Math 238 Review Problems for Final Exam

For problems #1 to #6, we define the followingpaths vector fields:

b(t) = the straight line in the xy planefrom the point (0,!4) to the point(2,0)

c(t) = the path along the circle x2 + y2 =25 in the xy plane from the point(!5,0) to the point (0,!5)

d(t) = (3cos(2t),3sin(2t),t) for 0 # t # 4B

F = 3x2y3i + 3x3y2j

G = (z3 ! y ! 2x)i + (x + y + z)j + xyzk

V = 3x3y2i ! 3x2y3j

1. Do each of the following for the path d(t):

(a) Find the velocity vector and theacceleration vector.

(b) Evaluate each of the velocity vectorand the acceleration vector at t=B/6.

(c) Find the arc length of the pathbetween t=7B/6 and t=3B/2.

2. Consider the line integral of G over thepath d(t). Write the integral in the form

, but do not attempt to evaluate this

integral.

3. Do each of the following for the vectorfield F:

(a) Decide whether or not this vectorfield could be the curl of anothervector field, and say why or why not.

(b) If possible, find a function f(x,y)where the vector field is Lf; if thisis not possible, give a reason why.

(c) If possible, find the line integralof the vector field over a path fromthe point (4,2) to the point (!7,8);if this is not possible, give areason why.

(d) If possible, find the line integralof the vector field over a path whichbegins and ends at the point (4,2).

4. Do each of the following for the vectorfield G:

(a) Decide whether or not this vectorfield could be the curl of anothervector field, and say why or why not.

(b) If possible, find a function f(x,y,z)where the vector field is Lf; if thisis not possible, give a reason why.

(c) If possible, find the line integralof the vector field over a path fromthe point (4,2,1) to the point(!7,8,3); if this is not possible,give a reason why.

(d) If possible, find the line integralof the vector field over a path whichbegins and ends at the point (4,2,1).

5. Do each of the following for the vectorfield V:

(a) Decide whether or not this vectorfield could be the curl of anothervector field, and say why or why not.

(b) If possible, find a function f(x,y)where the vector field is Lf; if thisis not possible, give a reason why.

(c) If possible, find the line integralof the vector field over a path fromthe point (4,2) to the point (!7,8);if this is not possible, give areason why.

(d) If possible, find the line integralof the vector field over a path whichbegins and ends at the point (4,2).

6. Parametrize each of the following paths bywriting each path as (x(t),y(t),z(t)) andstating the bounds a # t # b):

(a) b(t) (b) c(t)

7. Parametrize the surface consisting of thepart of the cone 3z2 = x2 + y2 which liesabove the xy plane and inside the spherex2 + y2 + z2 = 400, by writing the surface as(x(u,v),y(u,v),z(u,v)) and stating thebounds for u and v.

8. Consider the region in the xy planedetermined by x4 + y4 # 64. Use polarcoordinates to write a double integral whichwill give us the area of this region, but donot attempt to evaluate this integral.

9. Consider the surface area of the portion ofthe plane z ! x ! 2y = 0 which lies abovethe triangle in the xy plane determined byx = 0, y = 8, and y = !2x. Write a doubleintegral which will give us this surfacearea, but do not attempt to evaluate thisintegral.

10. Consider the surface area of the portionof the sphere x2 + y2 + z2 = 100 whichlies inside the cone z2 = 3x2 + 3y2 in thepart of R3 where y $ 0 and z $ 0.

(a) Recall that the sphere x2 + y2 + z2 = 100can be parametrized by

(10 cosu sinv , 10 sinu sinv , 10 cosv)for 0 # u < 2B and 0 # v # B .

Find limits on u and v which describedesired surface area. (Hint: it will behelpful to consider the angle the conemakes with the z axis in either the xzplane or the yz plane.)

(b) Write a double integral of analgebraically simplified function f(u,v)which will give us this surface area,but do not attempt to evaluate thisintegral.

11. Consider the volume of the region insideboth the sphere x2 + y2 + z2 = 25 and thecone z2 = (x2 + y2)/3 in the part of R3

where x # 0, y $ 0, and z $ 0. Usespherical coordinates to write a tripleintegral which will give us this volume,but do not attempt to evaluate thisintegral.

12. Consider the volume of the region insideboth the sphere x2 + y2 + z2 = 25 and thecylinder 9(x2 + y2) = 64 in the part of R3

where z # 0. Use cylindricalcoordinates to write a triple integralwhich will give us this volume, but donot attempt to evaluate this integral.

13. Consider the area in the first quadrantof R2 between the two circles x2 + y2 = 9and (x ! 2.5)2 + y2 = 6.25. Use polarcoordinates to write the sum of twodouble integrals which will give us thisarea, but do not attempt to evaluatethese integrals.

