contentsazhou/teaching/17f/hw-solutions.pdf · problem 5 (13.1.70). use vectors to prove that the...

MATH 32A-1 (17F) (L) R. Killip Calculus of Several Variables

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1 Homework 1 - Solutions 3






1


HOMEWORK 1 - SOLUTIONS

Problem 1 (13.1.44). Find the vector of length 3 in the direction of v = 4i + 3j.

Solution. The length of v is ‖v‖ =√

42 + 32 = 5, so the vector of length 3 in the same direction is

3

5v =

(12

5

)i +

(9

5

)j.

Problem 2 (13.1.46). Find the unit vector in the direction opposite to v =

(−2

4

).

Solution. The length of v is ‖v‖ =√

(−2)2 + 42 = 2√

5, so the unit vector in the opposite directionis

− 1

2√

5v =

(1/√

5

−2/√

5

).

Problem 3 (13.1.60). Sketch the parallelogram spanned by v =

(1

4

)and w =

(5

2

). Add the

vector u =

(2

3

)to the sketch and express u as a linear combination of v and w.

Solution. The first part is shown below.

v

w

u

3

Homework Solutions MATH 32A-1 (17F)

For the remainder of the problem, we need to find constants r and s such that u = rv+sw. Writingthis in components, (

2

3

)= r

(1

4

)+ s

(5

2

)=

(r + 5s

4r + 2s

).

Thus we must solve the system of equations

r + 5s = 2, 4r + 2s = 3.

The solution is (r, s) = (11/18, 5/18), so the desired linear combination is

u =

(11

18

)v +

(5

18

)w .

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Problem 4 (13.1.64). Determine the magnitude of the forces F1 and F2 in Figure 1a (omitted),assuming that there is no net force on the object.

Solution. We first create a force diagram in which every force vector has its tail at the origin.

F1

F2

Fgiven

45◦30◦

Let f1 = ‖F1‖ and f2 = ‖F2‖. The components of the vectors in the diagram are

F1 =

(0

f1

), F2 =

(f2/√

2

−f2/√

2

), Fgiven =

(−10√

3

−10

).

We are told that the forces sum to 0, so by looking at each component,

0 +f2√

2− 10

√3 = 0, f1 −

f2√2− 10 = 0.

From the first equation, we obtain f2 = 10√

6 . Then substituting this into the second equation,

we obtain f1 = 10 + 10√

3 .

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Problem 5 (13.1.70). Use vectors to prove that the segments joining the midpoints of oppositesides of a quadrilateral bisect each other.

Solution. Note: this solution is not what one would get by following the hint in the book.

Let ABCD be a quadrilateral, O be the origin, a =−→OA, and similarly define b, c, and d.

The midpoint of AB is 12 (a + b), while the midpoint of CD is 1

2 (c + d). Therefore, the midpointof the segment connecting these two points is given by

p =1

2

(1

2(a + b) +

1

2(c + d)

)=

1

4(a + b + c + d).

By a similar argument, the midpoint of the segment connecting the midpoints of BC and DA isalso p. Therefore the two segments intersect at their midpoints, so they bisect each other.

Problem 6 (13.2.44). Find the parametric equation for the line through (1,−1, 0) and (0,−1, 2).

Solution.

Problem 7 (13.2.50). Find the point of intersection of the lines

r(t) =

1

0

0

+ t

−3

1

0

and s(t) =

0

1

1

+ t

2

0

1

.

Problem 8 (13.2.67). A median of a triangle is a segment joining a vertex to the midpoint ofthe opposite side. Referring to Figure 2, prove that the medians of triangle ABC intersect at the

terminal point P of the vector1

3(u + v + w). The point P is the centroid of the triangle.

Hint: show, by parametrizing the segment AA′, that P lies two-thirds of the way from A to A′. Itwill follow similarly that P lies on the other two medians.

Figure 1: Centroid of a triangle (13.2.67)

Problem 9 (13.3.26). Find the angle between the vectors

3

1

1

and

2

−4

2

.

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Problem 10 (13.3.30). Find a vector that is orthogonal to

−1

2

2

.

Problem 11 (13.3.36). Simplify the expression

(v + w) · (v + w)− 2v ·w.

Problem 12 (13.3.62). Compute the component of u =

3

0

9

along v =

1

2

2

.

Problem 13 (13.3.68). Find the decomposition

a = a‖b + a⊥b

with respect to b for

a =

4

−1

5

and b =

2

1

1

.

Problem 14 (13.3.84). Let P and Q be antipodal points on a sphere of radius r centered at theorigin and let R be a third point on the sphere (Figure 3). Prove that PR and QR are orthogonal.

Figure 2: Right angle formed by antipodal points (13.3.84)

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Problem 1. Fix 0 < b ≤ a and consider the parametric curve

r(t) =

(a cos t

b sin t

),

which is an ellipse.

(a) Find the unit tangents T(t) and normals N(t) to the ellipse.

(b) Find the curvature of the ellipse as a function of t.

(c) Find the points of minimal and maximal curvature (if a > b) and mark them on a sketch.

(d) Find the osculating circles at the points of maximal and minimal curvature and mark themon a sketch. Include explicitly the locations of the centers.

(e) Defining c =√a2 − b2, show that the perimeter of the triangle with vertices

r(t), F1 =

(c

0

), F2 =

(−c0

)

is independent of t and determine its value. The points F1 and F2 are the foci of the ellipse.

(f) Show that the two vectors point to r(t) from each of the foci make equal angles with N(t).

Solution. (a) The plan here is to first find some tangent vector to the ellipse and some vectorwhich is normal (perpendicular) to this tangent vector. In the plane, this is sufficient to findthe unit tangent T and normal N.

For the tangent vector, we can take

r′(t) =

(−a sin t

b cos t

).

A vector which is normal to this is

M =

(b cos t

a sin t

).

Rescaling the tangent vector, we get the unit tangent

T(t) =1√

(−a sin t)2 + (b cos t)2

(−a sin t

b cos t

)=

1√b2 + c2 sin2 t

(−a sin t

b cos t

),

where c2 = a2 − b2 (see part e).

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If we rescale M, we obtain

M̃ =1√

b2 + c2 sin2 t

(b cos t

a sin t

),

which is a unit vector that is normal to T. However, there are two vectors in the plane whichsatisfy this, N and −N, so we have to figure out which one M̃ is. Note that when 0 < t < π/2,we are in the first quadrant and the curve is moving counterclockwise, so N points inwards(in the direction the curve is turning). Therefore, its components should both be negative,while the components of M̃ are both positive when 0 < t < π/2. Thus N = −M̃, or

N(t) =1√

b2 + c2 sin2 t

(−b cos t

−a sin t

).

(b) For this question, we will use the formula

κ(t) =‖r′(t)× r′′(t)‖‖r′(t)‖3

,

with the vectors realised in 3D by making the third components zero. We’ve computed r′(t)above, from which we get

‖r′(t)‖ =√b2 + c2 sin2 t,

r′′(t) =

(−a cos t

−b sin t

).

Then

r′(t)× r′′(t) =

−a sin t

b cos t

0

×−a cos t

−b sin t

0

=

0

0

ab

.

Thus

κ(t) =ab

(b2 + c2 sin2 t)3/2.

(c) The minimum curvature occurs where the denominator is as large as possible, in which casesin2 t = 1. For this value, we obtain

κ(t) =ab

(b2 + c2)3/2=

b

a2, t = π/2, 3π/2 .

These points are the “top” and “bottom” of the ellipse.

The maximum curvature occurs where the denominator is as small as possible, in which casesin2 t = 0. For this value, we obtain

κ(t) =ab

(b2)3/2=

a

b2, t = 0, π .

