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Math 215 Final Exam Solutions April 26, 2004
Problem 1. (10 pts.)
(a – 5 pts.) Find the distance from the point P (1, − 3 , 2) to the plane Π given by the equation y − z − 1 = z − x + 1.
Solution. Rewriting the equation of the plane in the standard form, x + y − 2 z = 2, we see that a normal vector
to the plane can be taken as n = 〈 1 , 1 , − 2 〉. Also, it is easy to see that a point Q(1, 1 , 0) is on the plane (since its
coordinates satisfy the equation of the plane). Then QP = 〈 0 , − 4 , 2 〉 and
dist(P, Π) =
∣ (^) comp n
QP
Answer: The distance from the point P to the plane Π is dist(P, Π) = 8/
(b – 5 pts.) Find an equation of the plane containing the line r(t) = 〈 2 − t, t, 7 t + 1〉 and the point A(− 1 , 1 , 3).
Solution. We can easily see that the line r(t) goes through the point P (2, 0 , 1) (at time t = 0) and has the
direction vector v = 〈− 1 , 1 , 7 〉. The vector v is a vector in the plane, for another vector in the plane we can take
PA = 〈− 3 , 1 , 2 〉. Then
n =
i j k
− 1 1 7
− 3 1 2
is a normal vector to the plane and then our equation of the plane is
−5(x + 1) − 19(y − 1) + 2(z − 3) = 0 or 5 x + 19y − 2 z = 8.
Answer: The equation of the plane is − 5 x + 19y − 2 z = 8. ♣
Problem 2. (15 pts.) Suppose that you have just reached the final level of the exciting computer game “ Horrible Monsters
Galore”. You find yourself in a large hall at the point (− 1 , 1), and a measure of your danger is given by f (x, y) = y
2 − xy +
sin
πx
(larger values of f indicate greater danger).
(a – 5 pts.) You decide to go directly north. Is it a good idea or a bad idea (briefly explain your reasoning)?
Solution. We have
∇f = 〈−y +
π
cos
πx
, 2 y − x〉, ∇f (− 1 , 1) = 〈− 1 , 3 〉.
Going directly north corresponds to going in the y-direction, and since fy (− 1 , 1) = 3 > 0 , danger increases in this
direction.
Answer: Definitely a bad idea. ♣
(b – 5 pts.) How does the danger change if you go in the direction of the vector v = 〈 2 , − 1 〉 instead (please give an actual
value for the rate of change)?
Solution. Computing the rate of change (the directional derivative) of f (x, y) in the direction of v, we get:
Dvf (− 1 , 1) = compv ∇f (− 1 , 1) =
Answer: The danger is decreasing at a rate of
(c – 5 pts.) The graph below shows the view of the hall together with the level curves and gradient vectors of the danger
function. What would be a safe place to be at (i.e., find a spot of least danger). Where would a monster
(monsters?) most likely be? Mark these points on the graph and explain very briefly. 1
y
x
Solution. Looking at the contour plot we can see that the absolute maximum of the danger function is attained at
the top left corner (− 3. 5 , 2 .5) (f (− 3. 5 , 2 .5) ≈ 15. 7 ). The danger is also increasing towards the bottom right corner
(3. 5 , − 1 .5) (f (3. 5 , − 1 .5) ≈ 6. 8 ). Local minimum of the danger function inside the region is attained at approximately
(− 1. 264 , − 0 .632) (f (− 1. 264 , − 0 .632) ≈ − 1. 35 ), but the absolute minimum on the region is attained on the boundary
at the point (3. 5 , 1 .75), f (3. 5 , 1 .75) ≈ − 3. 77. In fact, even the top right (f (3. 5 , 2 .5) ≈ − 3. 21 ) and lower left corners
(f (− 3. 5 , − 1 .5) ≈ 2. 29 ) are safer than the local minimum.
y
x –3 –^
–1 0
1 2
3
x
0
2
y
0
5
10
15
Figure 1. Locations of the local and global minima and maxima (left), and the graph of the danger function (right).
