boundary problem. Two parts: find the critical points onthe interior, and find the max/min on the boundary.

First, let’s do the critical points. We have fx = 2x−1 and fy = 2y. Solve fx = 0 and fy = 0,we find that the only possibility is (x, y) = (1/2, 0). So that’s our first candidate as a globalmax/min.

Now we need to find the max and min on the boundary. There are a couple ways to handlethis. First is the way we usually did it before: parametrize the boundary in terms of t,and find the values of t that are going to maximize or minimize. THe second way is to useLagrangle multipliers, since this is really a constrained optimization problem: we are tryingto find the max/min of f(x, y) = x2 − x+ y2 subject to the constraint x2 + y2 − 9 = 0.

First, the parametrization method: the curve is given by r(t) = 〈3 cos t, 3 sin t〉, where 0 ≤t ≤ 2π. Plugging that in to f , we get

g(t) = f(r(t)) = 9 cos2 t− 3 cos t+ 9 sin2 t = 9− 3 cos t.

What are the max and min of this, with respect to t? Well, take the derivative:

g′(t) = 3 sin t,

which is 0 when t = 0 or π. These are possible max/min on the boundary, and theycorrespond to the points r(0) = 〈3, 0〉 and r(π) = 〈−3, 0〉. The possible interesting pointsare therefore:

(x, y) f(x, y)(1/2, 0) −1/4

(3, 0) 6(−3, 0) 12

This shows that the max is 12, achieved at (−3, 0), and the min is −1/4, achieved at (1/2, 0).

2

We could also have used Lagrange to find the extreme points on the boundary. We want tomaximize f(x, y) = x2 − x+ y2 subject to the constraint g(x, y)x2 + y2 − 9 = 0.

∇f = 〈2x− 1, 2y〉∇g = 〈2x, 2y〉 .

The equation ∇f = λ∇g gives us the two equations 2x − 1 = λ(2x) and 2y = λ(2y). Thethird equation is x2 + y2 − 9 = 0. The second equation is the most promising. It says that2y(λ− 1) = 0, so either y = 0 or λ = 1. If y = 0, the third equation gives x = 3 or x = −3,giving two points (x, y) = (3, 0) and (x, y) = (−3, 0). In the other case, that λ = 1, we get2x−1 = 2x, which has no solutions. So we found the only two solutions. These are the samepoits we got with the paametrization method, so you’d do the remained of the problem thesame way.

Problem 4. Consider the surface S defined by z = 1 + x + 2y and above the rectangle[1, 2]× [2, 3].

a) Set up an integral to compute the volume below S and above the xy-plane.

We’d just use ˆ 2

x=1

ˆ 3

y=2

1 + x+ 2y dy dx = · · · = 15

2.

b) Set up an integral for the surface area of S.

This is more work. We want to compute

¨S

1 dS, which mean we have to go through all the

rigamarole of parametrization and so on.

The parametrization is x(u, v) = u, y(u, v) = v, and z(u, v) = 1 + u + 2v, with 1 ≤ u ≤ 2and 2 ≤ v ≤ 3. This gives

r(u, v) = 〈u, v, 1 + u+ 2v〉tu = 〈1, 0, 1〉tv = 〈0, 1, 2〉

tu × tv = 〈−1,−2, 1〉|tu × tv| =

√6.

The surface area becomes

SA =

¨S

1 dS =

ˆ 2

u=1

ˆ 3

v=2

1 |tu × tv| dv du

=

ˆ 2

u=1

ˆ 3

v=2

√6 dv du =

√6.

3

c) Set up an integral for the flux of 〈x2, y − z, 3〉 across S.

We’re going to use the stuff that we computed in the previous part of the problem. Tocompute flux, we substitute in the parametrization to our equation for the vector field:

¨S

F · n dS =

ˆ 2

u=1

ˆ 3

v=2

⟨u2, v − (1 + u+ 2v), 3

⟩· 〈−1,−2, 1〉 dv du

=

ˆ 2

u=1

ˆ 3

v=2

−u2 + 2u+ 2v + 5 dv du =32

3.

Problem 5. a) Consider the integral

ˆ 3

x=0

ˆ x3

y=0

xy dy dx. Sketch the region of integration,

and reverse the order of the integrals.

It’s the region underneath the graph of y = x3 and above the x-axis, for x between 0 and 3.The top right corner of the region is the point (3, 27), and so on this region we’re going tohave y from 0 to 27. For a given y, the lower bound on x is the graph, and the upper boundis x = 0. The graph is y = x3, which means x = 3

√y. So our integral is going to be:

ˆ 27

y=0

ˆ 3

x= 3√y

xy dx dy.

b) Consider the integral

ˆ 3

x=0

ˆ √9−x2

y=0

e−x2−y2 dy dx.. Convert this integral into polar coordi-

nates.

There are three things we need to do: convert the bounds to polar, convert the function topolar, and convert dy dx to polar. First the bounds. This region is the first quadrant partof a circle of radius 3, so we’re going to have 0 ≤ r ≤ 3 and 0 ≤ θ ≤ π/2. The function ise−x

2−y2 = e−r2. And as always, we’re going to use r dr dθ. So our answer is

ˆ 3

x=0

ˆ √9−x2

y=0

e−r2

r dr dθ.

4

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