These points are the “right” and “left” ends of the ellipse.

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(d) First we look at the osculating circle at the top. The radius of this circle is

R(π/2) =1

κ(π/2)=

a2

b,

while the center is

r(π/2) +R(π/2)N(π/2) =

(0

b

)+a2

b· 1√

b2 + c2

(0

−a

)=

(0

b− a2/b

)=

(0

−c2/b

).

The other osculating circles can be found similarly, and they are

Bottom Radius =a2

bCenter =

(0

c2/b

),

Left Radius =b2

aCenter =

(−c2/a

0

),

Right Radius =b2

aCenter =

(c2/a

0

).

(e) The perimeter of the triangle is

‖r(t)− F1‖+ ‖F1 − F2‖+ ‖F2 − r(t)‖

=√

(a cos t− c)2 + (b sin t)2 + 2c+√

(−c− a cos t)2 + (−b sin t)2

=√c2 cos2 t− 2ac cos t+ c2 + b2 + 2c+

√c2 cos2 t+ 2ac cos t+ c2 + b2

=√

(c cos t− a)2 + 2c+√

(c cos t+ a)2

= |c cos t− a|+ 2c+ |c cos t+ a|= (a− c cos t) + 2c+ (a+ c cos t) = 2a+ 2c,

which does not depend on t. Note that to simplify the absolute values, we used the fact that|c cos t| ≤ c < a.

(f) The angle θ1 between N and r(t)− F1 is given by

cos θ1 =(r(t)− F1) ·N‖r(t)− F1‖‖N‖

=1

b2 + c2 sin2 t

(a cos t− c)(−b cos t) + (b sin t)(−a sin t)

a− c cos t

=1

b2 + c2 sin2 t

bc cos t− aba− c cos t

=−b

b2 + c2 sin2 t.

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The angle θ2 between N and r(t)− F2 is similarly

cos θ2 =(r(t)− F2) ·N‖r(t)− F2‖‖N‖

=1

b2 + c2 sin2 t

(a cos t+ c)(−b cos t) + (b sin t)(−a sin t)

a+ c cos t

=−b

b2 + c2 sin2 t.

Thus cos θ1 = cos θ2, so θ1 = θ2.

Problem 2 (14.4.34). The Cornu spiral is the plane curve r(t) =

(x(t)

y(t)

), where

x(t) =

∫ t

0

sin

(u2

2

)du, y(t) =

∫ t

0

cos

(u2

2

)du.

Verify that κ(t) = |t|.

Solution. Using the fundamental theorem of calculus,

r′(t) =

(x′(t)

y′(t)

)=

(sin(t2/2)

cos(t2/2)

).

Observe that ‖r′(t)‖ = 1 for all t, so t is the arc length parameter. Then

κ(t) = ‖r′′(t)‖ = |t|.

Problem 3 (14.4.69). The angle of inclination at a point P on a plane curve is the angle θ betweenthe unit tangent vector T and the x-axis. Assume that r(s) is an arc length parameterization, andlet θ = θ(s) be the angle of inclination at r(s). Prove that

κ(s) =

∣∣∣∣dθds∣∣∣∣ .

Solution. Using the hint provided,

κ(s) =

∥∥∥∥dTds∥∥∥∥ =

∥∥∥∥∥ dds(

cos θ(s)

sin θ(s)

)∥∥∥∥∥ =

∥∥∥∥∥dθds(− sin θ(s)

cos θ(s)

)∥∥∥∥∥ =

∣∣∣∣dθds∣∣∣∣ .

Problem 4 (14.4.82). Let r(s) be an arc length parameterization of a closed curve C of length L.We call C an oval if dθ/ds > 0 (see Exercise 14.4.69 / Problem 3). Observe that −N points to theoutside of C. For k > 0, the curve C1 defined by r1(s) = r(s)− kN is called the expansion of C inthe normal direction.

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(a) Show that ‖r′1(s)‖ = ‖r′(s)‖+ kκ(s).

(b) As P moves around the oval counterclockwise, θ increases by 2π. Use this and a change of

variables to prove that

∫ L

0

κ(s) ds = 2π.

(c) Show that C1 has length L+ 2πk.

Solution. (a) We will need the result (which holds for all plane curves with non-zero curvature)

dN

ds= −κT.

To see this, since we have a plane curve, there are some coefficients a(s) and b(s) such that

N′ = aT + bN,

so a = T ·N′ and b = N ·N′. By differentiating N ·N = 1, we have

2N ·N′ = 0 =⇒ b = N ·N′ = 0,

and by differentiating T ·N = 0, we have

T′ ·N + T ·N′ = 0 =⇒ a = T ·N′ = −T′ ·N = −κ.

This proves the result.

Given r1(s) = r(s)− kN, we can now compute

r′1(s) = r′(s)− kN′ = T + kκT = (1 + kκ)T.

Thus‖r′1(s)‖ = ‖(1 + kκ)T‖ = 1 + kκ = ‖r′(s)‖+ kκ(s).

(b) The information in the problem tells us that θ(L)− θ(0) = 2π, and since C is an oval, we canreduce the expression from Problem 3 to

κ(s) =

∣∣∣∣dθds∣∣∣∣ =

dθ

ds.

Then ∫ L

0

κ(s) ds =

∫ L

0

dθ

dsds =

∫ θ(L)

θ(0)

dθ = θ(L)− θ(0) = 2π.

(c) Combining the previous two parts,

L(C1) =

∫ L

0

‖r′1(s)‖ ds =

∫ L

0

(1 + kκ(s)) ds = L+ k

∫ L

0

κ(s) ds = L+ 2πk.

13


Problem 5 (14.5.4). For r(t) = etj− cos(2t)k, calculate the velocity and acceleration vectors andthe speed at time t = 0.

Solution. We compute the velocity and acceleration directly as

r′(t) = etj + 2 sin(2t)k ,

r′′(t) = etj + 4 cos(2t)k .

Then the speed at time t = 0 is ‖r′(0)‖ = ‖j‖ = 1 .

Problem 6 (14.5.49). A space shuttle orbits the Earth at an altitude 400 km above the earth’ssurface, with constant speed v = 28 000 km · hr−1. Find the magnitude of the shuttle’s acceleration(in kilometers per square hour), assuming that the radius of the Earth is 6378 km.

Solution. Let R = 400 + 6 378 = 6 778 be the distance of the shuttle from the center of the Earth(in km). The magnitude of the shuttle’s acceleration is then

a =v2

R≈ 115 668 km · hr−2 .

Problem 7 (14.6.17). The total mechanical energy (kinetic energy plus potential energy) of aplanet of mass m orbiting a sun of mass M with position r and speed v = ‖r′‖ is

E =1

2mv2 − GMm

‖r‖.

(a) Prove the equations

d

dt

(1

2mv2

)= v · (ma),

d

dt

(GMm

‖r‖

)= v ·

(−GMm

‖r‖3r

).

(b) Then use Newton’s law F = ma to prove that energy is conserved, i.e. dE/dt = 0.

Solution. (a) For the first equation,

d

dt

(1

2mv2

)=

1

2md

dt(v · v) =

1

2m

(2v · dv

dt

)= v · (ma).

For the second,

d

dt

(GMm

‖r‖

)= GMm

d

dt

([r · r]−1/2

)= GMm

(−1

2[r · r]−3/2

d

dt(r · r)

)= −GMm

2r · r′

2[r · r]3/2

= v ·(−GMm

‖r‖3r

).

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(b) Observe that

F = ma = −GMm

‖r‖3r,

sodE

dt= v · (ma)− v ·

(−GMm

‖r‖3r

)= v · F− v · F = 0.

