Sample Final Exam #1 Math 3C

Math 3C
Sample Final Exam #1
Laney College, Fall 2010
Fred Bourgoin
1. Find the equation for the linear function with the contour diagram
shown below.
2. Consider the plane with equation z = 2x − y + 3.
(a) Find a unit vector normal to the plane.
(b) Find a unit vector in the plane.
3. The velocity of a flow at the point (x, y) is F~ (x, y) = ~i + x ~j. Parameterize the path of motion of an object in the flow that is at the point
(−2, 2) at time t = 0.
4. The contour diagram for f is given below. For each of the following,
determine whether it is positive, negative, or zero. Justify your answer.
(a) fx (P )
1
(b) fy (P )
(c) fxy (P )
5. Give an iterated integral you would use to evaluate
Z
f dA, where R
R
is the region pictured below. You do not need to evaluate any integrals
for this problem.
6. Find the minimum and maximum of the function z = 4x2 − xy + 4y 2
over the closed disk x2 + y 2 ≤ 2.
7. (a) Find parametric equations for the line passing through the points
(1, 2, 3) and (3, 5, 7).
(b) Parameterize the circle of radius 2 parallel to the xy-plane, centered at the point (0, 0, 1) and traversed clockwise when viewed
from above.
8. Let F~ (x, y) = y~i + x~j and let C be the square with opposite
Z corners
F~ · d~r.
at (0, 0) and (1, 1) traversed counterclockwise. Compute
C
Do not use Green’s Theorem, Stokes’ Theorem, or The Fundamental
Theorem of Line Integrals.
9. Let F~ (x, y, z) = x~i + y~j + (z 2 + 3) ~k, and let S be the rectangle z = 4,
0 ≤ x ≤ 2, 0 ≤ y ≤ 3, oriented in the positive z-direction. Calculate
the flux of F~ through S.
10. Let F~ (x, y, z) = x~i + y~j + z~k, and let S be the cube enclosing the
volume 0 ≤ x
Z ≤ 2, 0 ≤ y ≤ 2, 0 ≤ z ≤ 2. Use the divergence theorem
~
F~ · dA.
to calculate
S
2
Math 3C
Sample Final Exam #1 Solutions
Laney College, Fall 2010
Fred Bourgoin
1. Find the equation for the linear function with the contour diagram
shown below.
Solution. The x-slope is m = 2, the y-slope is n = −1, and the
plane goes through the point (0, −1, 4), so an equation for the plane
is z − 4 = 2(x − 0) − 1(x + 1), or z = 2x − y + 3.
2. Consider the plane with equation z = 2x − y + 3.
(a) Find a unit vector normal to the plane.
Solution. Rewrite the equation as 2x − y − z = −3, then a
~ ~
normal vector is ~v = 2~i −
√ j − k. To make it into a unit vector,
divide by its magnitude 6:
1
~u = √ (2~i − ~j − ~k).
6
(b) Find a unit vector in the plane.
Solution. There are many ways to do this. We could find two
points in the plane and go from there, or we can find a vector
perpendicular to ~u above—say, ~a = ~i + ~j + ~k. Then make it into
a unit vector:
~b = √1 (~i + ~j + ~k).
3
3
3. The velocity of a flow at the point (x, y) is F~ (x, y) = ~i + x~j. Parameterize the path of motion of an object in the flow that is at the point
(−2, 2) at time t = 0.
Solution. We are looking for
~r(t) = x~i + y~j = f (t)~i + g(t)~i
such that
dx
dt
= f 0 (t) = 1 and
f (t) = t + c1
dy
dt
and
= g0 (t) = x = f (t). We have
g(t) =
1 2
t + c1 t + c2 .
2
We also need to have f (0) = −2 and g(0) = 2, which we get by setting
c1 = −2 and c2 = 2. Hence, the path is parameterized by
1
2
~r(t) = (t − 2)~i +
t − 2t + 2 ~j.
2
4. The contour diagram for f is given below. For each of the following,
determine whether it is positive, negative, or zero. Justify your answer.
(a) fx (P )
Solution. fx (P ) > 0 since the z-values increase as we move in
the positive x-direction from P .
(b) fy (P )
Solution. fy (P ) > 0 since the z-values increase as we move in
the positive y-direction from P .
(c) fxy (P )
Solution. fxy (P ) = 0 because the contour diagram is that of a
plane.
4
5. Give an iterated integral you would use to evaluate
Z
f dA, where R
R
is the region pictured below. You do not need to evaluate any integrals
for this problem.