Answer: The safest place to be will be the (3. 5 , 1 .75) along the east wall (in fact, the whole northern part of
this wall is pretty safe). There are lots of big horrible monsters monsters in the NW corner at the point (− 3. 5 , 2 .5)
(and there are a few little hairy monsters in the SW corner). ♣
Problem 3. (10 pts.) Find absolute minimum and absolute maximum values of the function f (x, y) = xy on the region
D = {(x, y) | 9 x
2
2 ≤ 144 }. Where are these values attained?
Solution. We know that for a continuous function on a closed and bounded region absolute minimum and absolute maximum
always exist and are attained either at the critical points inside the region or on the boundary. Therefore, our problem spilts into
two parts.
Critical points. Since ∇f = 〈x, y〉 = 0 only at the origin, the only critical point is (0, 0). At that point the Hessian matrix is
H =
[
]
, det H = − 1 < 0 ,
0
1
2
z
0.2 0.4 0.6 0.8 1
r
–0.
0
1
x
0 0.^
1
y
0
1
2
z
Figure 3. Plot of the “slice” of the region in the rz-plane (left) and the 3d-region of integration (right).
Answer: The given integral represents the volume of the solid sketched on Figure 3 and it is equal to
E
1 dV = π/ 12.
Problem 5. (10 pts.)
Consider the surface S in the shape of the “african drum” (see the picture on the right).
The top and bottom parts are flat disks and the side is a part of the surface given by the
equation x
2
2 = z
2
- 1, − 2 ≤ z ≤ 2. Find the mass of this surface if the density function
for the top and bottom parts is 5 gr/cm
2 and the density of the side is 3 |z| gr/cm
2 .
0 1 2
0 1 2
0
1
2
Solution. When z = ± 2 , x
2
2 = 5, and so the top and bottom parts of the “drum” are (flat) disks of radius
- Then the
mass of each of them is given by
Mass =
S
σ dS =
S
5 dA = 5 Area(disk) = 5 · π(
2 = 25π,
and so the total mass of top and bottom is 50 π.
To compute the mass of the side we first note that since the density function is |z|, the mass of the part of the surface below the
xy-plane is the same as the mass above the xy-plane, so we shall compute the mass of just the top half of the surface. To do so
we first need to parametrize the surface. There are essentially two ways to do so.
We can choose θ and z coordinates to parametrize our surface as
r(θ, z) = 〈
z 2
z 2
- 1 sin θ, z〉, 0 ≤ θ ≤ 2 π, 0 ≤ z ≤ 2.
rθ = 〈−
z 2
z 2
rz = 〈
z
z 2
cos θ,
z √ z 2
sin θ, 1 〉
rθ × rz = 〈
z 2
z 2
dS = |rz × rθ | dA =
2 z 2
Mass (^) side =
0
∫ (^2) π
0
3 z
2 z 2
0
z
2 z 2
2 z
2
2
0
= 26π.
Alternatively, we can solve the equation x
2
2 = z
2
x 2
- y 2 − 1 (recall that we parametrize just the
top half of the surface). In this case the domain of the parametrization will be a ring between circles of radii r = 1 (corresponding
to z = 0) and r =
5 (corresponding to z = 2). Then using the formula for the surface element for the graph of a function and
switching to the polar coordinates we get
r(x, y) = 〈x, y,
x 2
dS =
1 + f 2 x +^ f^
2 y dA^ =
x 2
x 2
y 2
x 2
2 r 2 − 1
r 2 − 1
r dr dθ
Mass (^) side =
2 π
0
√ 5
1
r 2 − 1
2 r 2 − 1
r 2 − 1
r dr dθ = π
2 r
2 − 1
√ 5
1
= 26π,
and so the total mass of the side is 52 π.
Answer: The total mass of the surface is 102 π gramms.
Problem 6. (5 pts.)
Consider the vector field F = P (x, y)i + Q(x, y)j, where P = sin(x
2 ) − y
and Q = x
2 − e
y+
. Use Green’s Theorem to find the circulation of the
field F along the boundary C of the region D shown on the picture on the
right.