Problem 8 (14.R.26). A specially trained mouse runs counterclockwise in a circle of radius 0.6 mon the floor of an elevator with speed 0.3 m/s, while the elevator ascends from ground level (alongthe z-axis) at a speed of 12 m/s. Find the mouse’s acceleration vector as a function of time. Assumethat the circle is centered at the origin of the xy-plane and the mouse is at (0.6, 0, 0) at t = 0.

Solution. The position vector of the mouse is

r(t) =

0.6 cos(ωt)

0.6 sin(ωt)

12t

,

where ω is some constant to be determined. The velocity is then

v(t) =

−0.6ω sin(ωt)

0.6ω cos(ωt)

12

.

The fact that the mouse moves at speed 0.3 m/s in the elevator means that ω = 0.3/0.6 = 0.5 (inthe units of the problem). Then

a(t) =

−0.15 cos(0.5t)

−0.15 sin(0.5t)

0

.

Problem 9 (14.R.40). If a planet has zero mass (m = 0), then Newton’s laws of motion reduce tor′′(t) = 0 and the orbit is a straight line r(t) = r0 + tv0, where r0 = r(0) and v0 = r′(0). Show thatthe area swept out by the radial vector at time t is A(t) = 1

2‖r0 × v0‖t, and thus Kepler’s secondlaw continues to hold (the rate is constant).

Solution. The area swept out by the radial vector is that of a triangle with two of its sides givenbyr0 and tv0, so

A(t) =1

2‖r0 × (tv0)‖ =

1

2‖r0 × v0‖t.

ThendA

dt=

1

2‖r0 × v0‖

is constant, so Kepler’s second law still holds.

15



Problem 1 (15.1.18). Match each of graphs (A) and (B) with one of the following functions:

(i) f(x, y) = (cosx)(cos y) (ii) g(x, y) = cos(x2 + y2)

Solution. The function g is rotationally symmetric since x2 + y2 = r2 has no dependence on the(cylindrical) polar coordinate θ. Of the two graphs shown, (B) is rotationally symmetric while (A)is not, so g matches with (B) while f matches with (A).

Problem 2 (15.1.20). Match the functions (a)-(d) with their contour maps (A)-(D).

(a) f(x, y) = 3x+ 4y

(b) g(x, y) = x3 − y

(c) h(x, y) = 4x− 3y

(d) k(x, y) = x2 − y

17


Solution. The contour lines where the functions evaluate to zero are

f(x, y) = 0 y = (−3/4)x

g(x, y) = 0 y = x3

h(x, y) = 0 y = (4/3)x

k(x, y) = 0 y = x2.

These can be matched to (B), (A), (C), and (D), respectively.

Problem 3 (15.1.26). Sketch the graph of f(x, y) =1

x2 + y2 + 1and draw several horizontal and

vertical traces.

Solution. Omitted.

Problem 4 (15.1.56). The function f(x, t) = t−1/2e−x2/t models the temperature along a metal

bar after an intense burst of heat is applied at its center point.

(a) Sketch the vertical traces at times t = 1, 2, 3. What do these traces tell us about the way heatdiffuses through the bar?

(b) Sketch the vertical traces x = c for c = ±0.2,±0.4. Describe how temperature varies in timeat points near the center.

Solution. Omitted.

Problem 5. By repeatedly completing the square, find changes of variables of the form

X = x+ ay + b Y = y + c

that bring the paraboloids

z = x2 + 6xy + 10y2 + 4x+ 10y + 5

z = x2 − 2xy + 4y − 4

into standard position. Identify them as either elliptic or hyperbolic.

Solution. For the first one, the discriminant is 62 − 4 · 1 · 10 = −4 < 0, so the surface is an ellipticparaboloid. To confirm this, we complete the square as

z = x2 + 6xy + 10y2 + 4x+ 10y + 5 = [(x+ 3y + 2)2 − 9y2 − 12y − 4] + 10y2 + 10y + 5

= (x+ 3y + 2)2 + y2 − 2y + 1 = (x+ 3y + 2)2 + [(y − 1)2 − 1] + 1

= (x+ 3y + 2)2 + (y + 1)2.

Thus if we set X = x + 3y + 2 and Y = y + 1, we have our quadric surface in standard positionz = X2 + Y 2. This is an elliptic paraboloid in XY -space, hence also in xy-space.

18


For the second one, the discriminant is (−2)2 − 4 · 1 · 0 = 4 > 0, so the surface is a hyperbolicparaboloid. Completing the square,

z = x2 − 2xy + 4y − 4 = (x− y)2 − y2 + 4y − 4

= (x− y)2 − y2 + 4y − 4 = (x− y)2 − (y2 − 4y + 4)

= (x− y)2 − (y − 2)2.

If we set X = x− y and Y = y− 2, we have our quadric surface in standard position z = X2 − Y 2.This is a hyperbolic paraboloid in XY -space, hence also in xy−space.

Problem 6. By repeatedly completing the square, determine for each of the following equationswhether they describe a cone, ellipsoid, cylinder, or hyperboloid (of how many sheets):

2x2 + 2xy + 2xz + 3y2 − 4yz + 3z2 = 1 2x2 − 2xz + y2 + 2yz + z2 = 0

2xy + y2 + 2z2 − 4x− 12z = 14 2xy + y2 + 2z2 − 4x− 12z = −25.

Solution. For the first one, we find

0 = 2x2 + 2xy + 2xz + 3y2 − 4yz + 3z2 − 1 = 2

(x+

1

2y +

1

2z

)2

+5

2(y − z)2 − 1.

This describes a cylinder. Note that we could write the equation as

(x+ y)2 + (x+ z)2 + 2(y − z)2 − 1 = 0.

The reason we cannot conclude that the surface is an ellipsoid is that the transformation

X = x+ y; Y = x+ z; Z = y − z

is not actually invertible, i.e. we cannot uniquely recover x, y, z from X,Y, Z. Geometrically, thistransformation squashes everything into a plane, even if it was originally a 3-dimensional object.

For the second one, we find

0 = 2x2 − 2xz + y2 + 2yz + z2 = (z − x+ y)2 + (x+ y)2 − y2.

This describes a cone.

For the third one, we find

0 = 2xy + y2 + 2z2 − 4x− 12z − 14 = (y + x)2 − (x+ 2)2 + 2(z − 3)2 − 28.

This describes a hyperboloid of one sheet.

For the fourth one, we find

0 = 2xy + y2 + 2z2 − 4x− 12z + 25 = (y + x)2 − (x+ 2)2 + 2(z − 3)2 + 11.

This describes a hyperboloid of two sheets.

Problem 7 (13.6.32). Sketch the surface y2 + z2 = 1.

19


Solution. Omitted.

Problem 8 (13.6.36). Sketch the surface x2 − 4y2 = z.

Solution. Omitted.

Problem 9 (13.6.40). Find the equation of the elliptic cylinder passing through the points markedin Figure 4a.

Solution. The elliptic cross section shown lies in the xy-plane, so the quadric surface has no z-dependence. In the plane, the equation for the ellipse is

x2

4+y2

16= 1,

so this is also the equation for the elliptic cylinder.

Remark: the elliptic cylinder may not be unique, but the one given here is the simplest answer tocompute.

(a) Points on an elliptic cylinder (b) Quadric surface

Figure 1: (a) 13.6.40; (b) 13.6.42

Problem 10 (13.6.42). Find the equation of the quadric surface shown in Figure 4b.

