Math 261 – Solutions to Sample Final Exam Problems

1
Math 261 – Solutions to Sample Final Exam Problems
Math 261 – Solutions to Sample Final Exam Problems
~ = (y + x)~i + (y − x) ~j, where C1 is the curve consisting of the
1. Let F~ = 2xy 2 ~i + (2yx2 + 2y) ~j, and let G
circle of radius 2, centered at the origin and oriented counterclockwise, and where C2 is the curve consisting
of the line segment from (0, 0) to (1, 1) followed by the line segment from (1, 1) to (3, 1).
~ over C1 .
(a) Calculate the line integral of F~ over C1 .
(b) Calculate the line integral of G
~ over C2 .
(c) Calculate the line integral of F~ over C2 .
(d) Calculate the line integral of G
~ are conservative vector fields. Since
Solution. Before we begin, let us check to determine whether F~ or G
∂
2
∂y (2xy )
= 4xy =
∂
2
∂x (2yx
+ 2y), we see that F~ is a conservative vector field. To find a potential function f
for F~ , we integrate as follows:
Z
f (x, y) = 2xy 2 dx
Z
f (x, y) = (2yx2 + 2y) dy
= x2 y 2 + P (y)
= x2 y 2 + y 2 + Q(x)
Therefore, f (x, y) = x2 y 2 + y 2 is a potential function for F~ . On the other hand, since
~ is not a conservative vector field.
equal to ∂ (y − x) = −1, we see that G
∂
∂y (y
+ x) = 1 is not
∂x
R
(a) Since C1 is a closed curve and F~ is a conservative vector field, we know that C1 F~ · d~r = 0 without
doing any calculations. So our final answer is 0.
~ is not a conservative vector field, we must do this integral by parameterizing C1 . We can
(b) Since G
represent C1 by the parametric curve ~r(t) = 2 cos t~i + 2 sin t ~j, where 0 ≤ t ≤ 2π. Therefore, we have
Z 2π
Z
~ · d~r =
~ cos t, 2 sin t) · (−2 sin t~i + 2 cos t ~j) dt
G
G(2
C1
0
=
Z
2π
Z
2π
Z
2π
(2 sin t + 2 cos t)~i + (2 sin t − 2 cos t) ~j) · (−2 sin t~i + 2 cos t ~j) dt
0
=
[−4(sin2 t + cos2 t) − 4 sin t cos t + 4 sin t cos t] dt
0
=
(−4) dt,
0
so our final answer is −8π.
(c) Since F~ is conservative, we can use the potential function f (x, y) = x2 y 2 + y 2 that we calculated above,
and the Fundamental Theorem of Calculus for Line Integrals as follows:
Z
F~ · d~r = f (3, 1) − f (0, 0) = 10 − 0 = 10
C2
Therefore, our final answer is 10.
~ is not conservative, we must parameterize the two line segments comprising C2 and calculate
(d) Since G
this line integral using brute force. The line segment from (0, 0) to (1, 1) can be parameterized as
~r1 (t) = t~i + t ~j, where 0 ≤ t ≤ 1, and the line segment from (1, 1) to (3, 1) can be parameterized as
~r2 (t) = (1 + 2t)~i + ~j, where 0 ≤ t ≤ 1. Therefore, we have
Z 1
Z 1
Z
~ · d~r =
~ r1 (t)) · ~r1′ (t) dt +
~ r2 (t)) · ~r2′ (t) dt
G
G(~
G(~
C2
0
=
0
Z
1
Z
1
Z
1
(2t~i + 0~j) · (~i + ~j) dt +
0
=
0
(2t · 1 + 2(2 + 2t)) dt
0
=
0
Z
(6t + 4) dt
=
7,
1
((2 + 2t)~i + 2t ~j) · 2~i dt
2
Math 261 – Solutions to Sample Final Exam Problems
so our final answer is 7.
2. Set up but do not evaluate an iterated integral that gives the volume of the solid region that lies below the
sphere x2 + y 2 + z 2 = 2 and above the paraboloid z = x2 + y 2 .
Solution. We begin by finding the projection of the curve of intersection of the two surfaces onto the
xy-plane. Specifically, we substitute z = x2 + y 2 into the equation of the sphere to obtain:
z + z2 = 2
=⇒
z2 + z − 2 = 0
=⇒
(z + 2)(z − 1) = 0
=⇒
z = 1.