14. Find and classify all critical pointsfor each of the following functions:

(a) f(x,y) = x3 + y2 + 6xy

(b) f(x,y) = x3 + y3 + 3xy2

15. Find the extreme values off(x,y) = xy - 7x on the circlex2 + y2 = 15.

16. Consider the following integral:

(a) Sketch the region over which theintegral is taken.

(b) Rewrite the integral by reversing theorder of integration.

17. Consider the function f(x,y) = x2ln(y).

(a) Find the second degree Taylorpolynomial about (2,e) (but you donot have to algebraically simplifyit).

(b) Find the directional derivative atthe point (2,e) in the direction ofthe vector (3,4).

18. Find the equation of the plane tangentto the surface defined by5x + z2y ! z + 1 = 0 at the point(-3,4,2).

Answers

1. (a) dN(t) = (!6sin(2t),6cos(2t),1)dNN(t) = (!12cos(2t),!12sin(2t),0)

(b) dN(B/6) = (!3/3 , 3 , 1)dNN(B/6) = (!6 , !6/3 , 0)

(c)

2. The integral is

3. (a) Since div(F) = 6xy3 + 6x3y =/ 0, then Fcould not be the curl of anothervector field.

(b) Since curl(F) = 0, then F is agradient vector field, and F = Lfwhere f(x,y) = x3y3.

(c) Since F = Lf, the line integral overany path from the point (4,2) to thepoint (!7,8) must be f(!7,8) ! f(4,2)= !175616 ! 512 = ! 176128.

(d) Since F = Lf, the line integral overany path which begins and ends at thepoint (4,2) must be zero (0).

4. (a) Since div(G) = !1 + xy =/ 0, then Gcould not be the curl of anothervector field.

(b) Since curl(G) =(xz ! 1)i + (3z2 ! yz)j + 2k =/ 0, thenG cannot be a gradient vector field.

(c) Since G is not a gradient vectorfield, the line integral over a pathfrom the point (4,2,1) to the point(!7,8,3) cannot be found, unless itis specified which path is chosen.

(d) Since G is not be a gradient vectorfield, the line integral over a pathwhich begins and ends at the point(4,2,1) cannot be found, unless it isspecified which path is chosen.

5. (a) Since div(V) = 0, then V could be thecurl of another vector field.

(b) Since curl(V) = (! 6xy3 ! 6x3y)k =/ 0,then V cannot be a gradient vectorfield.

(c) Since V is not a gradient vectorfield, the line integral over a pathfrom the point (4,2) to the point(!7,8) cannot be found, unless it isspecified which path is chosen.

(d) Since V is not a gradient vectorfield, the line integral over a pathwhich begins and ends at the point(4,2) cannot be found, unless it isspecified which path is chosen.

6. (a) One possible answer isb(t) = (2t , 4t!4 , 0) for 0 # t # 1

(b) One possible answer isc(t) = (5cost , 5sint , 0)

for B # t # 3B/2

7. The sphere and the cone intersect whenx2 + y2 = 300 and z = 10. One possibleanswer is(u cosv , u sinv , u//3)

for 0 # u # 10/3 and 0 # v # 2B

8.

9. First, we recognize that the planez ! x ! 2y = 0 is just the functionf(x,y) = x + 2y. The triangle in the xyplane determined by x = 0, y = 8, andy = !2x can be described by either

!4 # x # 0 or by 0 # y # 8!2x # y # 8 !y/2 # x # 0

The desired surface area is

or .

10. (a) 0 # u < B and 0 # v # B/6

(b) Mu(u,v) =(!10 sinu sinv , 10 cosu sinv , 0)

Mv(u,v) =(10 cosu cosv , 10 sinu cosv , !10 sinv)

Mu(u,v)×Mv(u,v) = (!100 cosu sin2v, !100 sinu sin2v, !100 sinv cosv)

|| Mu(u,v)×Mv(u,v) || = 100 sinv

The desired surface area is .

11.

12.

13.

14. (a) (0,0) is a saddle point(6,!18) is a local minimum

(b) (0,0) is the only critical point, butthe Second Derivative Test fails toclassify this point.However, once we realize thatf(0,0)=0 and f(x,y) can be negativeor positive, we then see that (0,0)must be a saddle point.

15. The extrema are at (/12.75 , !1.5) whichgives the minimum and (!/12.75 , !1.5)which gives the maximum.

16. (a)

(b)

17. (a) 4 + 4(x ! 2) + (4/e)(y ! e) + (x ! 2)2

! (2/e2)(y ! e)2 + (4/e)(x ! 2)(y ! e)

(b) 12/5 + 16/(5e)

18. 5x + 4y +15z = 31