Solution. The quadric surface shown is a cone, so it has the form (x/a)2 + (y/b)2 = (z/c)2 forsome a, b, c > 0. We know that (6, 0, 5) and (0, 8, 5) lie on the cone, so (6/a)2 = (5/c)2 and(8/b)2 = (5/c)2. A convenient choice which satisfies these equations is a = 6, b = 8, and c = 5, sothe equation for the cone is

x2

36+y2

64=z2

25.

Problem 11 (13.6.47). Let S be the hyperboloid x2 + y2 = z2 + 1 and let P = (α, β, 0) be a pointon S in the xy-plane. Show that there are precisely two lines through P entirely contained in S.

20


Solution. Note that no line lying solely in the xy-plane will be entirely contained in S, so we canassume that a line which is contained in S has the form (x, y, z) = (α + at, β + bt, t) for someconstants a, b. Substituting this into the equation for the hyperboloid,

(α+ at)2 + (β + bt)2 = t2 + 1

(a2 + b2)t2 + 2(αa+ βb)t+ (α2 + β2) = t2 + 1.

This must hold for all t, so each coefficient must agree. Since P = (α, β, 0) lies on S, α2 + β2 = 1,so the constant terms match up. Thus we have a2 + b2 = 1 and αa + βb = 0. The first conditiontells us that (a, b) lies on the unit circle in the xy-plane, while the second tells us that the vectorsαi+βj and ai+bj are perpendicular. This gives us exactly two solutions, corresponding to rotationof (α, β) around the unit circle by ±π/2. These two solutions correspond to the two lines.

Remark: the direction vectors of the lines are −βi+αj+k and βi−αj+k, which are perpendicular.(Test this by computing the dot product.)

21



Problem 1. Use the ε-δ definition to show that the following are continuous at (x, y) = (0, 0):

f(x, y) = 7, f(x, y) = x2 − y2.

Solution. For the first one, let ε > 0 be arbitrary. To show that f(x, y) = 7 is continuous at(x, y) = (0, 0), we must find δ > 0 such that√

(x− 0)2 + (y − 0)2 =√x2 + y2 < δ =⇒ |f(x, y)− f(0, 0)| = 0 < ε.

The latter inequality holds no matter what (x, y) is, so we could take any positive δ, say δ = 1.

For the second one, let ε > 0 be arbitrary. To show that f(x, y) = x2 − y2 is continuous at(x, y) = (0, 0), we must find δ > 0 such that√

x2 + y2 < δ =⇒ |f(x, y)− f(0, 0)| = |x2 − y2| < ε.

For this, we use the triangle inequality to write

|x2 − y2| = |x2 + (−y2)| ≤ |x2|+ |−y2|= x2 + y2 < δ2,

so if we choose δ =√ε, then |x2 − y2| < δ2 = ε, as required.

Problem 2 (15.2.14). Let f(x, y) = xy/(x2 + y2). Show that f(x, y) approaches zero along the xand y axes. Then prove that lim

(x,y)→(0,0)f(x, y) does not exist by showing that the limit along the

line y = x is non-zero.

Solution. Along the x-axis, y = 0, so the limit along the x-axis is

limx→0

f(x, 0) = limx→0

x · 0x2 + 02

= 0.

Similarly, the limit along the y-axis is 0. Along y = x, we have the parameterization (x, y) = (t, t),and the limit is

limt→0

f(t, t) = limt→0

t · tt2 + t2

=1

2,

so different paths can give different limits. Thus the limit lim(x,y)→(0,0)

f(x, y) does not exist.

Problem 3 (15.2.20). Evaluate the limit lim(x,y)→(0,0)

x2 − y2

x2 + y2or show that it does not exist.

Solution. Along the x-axis, the limit is 1, while along the line y = x, the limit is 0, so the limitdoes not exist.

Problem 4 (15.2.28). Evaluate the limit lim(z,w)→(−1,2)

(z2w − 9z) or show that it does not exist.

23


Solution. The function (z, w) 7→ z2w − 9z is continuous since it is a polynomial, so the limit is(−1)2 · 2− 9(−1) = 11.

Problem 5 (15.2.40). Evaluate the limit lim(x,y)→(1,1)

x2 + y2 − 2

|x− 1|+ |y − 1|or show that it does not exist.

Solution. Let u = x− 1 and v = y − 1 to simplify to

lim(u,v)→(0,0)

(u+ 1)2 + (v + 1)2 − 2

|u|+ |v|= lim

(u,v)→(0,0)

u2 + 2u+ v2 + 2v

|u|+ |v|.

Approaching the origin from above, where v > 0 and u = 0, we have |v| = v, so the limit is that ofv + 2 as v → 0+, which is 2. Approaching the origin from below, we have |v| = −v, so the limit isthat of −v − 2 as v → 0−, which is −2. Hence the limit does not exist.

Problem 6 (15.3.16). Compute the first-order partial derivatives of V = πr2h.

Solution. The partials are∂V

∂r= 2πrh,

∂V

∂h= πr2.

Problem 7 (15.3.32). Compute the first-order partial derivatives of P = e√y2+z2 .

Solution. The partials are

∂P

∂y= e√y2+z2 ∂

∂y

(√y2 + z2

)=ye√y2+z2√

y2 + z2,

∂P

∂z= e√y2+z2 ∂

∂z

(√y2 + z2

)=ze√y2+z2√

y2 + z2.

Problem 8 (15.3.34). Compute the first-order partial derivatives of z = yx.

Solution. Write z = ex ln y. The partials are

∂z

∂x= ex ln y · ln y = yx ln y,

∂z

∂y= ex ln y · x

y= xyx−1.

Problem 9 (15.3.52). Calculate ∂P/∂T and ∂P/∂V , where pressure P , volume V , and temperatureT are related by the ideal gas law PV = nRT (here R and n are constants).

24


Solution. Write P = nRT/V . The partials are

∂P

∂T=nR

V,

∂P

∂V= −nRT

V 2.

The latter can also be found by implicit differentiation from the ideal gas law in its original form.Noting that T is held fixed while P and V can vary,

∂

∂V(PV ) =

∂

∂V(nRT ) =⇒ ∂P

∂VV + P = 0 =⇒ ∂P

∂V= −P

V.

The agreement of these two answers comes from substituting P = nRT/V .

Problem 10 (15.3.68). Compute the derivative uxx for u(x, t) = t−1/2e−x2/4t.

Solution. We have

ux = t−1/2e−x2/4t · −2x

4t= −1

2t−3/2xe−x

2/4t,

uxx = −1

2t−3/2e−x

2/4t − 1

2t−3/2xe−x

2/4t · −2x

4t

=

(1

4x2 − 1

2t

)t−5/2e−x

2/4t.

Problem 11 (15.3.76). Show that u(x, t) = sin(nx)e−n2t satisfies the heat equation for any n:

∂u

∂t=∂2u

∂x2.

Solution. We have

∂u

∂t= −n2 sin(nx)e−n

2t,

∂u

∂x= n cos(nx)e−n

2t,

∂2u

∂x2= −n2 sin(nx)e−n

2t,

so the required equality holds.

Problem 12 (15.4.8). Find an equation of the tangent plane to g(x, y) = ex/y at (2, 1).

Solution. The gradient of g at (2, 1) is

∇g(2,1) =

(∂g/∂x

∂g/∂y

)∣∣∣∣∣(2,1)

=

(e2

−2e2

).

Therefore, the tangent plane at (2, 1) is

z = g(2, 1) +∇g(2,1) ·

(x− 2

y − 1

)= e2 + e2(x− 2)− 2e2(y − 1).

25


Problem 13 (15.4.14). Write the linear approximation to f(x, y) = x(1 + y)−1 at (a, b) = (8, 1) inthe form

f(a+ h, b+ k) ≈ f(a, b) + fx(a, b)h+ fy(a, b)k.