Therefore, the intersection of these two surfaces takes place when z = 1; substituting z = 1 into the equation of the paraboloid yields the circle x2 + y 2 = 1.
It follows that the projection of this solid onto the xy-plane is given by the region
D = {(r, θ) : 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π} (see diagram to the right), so we have
Volume
=
Z
2π
Z
1
0
=
−1
Z
Z
1
0
√
Z
√
z
2−r 2
r2
1−x2
√
− 1−x2
r dz dr dθ
(in cylindrical coordinates)
√
Z 2−x2 −y2
dz dy dx
(in rectangular coordinates)
x
y
x2 +y 2
R
3. Let F~ = x2 ~i+z sin(yz) ~j +y sin(yz) ~k. Calculate C F~ · d~r,
where C is the path from A = (0, 0, 1) to B = (3, 1, 2)
shown in the figure to the right.
B
z
A
x
y
Solution. Calculating, we observe that curl F~ = ~0, so the vector field F~ is conservative and we can use the
Fundamental Theorem of Calculus for Line Integrals, provided
we do below:
Z
f (x, y, z) =
x2 dx
=
Z
f (x, y, z) =
z sin(yz) dy =
Z
f (x, y, z) =
y sin(yz) dz =
that we can find a potential function, which
x3
+ p(y, z)
3
− cos(yz) + q(x, z)
− cos(yz) + r(x, y)
3
Therefore, f (x, y, z) = x3 − cos(yz) is a potential function for F~ , so by the Fundamental Theorem of Calculus
for Line Integrals, we have
Z
F~ · d~r = f (3, 1, 2) − f (0, 0, 1) = (9 − cos 2) − (0 − cos 0) = 10 − cos 2.
C
R
~
~
~ = 2x~i − 3xy ~j + xz 2 ~k. Calculate
4. Let H
S H · dA, where S is the surface of the cube with corners at
(0, 0, 0), (1, 0, 0), (0, 1, 0), (1, 1, 0), (0, 0, 1), (1, 0, 1), (0, 1, 1), and (1, 1, 1), oriented outward.
Solution. Note that S is the surface of the solid region given by
E = {(x, y, z) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1} .
Therefore, by the Divergence Theorem, we have
Z
Z
~ dV =
~
~
div H
H · dA =
S
Z
1
Z
1
Z
1
=
1
=
Z
E
=
0
1.
Z
1
0
0
=
1
0
0
0
Z
Z
0
1
(2 − 3x + 2xz) dy dz dx
(2 − 3x + 2xz) dz dx
z=1
(2z − 3xz + xz 2 )z=0 dx
(2 − 2x) dx
3
Math 261 – Solutions to Sample Final Exam Problems
5. Let W be the solid region in the first octant that lies below the plane z = 6 and inside the cylinder x2 +y 2 = 4
(see figure to the right). Let F~ = z ~i + (xy + 1) ~j +(yez ) ~k.
z
R
~ where S1 is the left surface of the solid region W (that
(a) Calculate S1 F~ · dA
is, the portion of W that lies in the plane y = 0), oriented in the positive y
direction.
R
~ where S2 is the curved portion of W that lies on the
(b) Calculate S2 F~ · dA,
y
x
cylinder, oriented away from the origin.
Solution.
(a) On the left surface of the region, note that ~n = ~j, and since y = 0 on the left surface of the region, we
have F~ = z ~i + (xy + 1) ~j + yez ~k = z ~i + ~j. Therefore,
Z
Z
Z
Z
~=
dA
((z~i + ~j) · ~j)dA =
F~ · ~n dA =
F~ · dA
S1
S1
S1
S1
=
Surface Area of S1
=
12.
(b) For the cylindrical surface of the region, we have R = 2, x = R cos θ = 2 cos θ, and y = R sin θ = 2 sin θ.
Therefore, on this portion of the surface, our vector field becomes
F~ = z ~i + (xy + 1) ~j + yez ~k = z ~i + (4 sin θ cos θ + 1) ~j + 2 sin θ ez ~k.