Solution. To do this in one double integral,
Z 1Z 2−y
f (x, y) dxdy.
0 y
If you prefer two integrals:
Z 1Z x
Z 2Z
f (x, y) dydx +
0 0
2−x
f (x, y) dydx.
1 0
6. Find the minimum and maximum of the function z = 4x2 − xy + 4y 2
over the closed disk x2 + y 2 ≤ 2.
Solution. First, we look for critical points on the boundary using
Lagrange multipliers. The system to solve is
8x − y = λ2x
.
−x + 8y = λ2y
Soling λ in both equations and setting them equal to each other yields
8x − y
−x + 8y
=
=⇒ 16xy − 2y 2 = −2x2 + 16xy =⇒ y 2 = x2 .
2x
2y
Plugging that into the constraint equation, we get 2x2 = 2; that is,
x = ±1. So we have four critical points on the boundary: (−1, −1),
(−1, 1), (1, −1), and (1, 1). Now for the interior:
∂z/∂x = 8x − y = 0
∂z/∂x = −x + 8y = 0
5
=⇒
x = y = 0.
The only critical point there is (0, 0). Now,
f (−1, −1) = f (1, 1) = 7
f (−1, 1) = f (1, −1) = 9
f (0, 0) = 0
We have maxima at (−1, 1) and (1, −1), and a minimum at (0, 0).
7. (a) Find parametric equations for the line passing through the points
(1, 2, 3) and (3, 5, 7).
Solution. The line goes through the point (1, 2, 3) and has the
direction of ~v = 3~i + 5~j + 7~k, so an equation for it is
~r(t) = ~i+2~j +3~k +t(3~i+5~j +7~k) = (1+3t)~i+(2+5t)~j +(3+7t)~k.
(b) Parameterize the circle of radius 2 parallel to the xy-plane, centered at the point (0, 0, 1) and traversed clockwise when viewed
from above.
Solution. One possibility is ~r(t) = 2 sin t~i + 2 cos t~j + ~k.
8. Let F~ (x, y) = y~i + x~j and let C be the square with opposite
Z corners
F~ · d~r.
at (0, 0) and (1, 1) traversed counterclockwise. Compute
C
Do not use Green’s Theorem, Stokes’ Theorem, or The Fundamental
Theorem of Line Integrals.
Solution. If C1 is the side of the square on the x-axis, and C2 , C3 ,
C4 are the other sides proceeding counterclockwise, the sides can be
parameterized as follows.
C1
C2
C3
C4
:
:
:
:
~r1 (t) = t~i,
~r2 (t) = ~i + t~j,
~r3 (t) = (1 − t)~i + ~j,
~r4 (t) = (1 − t)~j,
6
0≤t≤1
0≤t≤1
0≤t≤1
0≤t≤1
Then, we have
Z 1
Z
~ · d~r = (t~j) · (~i) dt = 0,
F
C1
0
1
~ · d~r =
F
Z
Z
Z
1
~ · d~r =
F
Z
Z
1
~ · d~r =
F
Z
C2
C3
C4
Hence,
vative.
Z
0
0
0
Z
(t~i + ~j) · (~j) dt =
1
1 dt = 1,
0
(~i + (1 − t)~j) · (−~i) dt =
Z
0
1
−1 dt = −1,
((1 − t)~i) · (−~j) dt = 0.
F~ · d~r = 0. That should not be surprising since F~ is conser-
C
9. Let F~ (x, y, z) = x~i + y~j + (z 2 + 3) ~k, and let S be the rectangle z = 4,
0 ≤ x ≤ 2, 0 ≤ y ≤ 3, oriented in the positive z-direction. Calculate
the flux of F~ through S.
Solution. The area vector is 6~k. (Its direction is that of the orientation of the surface, and its magnitude is the area of the surface.)
On the given surface, the field is F~ = x~i + y~j + 19 ~k, but only the ~k
component matters for this surface.
Z
~ = 19 · 6 = 114.
F~ · dA
S
10. Let F~ (x, y, z) = x~i + y~j + z~k, and let S be the cube enclosing the
volume 0 ≤ x
Z ≤ 2, 0 ≤ y ≤ 2, 0 ≤ z ≤ 2. Use the divergence theorem
~
F~ · dA.
to calculate
S
~ = 1 + 1 + 1 = 3,
Solution. Since div F
Z
Z
Z
~
~
~
div F dV =
F · dA =
S
W
W
7
3 dV = 3 · 8 = 24.