Use it to estimate 7.98/2.02 and compare with the value obtained using a calculator.

Solution. We evaluate

f(8, 1) = 8 · (1 + 1)−1 = 4,

fx(8, 1) = (1 + y)−1∣∣(8,1)

= 1/2,

fy(8, 1) = −x(1 + y)−2∣∣(8,1)

= −2.

Thus the linear approximation is

f(8 + h, 1 + k) ≈ 4 +1

2h− 2k.

To estimate 7.98/2.02 = f(7.98, 1.02), we use this with h = −0.02 and k = 0.02 to get

7.98

2.02≈ 4 +

1

2· (−0.02)− 2 · 0.02 = 3.95.

For comparison, the true value is 3.9504.

Problem 14 (15.4.39). The volume V of a right circular cylinder is computed using the values 3.5m for diameter and 6.2 m for height. Use the linear approximation to estimate the maximum errorin V if each of these values has a possible error of at most 5%. Recall that V = πr2h.

Solution. The radius of the cylinder is r = 1.75 m, so when computing the linear approximation,we have (dropping units)

V (1.75 + h, 6.2 + k) ≈ V (1.75, 6.2) +∇V(1.75,6/2) ·

(h

k

)= π · 1.752 · 6.2 + 2π · 1.75 · 6.2h+ π · 1.752k.

The minimum and maximum values of h are −0.05 · 1.75 and 0.05 · 1.75, while the minimum andmaximum values of k are −0.05 · 6.2 and 0.05 · 6.2. Trying different combinations of these whencomputing the error term 2π ·1.75 ·6.2h+π ·1.752k, the largest error occurs when h and k have thesame sign and have as large magnitude as possible. This corresponds to a maximum actual errorof ≈ 2.185, corresponding to a percent error of ≈ 3.66%.

26


Problem 15 (15.5.12). Use the chain rule to calculate ddtf(r(t)) for

f(x, y) = x2 − 3xy, r(t) = cos ti + sin tj, t = π/2.

Solution. We compute

d

dtf(r(t)) = ∇fr(π/2) ·

dr

dt

∣∣∣∣t=π/2

=

(2x− 3y

−3x

)∣∣∣∣∣(x,y)=(0,1)

·

(− sin t

cos t

)∣∣∣∣∣t=π/2

=

(−3

0

)·

(−1

0

)= 3.

Problem 16 (15.5.30). Calculate the directional derivative of g(x, y, z) = x ln(y+z) in the directionof v = 2i− j + k at P = (2, e, e). Remember to normalize the direction vector.

Solution. The unit vector in this direction is

u =1√

22 + (−1)2 + 12

2

−1

1

=1√6

2

−1

1

,

so the directional derivative is

Duf = ∇f(2,e,e) · u

=

ln(y + z)

x/(y + z)

x/(y + z)

∣∣∣∣∣∣∣(2,e,e)

· 1√6

2

−1

1

=

1√6

(2 ln(2e)− 1

e+

1

e

)=

2 ln 2 + 2√6

.

Problem 17. Consider the surface z =√x2 + y2, similar to that in Exercise 15.5.48.

(a) Show that the plane tangent to this surface at any point (x, y, z) 6= (0, 0, 0) passes throughthe origin.

(b) Explain why the point (0, 0, 0) was excluded.

Solution. (a) At the point (a, b, c), with c =√a2 + b2, the tangent plane is

z = c+ zx · (x− a) + zy · (x− b) = c+a

c(x− a) +

b

c(y − b).

27


Passing through the origin means that (0, 0, 0) is a solution to this equation. When x = 0and y = 0, we have

z = c− a2

c− b2

c=c2 − a2 − b2

c= 0.

(b) At (0, 0), the partial derivatives of the function f(x, y) =√x2 + y2 (of which the surface is

the graph) do not exist, so the function cannot be differentiable there, i.e. there is no planetangent to the surface at (0, 0, 0).

Problem 18. Consider a gas whose pressure P , volume V , and temperature T are related by(P +

1

V 2

)V = T,

as well as the following path, defined for t > 0,(V (t)

T (t)

)=

(t

t−3/2

).

(a) Writing P as a function of V and T , determine

∂P

∂V(V (t), T (t)).

(b) Now computed

dtP (V (t), T (t)).

Solution. (a) Solving for P from the equation given gives

P =T

V− 1

V 2,

so∂P

∂V= − T

V 2+

2

V 3.

(b) The other partial derivative is∂P

∂V=

1

V,

so applying the chain rule,

d

dtP (V (t), T (t)) =

(−T (t)/V (t)2 + 2/V (t)3

1/V (t)

)·

(V ′(t)

T ′(t)

)

=

(−t7/2 + 2t−3

t−1

)·

(1

(−3/2)t−5/2

)

= −5

2t−7/2 + 2t−3.

28


Alternatively, we can directly compute

P (V (t), T (t)) =T (t)

V (t)− 1

V (t)2= t−5/2 − t−2

and differentiate to get the same answer.

Remark: The equation given is a modification of the ideal gas law known as the van der Waalsequation (for a particular choice of units and constants which depend on the gas).

29



Problem 1 (15.5.32). Find the directional derivative of f(x, y) = x2 + 4y2 at the point P = (3, 2)in the direction pointing to the origin.

Solution. The gradient is

∇f(x,y) =

(2x

8y

),

while the appropriate direction vector is

u =

(−3/√

13

−2/√

13

),

so the desired directional derivative is

∇f(3,2) · u =

(6

16

)·

(−3/√

13

−2/√

13

)= − 50√

13.

Problem 2 (15.5.34). Suppose you are hiking on a terrain modeled by z = x2 + y2 − y. You areat the point (1, 2, 3).

(a) Determine the slope you would encounter if you headed due east (in the positive x-direction)from your position. What angle of inclination does that correspond to?

(b) Determine the slope you would encounter if you headed due north (in the positive y-direction)from your position. What angle of inclination does that correspond to?

(c) Determine the slope you would encounter if you headed due northeast from your position.What angle of inclination does that correspond to?

(d) Determine the steepest slope you could encounter from your position.

Solution. For each part, we need the gradient at (x, y) = (1, 2), which is given by

∇z(x,y) =

(2x

2y − 1

), ∇z(1,2) =

(2

3

).

(a) Here we take u = i, and∇z(1,2) · u = 2

is the slope. This corresponds to an angle arctan 2 ≈ 63.43◦.

(b) Here we take u = j, and∇z(1,2) · u = 3

is the slope. This corresponds to an angle arctan 3 ≈ 71.57◦.

31


(c) Here we take u =(1/√

2)

(i + j), and

∇z(1,2) · u =5√2

is the slope. This corresponds to an angle arctan(5/√

2)≈ 74.21◦.

(d) The maximum possible slope is ‖∇z(1,2)‖ =√

13.

Problem 3 (15.5.58). Find a unit vector n that is normal to the surface z2 − 2x4 − y4 = 16 atP = (2, 2, 8) that points in the direction of the xy-plane. (In other words, if you travel in thedirection of n, you will eventually cross the xy-plane.)

Solution. Letting f(x, y, z) = z2 − 2x4 − y4, the surface given is the level surface f(x, y, z) = 16.A vector normal to the level surface at P is given by the gradient ∇fP = −64i− 32j + 16k, whichpoints away from the xy-plane. Thus the vector we want is

n = − ∇fP‖∇fP ‖

=1√21

4

2

−1

.

Problem 4. Consider the surface z2 = 1 + x2 + y2.