~ = (cos θ ~i + sin θ ~j)2 dθ dz. Putting all of these facts together, we obtain
In addition, we have dA
Z π2 Z 6
Z
~ =
F~ · dA
(z ~i + (4 sin θ cos θ + 1) ~j + 2 sin θ ez ~k) · (cos θ ~i + sin θ ~j) 2 dz dθ
S2
0
0
Z
π
2
π
2
=
Z
π
2
=
Z
=
0
Z
6
(2z cos θ + 8 sin2 θ cos θ + 2 sin θ) dz dθ
0
z=6
z 2 cos θ + (8 sin2 θ cos θ + 2 sin θ)z dθ
z=0
0
(36 cos θ + 48 sin2 θ cos θ + 12 sin θ) dθ
0
=
=
π2
36 sin θ + 16 sin3 θ − 12 cos θ 0
64.
6. A plane passes through the points (1, 3, −2), (−1, −3, 4), and (2, 0, −2).
(a) Find the equation of this plane.
(b) At what point does the line ~r(t) = (1 + 2t)~i + (2 − 3t) ~j + (−1 + t) ~k intersect this plane?
Solution. Let P = (1, 3, −2), let Q = (−1, −3, 4), and let R = (2, 0, −2).
(a) Let ~u = P~R = ~i − 3~j and ~v = P~Q = −2~i − 6~j + 6~k. Then the vector
v
Q(−1,−3,4)
P(1,3,−2)
~n = ~u × ~v = −18~i − 6~j − 12~k
u
R(2,0,−2)
is a normal vector to the plane. Therefore, since (1, 3, −2) is a point on the plane,
the equation of the plane is given by −18(x − 1) − 6(y − 3) − 12(z + 2) = 0, or,
after simplifying, we can also write the equation of the plane as 3x + y + 2z = 2.
(b) Since we can rewrite ~r(t) parametrically as x = 1 + 2t, y = 2 − 3t, z = −1 + t, we can substitute into the
equation of our plane from part (a) to obtain
a
3(1 + 2t) + (2 − 3t) + 2(−1 + t) = 2 =⇒ 3 + 6t + 2 − 3t − 2 + 2t = 2
=⇒ 5t = −1
=⇒ t = −1/5.
4
Math 261 – Solutions to Sample Final Exam Problems
Substituting t = −1/5 for t in the equation for ~r(t), we obtain x = 3/5, y = 13/5, and z = −6/5, so the
point of intersection is given by (3/5, 13/5, −6/5), which is our final answer.
7. If w
~ = ~i − ~j + 2 ~k, ~u = 2~i + a~j + 3~k, and ~v = ~i + ~j, find:
(a) A unit vector parallel to w.
~
(b) The value of a making w
~ perpendicular to ~u.
(c) A unit vector perpendicular to both w
~ and ~v .
Solution.
~i − ~j + 2~k
1
1
2
w
~
√
= √ ~i − √ ~j + √ ~k
=
kwk
~
6
6
6
6
(b) w
~ · ~u = 0 =⇒ 2 − a + 6 = 0 =⇒ a = 8.
(c) First, note that w
~ × ~v = −2~i + 2~j + 2~k, so one such vector is given by
(a)
w
~ × ~v
kw
~ × ~v k
=
−2~i + 2~j + 2~k
√
12
=
−2~i + 2~j + 2~k
√
2 3
=
1
1
1
− √ ~i + √ ~j + √ ~k.
3
3
3
8. A cylindrical tube of radius 2 cm and length 3 cm contains a gas. Since the tube is spinning around its axis,
the density of the gas increases with its distance from the axis. The density, D, at a distance of r cm from
the axis is D(r) = 1 + r grams per cubic centimeter. Find the mass of the gas in the tube.
Solution. We first note that D(r) has units of grams/cm3, and dV has units of cm3 , so the product
RD(r) · dV has units of grams, which yields a mass. Therefore, the mass we are seeking is given by the integral
D(r) dV, where W = {(r, θ, z) : 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 3} , so we have
W
Mass =
Z
W
D(r) dV =
Z
0
2
Z
0
3
Z
0
2π
(1 + r)r dθ dz dr
= 6π
Z
2
(r + r2 ) dr
0
= 28π grams.