D
C
0
2
4
6
8
10
–6 –4 –2 2 4
Solution. By Green’s theorem, ∮
C
F • dr =
D
curl F dA =
D
(Qx − Py ) dA =
D
(2x + 1) dA.
Using the slope and the x-intercept we see that the equation of the hypothenuse of our triangle is y = 3/2(x + 3) = (3x + 9)/ 2 ,
and so we get
∮
C
F • dr =
− 5
∫ 3 x+ 2
− 3
(2x + 1) dy dx =
− 5
(3x
2
3
x
2
3
− 5
Answer: The circulation of the field F along the curve C is 80 units. ♣
Problem 7. (5 pts.) Find the flux of a vector field F = (x
2 − y
2 )i + 2xzj + (z
2 − 2 xz)k across the surface S of the sphere
of radius 3 centered at the origin (with the outward orientation).
Solution. Since the surface S is closed and is the boundary of the ball B of radius 3 , we can use the Divergence Theorem to get ∫ ∫
S
F • dS =
B
div F dV =
B
(2x + 2z − 2 x) dV =
B
2 z dV = 0
by symmetry.
Answer: The flux of F across S is 0. ♣
Solution. Solving the equation of the plane for z and using the variables x and y as our parameters, we get
r(x, y) = 〈x, y, 3 x + y − 6 〉, x
2
2 ≤ 1.
Answer: r(x, y) = 〈x, y, 3 x + y − 6 〉, x
2
2 ≤ 1. ♣
–3 –^
–1 0
1 2
x
0
2
y
0
z
Figure 4. The surface S and its vectorial area element dS.
(b – 5 pts.) Find the flux of the vector field F(x, y, z) = xyi + zyj − x
2 k across S.
Solution. Using the parametrization from part (a) we get ∫ ∫
S
F • dS =
D
〈xy, (3x + y − 6)y, −x
2 〉 • 〈− 3 , − 1 , 1 〉 dA =
D
(− 6 xy − 6 y − y
2 − x
2 ) dA
noticing that integrals of 6 xy and 6 y will cancel by symmetry and switching to polar coordinates
∫ (^2) π
0
0
−r
2 · r dr dθ = −
π
Answer:
S
F • dS = −π/ 2. ♣
Problem 10. (10 pts.) Consider the plots of two vector fields: F(left) and G (right).
1
2
y
–2 –1 1 2
x
0
1
2
y
–2 –1 1 2
x
Vector field F(x, y) Vector field G(x, y)
(a – 2 pts.) On either of the plots carefully mark a point at which the field has a non-zero divergence and indicate if it is
positive or negative.
(b – 2 pts.) On either of the plots carefully mark a point at which the field has a non-zero curl and indicate if it is positive
or negative.
(c – 2 pts.) On either of the plots sketch a closed trajectory such that the circulation of the vector field along it is non-zero
and indicate if it is positive or negative.
(d – 2 pts.) On either of the plots give an example of two trajectories having the same starting and ending points such that
the work of the field along these trajectories is different
(e – 2 pts.) Comment on the possibility of each vector field being a gradient vector field.
Solution.
A
1
2
y
–2 –1 1 2
x
C
T S
B
0
1
2
y
–2 –1 1 2
x
Figure 5. Vector fields F(x, y) and G(x, y)
(1) Answer: The field F has a positive divergence at the point A (A is a source), the field G has a negative divergence
at the point B (B is a sink).
(2) Answer: The field G has a positive (counter-clockwise) curl at the point B.
(3) Answer: The field G has a positive circulation along the curve C.
(4) Answer: The work of the field G on the top trajectory from the point S to the point T is clearly positive, and
the work on the bottom trajectory is clearly negative
(5) Answer: Since the vector field G has an obvious curl, it can not be a gradient vector field. On the other hand,
vector field F has no obvoious curl, and so it may be a gradient vector field (in fact, F = ∇(sin(1. 3 x)/2 + y
2 /4) ).