(a) Find all points on the surface where the tangent plane is perpendicular to

1

1

2

.

(b) Find all points on the surface whose tangent plane passes through (0, 0, 1/2).

Solution. Let f(x, y, z) = x2 + y2 − z2. The surface given is the level surface f(x, y, z) = −1, so toget a tangent plane at P = (a, b, c), we compute

∇fP = 2ai + 2bj− 2ck,

and then the equation for the tangent plane is

2ax+ 2by − 2cz = 2a2 + 2b2 − 2c2 = −2,

where we have noted that f(a, b, c) = −1 by assumption.

(a) For the tangent plane to be perpendicular to i + j + 2k, we require c = −2a = −2b, so

(−2a)2 = 1 + a2 + a2 =⇒ a = ±1/√

2.

Thus the two points are (1√2,

1√2,−√

2

),

(− 1√

2,− 1√

2,√

2

).

32


(b) For the tangent plane to pass through (0, 0, 1/2), we require −c = −2, so c = 2. Thena2 + b2 = 3, so any point of the form

(√3 cos θ,

√3 sin θ, 2

)will work.

Problem 5 (15.6.4). Calculate ∂f/∂r and ∂f/∂t for

f(x, y, z) = xy + z2; x = r + s− 2t, y = 3rt, z = s2.

Solution. By the chain rule,

∂f

partialr=∂f

∂x

∂x

∂r+∂f

∂y

∂y

∂r+∂f

∂z

∂z

∂r

= y · 1 + x · 3t+ 2z · 0= 6rt+ 3ts− 3t2,

∂f

∂t=∂f

∂x

∂x

∂t+∂f

∂y

∂y

∂t+∂f

∂z

∂z

∂t

= y · (−2) + x · 3r − 2z · 0= 3r2 + 3rs− 12rt.

Problem 6 (15.6.13). Evaluate ∂g/∂θ at (r, θ) =(2√

2, π/4)

for

g(x, y) =1

x2 + y2; x = r cos θ, y = r sin θ.

Solution. By the chain rule,

∂g

∂θ=∂g

∂x

∂x

∂θ+∂g

∂y

∂y

∂θ

= − 2x

(x2 + y2)2· (−r sin θ)− 2y

(x2 + y2)2· (r cos θ)

=2r cos θ(r sin θ)

r4− 2r sin θ(r cos θ)

r4= 0.

Alternatively, note that g(r, θ) = 1/r2, so gθ = 0.

This answer does not depend on the choice of r and θ.

Problem 7 (15.6.20). The law of cosines states that c2 = a2 + b2 − 2ab cos θ, where a, b, c are thesides of a triangle and θ is the angle opposite the side of length c.

(a) Compute ∂θ/∂a, ∂θ/∂b, and ∂θ/∂c using implicit differentiation.

(b) Suppose that a = 10, b = 16, and c = 22. Estimate the change in θ if a and b are increasedby 1 and c is increased by 2.

33


Solution. (a) Fixing b and c,

∂

∂a(c2) =

∂

∂a(a2 + b2 − 2ab cos θ)

0 = 2a− 2b cos θ + 2ab sin θ∂θ

∂a∂θ

∂a=b cos θ − aab sin θ

.

Similarly,∂θ

∂b=a cos θ − bab sin θ

,

while∂θ

∂c= − c

ab sin θ.

(b) For the given values of a, b, and c, the law of cosines gives us θ = arccos(2/5) ≈ 66.42◦. Then

∆θ ≈ ∂θ

∂a∆a+

∂θ

∂b∆b+

∂θ

∂c∆c

= 2 ·(− 22

10 · 16 · sin θ

)= − 11

8√

21≈ −0.300.

Problem 8 (15.6.21). Let u = u(x, y) and let (r, θ) be polar coordinates. Verify the relation

‖∇u‖2 = u2r +1

r2u2θ.

Solution. We compute with the chain rule

∂u

∂r=∂u

∂x

∂x

∂r+∂u

∂y

∂y

∂r

= ux cos θ + uy sin θ,

∂u

∂θ=∂u

∂x

∂x

∂θ+∂u

∂y

∂y

∂θ

= −rux sin θ + ruy cos θ.

Then the result follows by substitution and expanding (note that ‖∇u‖2 = u2x + u2y).

Problem 9 (15.6.30). Compute ∂r/∂t and ∂t/∂r using implicit differentiation if r2 = tes/r.

Solution. Fixing s, we have

∂

∂t(r2) =

∂

∂t(tes/r)

2r∂r

∂t= es/r − st

r2es/r

∂r

∂t

∂r

∂t=

es/r

2r + stes/r/r2.

34


Similarly, we can find

∂t

∂r=

2r + stes/r/r2

es/r=

(∂r

∂t

)−1.

Problem 10 (15.6.44). A function f(x, y, z) is homogeneous of degree n if

f(λx, λy, λz) = λnf(x, y, z)

for all λ ∈ R.

Prove that if f(x, y, z) is homogeneous of degree n, then

x∂f

∂x+ y

∂f

∂y+ z

∂f

∂z= nf.

Solution. Let F (t) = f(ta, tb, tc). By the chain rule with (x, y, z) = (ta, tb, tc),

F ′(1) =∂f

∂x

dx

dt+∂f

∂y

dy

dt+∂f

∂z

dz

dt= a

∂f

∂x+ b

∂f

∂y+ c

∂f

∂z.

On the other hand, since f is homogeneous with degree n, we have F (t) = tnf(a, b, c), so F ′(1) =nf(a, b, c). Thus

a∂f

∂x+ b

∂f

∂y+ c

∂f

∂z= nf(a, b, c),

so as functions,

x∂f

∂x+ y

∂f

∂y+ z

∂f

∂z= nf.

Problem 11 (15.6.46). Suppose that f is a function of x and y, where x = g(t, s) and y = h(t, s).Show that

ftt = fxx

(∂x

∂t

)2

+ 2fxy

(∂x

∂t

)(∂y

∂t

)+ fyy

(∂y

∂t

)2

+ fx∂2x

∂t2+ fy

∂2y

∂t2.

Solution. A first application of the chain rule gives

ft = fx∂x

∂t+ fy

∂y

∂t.

Differentiating by t a second time,

ftt = fxt∂x

∂t+ fx

∂2x

∂t2+ fyt

∂y

∂t+ fy

∂2y

∂t2.

By the chain rule again,

fxt = fxx∂x

∂t+ fxy

∂y

∂t,

fyt = fyx∂x

∂t+ fyy

∂y

∂t.

Substituting this in and expanding, we get the claimed result (conditional on fxy = fyx, which canusually be assumed for this course).

35


Problem 12. Compute all partial derivatives of

F (x, y, z) =

∫ y

x

ezs2 − 1

sds,

and so determined

dt

∫ 2t

t

ets2 − 1

sds.

Solution. The fundamental theorem of calculus gives us

∂F

∂x= −e

zx2 − 1

x,

∂F

∂y=ezy

2 − 1

y,

while differentiating under the integral sign (can be assumed for this course) gives

∂F

∂z=

∫ y

x

sezs2

ds =1

2z

(ezy

2

− ezx2).

Letting x = t, y = 2t, and z = t, we have

d

dt

∫ 2t

t

ets2 − 1

sds =

d

dtF (t, 2t, t)

=∂F

∂x

dx

dt+∂F

∂y

dy

dt+∂F

∂z

dz

dt

= −et3 − 1

t+ �2 ·

e4t3 − 1

�2 · t+e4t

3 − et3

2t

=3

2· e

4t3 − et3

t.

Problem 13 (15.7.4). Use the contour map in Figure 1 to determine whether the critical pointsA,B,C,D are local minima, local maxima, or saddle points.