9. The kinetic energy, E, of a moving object having mass m kilograms and speed v meters per second is given by
E = f (m, v) = 12 mv 2 Joules. Evaluate and give a complete sentence interpretation of each of the following
quantities.
(a) f (2, 3)
(b) fv (2, 3)
(c) fm (2, 3)
Solution.
(a) We have f (2, 3) = 21 (2)(3)2 = 9 Joules. Thus, f (2, 3) gives the kinetic energy, in Joules, of an object of
mass 2 kilograms moving at a speed of 3 meters per second.
(b) We have fv (m, v) = mv, so that fv (2, 3) = 6 Joules per meter per second. Thus, the fact that
fv (2, 3) = 6 indicates the following: for an object having a mass of 2 kilograms, the kinetic energy
increases at a rate of about 6 Joules for each meter per second that the speed of the object is increased
beyond a speed of 3 meters per second.
(c) We have fm (m, v) = 12 v 2 . Therefore, fm (2, 3) = 21 (9) = 4.5 Joules per kilogram. So the fact that
fm (2, 3) = 4.5 indicates the following: for an object moving at a speed of 3 meters per second, the
kinetic energy increases at a rate of about 4.5 Joules for each kilogram of mass added to the object
beyond a mass of 2 kilograms.
5
Math 261 – Solutions to Sample Final Exam Problems
10. Given below are level curve diagrams for two functions, f
and g.
z = f (x, y)
20
(a) Which of the two functions above is linear? Why?
Find a formula for the linear function.
8
15
y
10
(b) Estimate gx (10, 5) and gy (10, 5).
z = g(x, y)
20
15
6
y
4
10
2
5
6
5
0
5
8
4
10 15
x
20
2
5
x
10
0
15
20
Solution.
(a) The function z = f (x, y) is linear because its level curves are equally spaced straight lines whose z-values
decrease at a constant rate as you move from one side of the diagram to the other. To find an equation
for f, we first use the diagram to deduce that (0, 0, 2), (10, 0, 0), and (0, 5, 4) are all points on the graph
of f. Therefore, we have
m = slope of f in x-direction =
Similarly,
0−2
1
∆z
=
=− .
∆x
10 − 0
5
n = slope of f in y-direction =
Thus, the equation for our plane is given by
∆z
4−2
2
=
= .
∆y
5−0
5
1
2
z = z0 + m(x − x0 ) + n(y − y0 ) = 2 − (x − 0) + (y − 0),
5
5
which yields a final answer of f (x, y) = 2 − 15 x + 25 y.
(b) Using the level curve diagram, we see that (10, 5, 4) and (20, 5, 2) are points on the graph of g. Therefore,
we have
∆z
2−4
1
gx (10, 5) ≈
=
=− ,
∆x
20 − 10
5
so our approximation for gx (10, 5) is −1/5. Similarly, since the point (10, 10, 6) also lies on the graph of
g, we have
6−4
2
∆z
=
= ,
gy (10, 5) ≈
∆y
10 − 5
5
so our approximation of gy (10, 5) is 2/5.
11. Suppose that the temperature at a point (x, y) is given by the formula T (x, y) =
where x and y are measured in meters.
x2
256
degrees Celsius,
+ 2y 2 + 1
(a) If you move away from the point (1, 1) in the direction of the point (2, 3), is the temperature increasing
or decreasing? At what rate?
(b) In what direction should you move away from the point (1, 1) for the temperature to remain constant?
Solution. Before doing parts (a) and (b), we calculate a formula for the gradient function since we know
that we will need this for both parts of the problem. We have T (x, y) = 256(x2 + 2y 2 + 1)−1 , so that
∇T (x, y) = −256(x2 + 2y 2 + 1)−2 · 2x~i − 256(x2 + 2y 2 + 1)−2 · 4y ~j = −256(x2 + 2y 2 + 1)−2 (2x~i + 4y ~j).
Therefore, ∇T (1, 1) = −256(1 + 2 + 1)−2 (2~i + 4~j) = −32~i − 64~j.