Figure 1: Contour map

36


Solution. Point A is a local maximum, point C is a local minimum, and points B and D are saddlepoints.

Problem 14 (15.7.12). Find the critical points of the function f(x, y) = x3 + y4− 6x− 2y2. Thenuse the second derivative test to determine whether they are local minima, local maxima, or saddlepoints (or state that the test fails).

Solution. To find the critical points, we solve ∇f = 0, i.e.

fx = 3x2 − 6 = 0,

fy = 4y3 − 4y = 0.

There are six critical points, given by x = ±√

2 and y ∈ {0,±1}.For the second derivative test, we compute the Hessian matrix and discriminant

Hf =

(6x 0

0 12y2 − 4

), D = 6x(12y2 − 4).

The results of the second derivative test are tabulated below.

x = −√

2 x =√

2

y = −1 Saddle Local min

y = 0 Local max Saddle

y = 1 Saddle Local min

Problem 15 (15.7.24). Show that f(x, y) = x2 has infinitely many critical points (as a functionof two variables) and that the second derivative test for all of them. What is the minimum valueof f? Does f(x, y) have any local maxima?

Solution. Critical points are located where x = 0, which happens at (0, y) for all y ∈ R. TheHessian and discriminant are

Hf =

(2 0

0 0

), D = 0

at all points, in particular at all critical points, so the second derivative test never works. Theminimum value of f is 0, attained at all critical points, and f has no local maxima. (Moving in adirection of increasing |x| always increases f .)

Problem 16 (15.7.26). Let f(x, y) = (x2 + y2)e−x2−y2 .

(a) Where does f take on its minimum value? Do not use calculus to answer this question.

(b) Verify that the set of critical points of f consists of the origin (0, 0) and the unit circlex2 + y2 = 1.

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(c) The second derivative test fails for points on the unit circle (this can be checked by somelengthy algebra). Prove, however, that f takes on its maximum value on the unit circle byanalyzing the function g(t) = te−t for t > 0.

Solution. (a) At (0, 0), we have f(0, 0) = 0, but for all other (x, y), both x2 + y2 and e−x2−y2 are

positive, so f(x, y) > 0 for (x, y) 6= (0, 0). Thus f takes its minimum value at (0, 0).

(b) Critical points are given by

fx = 2xe−x2−y2 − 2x(x2 + y2)e−x

2−y2 = 0,

fy = 2ye−x2−y2 − 2y(x2 + y2)e−x

2−y2 = 0.

These equations reduce to

x(1− x2 − y2) = 0, y(1− x2 − y2) = 0,

so the solutions are (0, 0) and x2 + y2 = 1.

(c) Let t = x2 + y2, so then f(x, y) = g(t) = te−t. Then g′(t) = e−t − te−t = 0 for t = 1, whichcorresponds to x2 + y2 = 1, and g′′(t) = te−t − 2e−t < 0 for t = 1, so g has a maximum att = 1.

Problem 17. In class, we introduced the paraboloid of best approximation to f(x, y) at (0, 0) as

p(x, y) = f(0, 0) +∇f(0, 0) ·

(x

y

)+

1

2

∂2f

∂x2(0, 0)x2 +

∂2f

∂x∂y(0, 0)xy +

1

2

∂2f

∂y2(0, 0)y2.

Show that for any vector u =

(a

b

), we have

d

dt

∣∣∣∣t=0

f(tu) =d

dt

∣∣∣∣t=0

p(tu)

andd2

dt2

∣∣∣∣t=0

f(tu) =d2

dt2

∣∣∣∣t=0

p(tu).

Solution. For the first statement, we have by the chain rule that

d

dt

∣∣∣∣t=0

f(tu) = ∇f(0, 0) · u,

d

dt

∣∣∣∣t=0

p(tu) = ∇p(0, 0) · u,

so it suffices to show that ∇f(0, 0) = ∇p(0, 0). For this, we expand the dot product and compute

px(0, 0) = fx(0, 0) + fxx(0, 0) · 0 + fxy(0, 0) · 0 = fx(0, 0),

py(0, 0) = fy(0, 0) + fxy(0, 0) · 0 + fyy(0, 0) · 0 = fy(0, 0),

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which is the required result.

For the second statement, we use 15.6.46 to get, noting that x = ta and y = tb,

ftt = fxx(xt)2 + 2fxyxtyt + fyy(yt)

2 + fxxtt + fyytt

= a2fxx(0, 0) + 2abfxy(0, 0) + b2fyy(0, 0),

ptt = pxx(xt)2 + 2pxyxtyt + pyy(yt)

2 + pxxtt + pyytt

= a2fxx(0, 0) + 2abfxy(0, 0) + b2fyy(0, 0),

so ftt = ptt at t = 0.

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Problem 1. (a) Find all critical points of the function

f(x, y, z) = (x+ 4y + z) exp(1/2− x2 − y2 − z2).

(b) Pick one and determine the approximating paraboloid at that point.

(c) Complete squares to determine if the critical point is a maximum, a minimum, or a saddle.

Solution. (a) We compute

∇f = exp

(1

2− x2 − y2 − z2

)1− 2x(x+ 4y + z)

4− 2y(x+ 4y + z)

1− 2z(x+ 4y + z)

.

At a critical point, ∇f = 0. One can check that if any of x, y, z is 0, then ∇f 6= 0, so we canthen say that at a critical point,

x+ 4y + z =1

2x=

4

2y=

1

2z.

From these equations, z = x and y = 4x. Setting the first component to 0, we have

1− 2x(x+ 4(4x) + x) = 0 =⇒ x = ±1/6.

If x = 1/6, then y = 2/3 and z = 1/6, so we get the critical point (1/6, 2/3, 1/6).

If x = −1/6, then we get the critical point (−1/6,−2/3,−1/6).

(b) For (b) and (c), we use the critical point (x0, y0, z0) = (1/6, 2/3, 1/6).

Write (X,Y, Z) = (x− x0, y − y0, z − z0). The approximating paraboloid is

p(x, y, z) = f(x0, y0, z0) +

��

��

∇f(x0, y0, z0) ·

XYZ

+

1

2

(X Y Z

)fxx fxy fxz

fyx fyy fyz

fzx fzy fzz

(x0,y0,z0)

XYZ

= 3 +1

2

(X Y Z

)−19/3 −4/3 −1/3

−4/3 −34/3 −4/3

−1/3 −4/3 −19/3

XYZ

= 3− 1

6(19X2 + 34Y 2 + 19Z2 + 8XY + 8Y Z + 2ZX).

41


(c) Completing the square, we have

p(x, y, z) = 3− 19

6

(X +

4

19Y +

1

19Z

)2

− 105

19

(Y +

4

35Z

)2

− 2052

665Z2.

This means that the critical point is a local maximum.

Problem 2 (15.7.35). Find the maximum of

f(x, y) = x+ y − x2 − y2 − xy

on the square 0 ≤ x ≤ 2, 0 ≤ y ≤ 2.

Solution. On each segment of the boundary, we have one-variable quadratic optimization (withboundary endpoints), from which we get possible maxima at

(0, 2), (2, 2), (2, 0), (1/2, 0), (0, 1/2).

In the interior, a local maximum has

∇f =

(1− 2x− y1− 2y − x

)= 0.

The unique solution is (x, y) = (1/3, 1/3).

Testing these six points, we find that the maximum occurs at (1/3, 1/3), with value 1/3.

Problem 3 (15.7.44). Determine the global extreme values of the function

f(x, y) = (4y2 − x2)e−x2−y2

on the domain x2 + y2 ≤ 2.

Solution. On the boundary x2 + y2 = 2, we have

f(x, y) = (5y2 − 2)e−2,

which is maximized at y2 = 2 and minimized at y2 = 0, with values 8e−2 and −2e−2 respectively.

Inside the boundary, the extrema occur at critical points, so we solve

∇f(x, y) = e−x2−y2

(−2x− 2x(4y2 − x2)

8y − 2y(4y2 − x2)

)= 0.

If x 6= 0, then 4y2 − x2 = −1 from the first component, and if y 6= 0, then 4y2 − x2 = 4 from thesecond component. Thus at least one of x or y is 0, and from there we can find that the solutionsinside the boundary are

(0, 0), (0, 1), (0,−1), (1, 0), (−1, 0).

Testing these points, we get values 0, 4e−1, and −e−1. Comparing, either by calculator/computeror using the fact that e > 2, the global maximum is 4e−1 and the global minimum is −e−1.

42


Problem 4 (15.7.54). A box with a volume of 8 m3 is to be constructed with a gold-plated top,silver-plated bottom, and copper-plated sides. If gold plate costs $120 per square meter, silver platecosts $40 per square meter, and copper plate costs $10 per square meter, find the dimensions thatwill minimize the cost of the materials for the box.

Solution. Let x, y > 0 the dimensions of the base (bottom) of the box, and let z = 8/xy be theheight of the box. The total cost of materials is

c(x, y) = 120xy + 40xy + 10(2xz + 2yz) = 160xy +20

x+

20

y.

To find the critical points, we solve

∇c =

(160y − 20/x2

160x− 20/y2

)= 0.

These equations give 8x2y = 1 and 8xy2 = 1. Multiplying and taking cube roots gives 4xy = 1,and from this we get x = y = 1/2, so z = 32.

Problem 5. Consider three fixed points x1, x2, and x3 in space. Find the point x that minimizesthe sum of distances squared, that is, which minimizes

f(x) = ‖x− x1‖2 + ‖x− x2‖2 + ‖x− x3‖2.

Incidentally, if x is tied to the three points via (idealized) elastic bands, this minimum would bethe equilibrium position (i.e. the position of least energy).

Hint: Use vectors as much as possible - writing out all the components gets very messy.

Solution. We solve∇f = 2(x− x1) + 2(x− x2) + 2(x− x3) = 0

to get

x =1

3(x1 + x2 + x3).

(To compute the gradient, go one term at a time, and for each term, write things in terms ofcomponents, then take the result and put it back in vector form.)

Problem 6 (15.8.2). Find the extreme values of f(x, y) = x2 + 2y2 subject to the constraintg(x, y) = 4x− 6y = 25.

(a) Show that the Lagrange equations yield 2x = 4λ and 4y = −6λ.

(b) Show that if x = 0 or y = 0, then the Lagrange equations give x = y = 0. Since (0, 0) doesnot satisfy the constraint, you may assume that x and y are non-zero.

(c) Use the Lagrange equations to show that y = (−3/4)x.

(d) Substitute in the constraint equation to show that there is a unique critical point P .

(e) Does P correspond to a minimum or a maximum value of f?

43


Solution. (a) The Lagrange equations are

∇f = λ∇g =⇒

(2x

4y

)= λ

(4

−6

).

(b) If x = 0, then 4λ = 0, so λ = 0 and y = 0.

Similarly, if y = 0, then −6λ = 0, so λ = 0 and x = 0.

(c) Solving the Lagrange equations for λ, we haveλ = x/2 = −2y/3, so y = (−3/4)x.

(d) In the constraint equation, we have

4x− 6

(−3

4

)x =

17

2x = 25,

so the unique critical point is (x, y) = (50/17,−75/34).

(e) Since the line g = 0 does not meet contours of f corresponding to smaller values than at P ,it must be that P is a minimum.

Problem 7 (15.8.10). Find the extreme values of f(x, y) = x2y4 subject to x2 + 2y2 = 6.

Solution. Let g(x, y) = x2 + 2y2 − 6, so the constraint is g = 0.

If ∇g = 0, then 2x = 0 and 4y = 0, so (x, y) = (0, 0). This does not satisfy the constraint equation.

The Lagrange equations give ∇f = λ∇g, or

2xy4 = 2xλ, 4x2y3 = 4yλ.

If x = 0 or y = 0, then these equations can be satisfied and f = 0. Note that f(x, y) = x2y4 =(xy2) ≥ 0 for all x, y, and 0 is attained, so the minimum value of f is 0.

Otherwise, we get λ = y4 = x2y2, so x2 = y2. In this case we have x2 = y2 = 2, so

f(x, y) = x2y4 = (x2)(y2)2 = 8.

This must be the maximum value.

Problem 8 (15.8.22). Use Lagrange multipliers to find the maximum area of a rectangle inscribedin the ellipse

x2

a2+y2

b2= 1.

Solution. We wish to maximize f(x, y) = 4xy subject to g(x, y) = x2/a2 + y2/b2 = 1.

As in the previous problem, ∇g = 0 gives (x, y) = (0, 0), which does not satisfy the constraint.

The Lagrange equations are 4y = 2λx/a2 and 4x = 2λy/b2. If y = 0, then x = 0, but (0, 0) doesnot satisfy the constraint, so solutions must satisfy y 6= 0. The second equation gives λ = 2xb2/y,and substituting into the first equation yields

4y =2x

a2· 2xb2

y=⇒ y2/b2 = x2/a2.

44


Substituting this into the constaint equation, x/a = y/b = 1/√

2, so x = a/√

2 and y = b/√

2.

The corresponding maximum area is 2ab.

Problem 9 (15.8.52). This exercise shows that the multiplier may be interpreted as a rate ofchange in general. Assume that the maximum of f(x, y) subject to g(x, y) = c occurs at a point P .Then P depends on the value of c, so we may write P = (x(c), y(c)) and we have g(x(c), y(c)) = c.

(a) Show that

∇g(x(c), y(c)) ·

(x′(c)

y′(c)

)= 1.

(b) Use the chain rule and the Lagrange condition ∇fP = λ∇gP to show that

d

dcf(x(c), y(c)) = λ.

(c) Conclude that λ is the rate of increase in f per unit increase in the “budget level” c.

Solution. Differentiating g(x(c), y(c)) = c, we have

d

dcg(x(c), y(c)) = ∇g(x(c), y(c)) ·

(x′(c)

y′(c)

)=

d

dc(c) = 1.

Thend

dcf(x(c), y(c)) = ∇fP ·

(x′(c)

y′(c)

)= λ∇gP ·

(x′(c)

y′(c)

)= λ.

Problem 10. Fix a unit vector n as well as two points x1 and x2 in the plane.

(a) Show that any point x on the linen · x = 0

that minimizesf(x) = ‖x− x1‖+ ‖x− x2‖

must have the property (x− x1

‖x− x1‖

)⊥n

+

(x− x2

‖x− x2‖

)⊥n

= 0.

We assume here that neither x1 nor x2 lie on the line n · x = 0.

(b) Observe that the previous relation can be expressed as follows: “the angle of incidence isequal to the angle of reflection,” at least if both points are on the same side of the ‘mirror’occupying the line n · x = 0.

Solution. We wish to minimize f subject to g(x) = n · x = 0. First note that ∇g = n 6= 0 for anyx, so we just need to look at the Lagrange equation

∇f =x− x1

‖x− x1‖+

x− x2

‖x− x2‖= λn.

This is parallel to n, so the perpendicular part of the left hand side is 0.

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contentsazhou/teaching/17f/hw-solutions.pdf · problem 5 (13.1.70). use vectors to prove that the...

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