(a) We first note that we will be moving from (1, 1)√to (2, 3), which
means we are moving in the direction of
√
the vector ~i + 2~j. Therefore, the vector ~u = (1/ 5)~i + (2/ 5) ~j describes a unit vector in the direction
of our motion. Therefore, we have
Tu~ (1, 1) = ∇T (1, 1) · ~u
1
2
= (−32~i − 64~j) · ( √ ~i + √ ~j)
5
5
160
≈ −71.6
= −√
5
6
Math 261 – Solutions to Sample Final Exam Problems
and we therefore see that the temperature is decreasing at a rate of about 71.6 degrees Celsius per meter.
(b) Here, we are looking for a unit vector ~u = u1 ~i + u2 ~j such that Tu~ (1, 1) = 0. Thus, we have
Tu~ (1, 1) = 0
=⇒
∇T (1, 1) · ~u = 0
=⇒
=⇒
=⇒
(−32~i − 64~j) · (u1 ~i + u2 ~j) = 0
−32u1 − 64u2 = 0
u1 = −2u2 .
Since we also know that ~u must be a unit vector, we also know that k~uk = 1, so
5u22 = 1
2
1
and u1 = −2u2 = − √
=⇒
u2 = √
5
5
√
√
Therefore, we must move in the direction of (−2/ 5)~i + (1/ 5) ~j in order for the temperature to remain
constant.
u21 + u22 = 1
(−2u2 )2 + u22 = 1
=⇒
=⇒
12. Find the equation of the line containing the point (1, 3, −2) that is parallel to both of the planes x+2y −z = 4
and 2x − y + z = 1.
Solution. Note that ~n1 = ~i + 2~j − ~k and ~n2 = 2~i − ~j + ~k are normal vectors to the 2 respective planes.
Thus, since the desired line is parallel to both planes, it is perpendicular to both normal vectors. Therefore,
~n1 × ~n2 = ~i − 3~j − 5~k is parallel to the line, so the line is given by the vector equation
~r(t) = (1 + t)~i + (3 − 3t) ~j + (−2 − 5t) ~k.
We could also write the answer in parametric form as
x = 1 + t, y = 3 − 3t, z = −2 − 5t.
13. Let W be the solid object consisting of two solid cylinders meeting at right angles at the origin. One
cylinder is centered on the y-axis, between y = −5 and y = 5 with radius 2 and the other is centered on the
x-axis between x = −5 and x = 5 with radius 2 (see figure below). Let S be the whole surface of W except
for the circular end of the cylinder centered at (0, 5, 0). The boundary of S is a circle, C, and the surface S
is oriented outward. Let F~ = −xy ~i+3~j +yz ~k, and suppose you are also told that F~ = curl (3z ~i+xyz ~j +2 ~k).
R
~ Write down two other in(a) Suppose that you want to calculate S F~ · dA.
R
~
tegrals that have the same value as S F~ · dA.
R
~ using whatever method is easiest.
(b) Calculate S F~ · dA
z
y
x
Solution.
Before we do part (a) and part (b), let’s make some preliminary observations. By Stokes’
R
R
~ = (3z ~i + xyz ~j + 2 ~k) · d~r, where C is oriented as shown
Theorem, we have S F~ · dA
C
in the diagram to the right. Furthermore, letting S2 be the portion of the plane y = 5
that lies inside C, we see that S2 has the same boundary curve as S, so that by Stokes’
Theorem, we have
Z
Z
Z
~
~
~
~
~
~
(3z i + xyz j + 2 k) · d~r =
F · dA =
F~ · dA,
S2
C
where S2 is oriented in the negative y direction.
n
S
C
7
Math 261 – Solutions to Sample Final Exam Problems
R
R
~ is equal to the line integral (3z ~i + xyz ~j + 2 ~k) · d~r,
(a) By the observations above, we see that S F~ · dA
R C
~ is also equal to
where C is oriented as shown in the diagram above. Similarly, we see that S F~ · dA
R
~
~
F · dA, where S2 is the disk of radius 2 on the interior of C, oriented in the negative y direction.
S2
(b) Since the surface S2 has equation y = 5, and since ~n = −~j is orthogonal to S2 at every point on S2 with
the orientation described above, we have
Z
Z
Z
Z
~=
(−3) dA
(−xy ~i + 3 ~j + yz ~k) · (−~j) dA =
F~ · ~n dA =
F~ · dA
S
S2
S2
S2
= (−3)(Area of S2 )
= −12π.