Mathematics 2210 EXAM II SOLUTIONS Fall 2015
(20 points) 1. Let the pollen concentration in R3 be given by P (x, y, z) = e−(x
Suppose a bee is flying on a trajectory given by the helix
2 +y2 +z2 )
r(t) = (x(t), y(t), z(t)) = (0, 4 − 2t, 2 − t).
(a) Use the chain rule to find the rate of change dP
dt of the pollen concentration as the bee approaches the flower. Express your answer
in terms of t only and simplify it.
Answer.
dP
dt
= 20e−((4−2t)
2 +(2−2t)2 )
− 12te−((4−2t)
2 +(2−2t)2 )
(b) Find the level sets of P (x, y, z) in (a), and be sure to show your
work.
2
2
2
Answer. P (x, y, z) = c ∈ R so e−(x +y +z ) = c. Since (x2 +
y 2 + z 2 ) ≥ 0 we have x2 + y 2 + z 2 = − ln(c) = ln(1/c) ≥ 0. So we
have spheres of radius ln(1/c) as the level sets, where 0 < c ≤ 1.
(10 points) 2. Find the following limit. If it does not exist, demonstrate why not.
(x + y)2
.
(x,y)→(0,0) x2 + y 2
lim
Answer. The limit does not exist. This can be seen, for example,
by approaching along the line y = x and along the y-axis, and getting
different limits.
(15 points) 3. Find the directional derivative of f (x, y, z) = x2 y + 2yz + z 2 at the
point P = (1, 1, 1) in the direction d~ = i − 2j + 3k.
~ This is given
Answer. We need a unit vector in the direction of d.
1
by u
~ = √14 < 1, −2, 3 > . The directional derivative at the point P is
given by ∇f |(1,1,1) · ~u = (2xy, x2 + 2z, 2y + 2z)(1,1,1) · ~u = (2, 3, 4) · ~u =
√
√
(1/ 14)(2 − 6 + 12) = (8/ 14).
(15 points) 4. (a) Consider the saddle function f (x, y) = x2 − y 2 .
i. Show that this function is harmonic.
1
.
Solution. A harmonic function must sastify Laplace’s equa∂2f
∂2f
+
= 0.
tion, ∇2 f = 0, or rewritten,
∂x2
∂y 2
∂ 2f
=2
∂x2
2
∂ f
= −2
∂y 2
2 − 2 = 0.
ii. Find the maximum value of f on the disk D, and where it
occurs. Be sure to justify your answer.
Solution. A harmonic function on a closed, bounded set
always achieves its extrema on the boundary, in the case the
unit circle x2 + y 2 = 1. Changing f to polar coordinates,
f = (r cos θ)2 − (r sin θ)2
f = r 2 (cos2 θ − sin2 θ)
f = r 2 (2 cos2 θ − 1)
The unit circle in polar coordinates is the equation r = 1,
thus we want to find the extrema of f over all values of θ.
f = 2 cos2 θ − 1
This is maximized when cos θ = 1 or −1, so when θ = 0 or
π. The maximum attained at these points (corresponding to
the Cartesian labeling of (−1, 0) and (1, 0)) is 1.
(15 points) 5. Consider the function z = f (x, y) = 2 + x2 + y 2 whose graph is a
paraboloid in R3 . Sketch in R2 the
√ two
√ level curves defined by z = 6
~ = ( 2, 2), include in your sketch both
and z = 11. At the point P
the gradient vector ∇f and a tangent vector to the level curve. What
~?
is the angle between the gradient vector and the tangent vector at P
Answer.
The level sets for z = 6 and z = 11 correspond to circles. The inner
circle has radius 2 and the outer circle has radius 3. The point P
lies on the z = 6 level curve (circle radius 2, since 2+2+2=6). The
gradient from that point forms an angle of 90 degrees with the tangent
from that point.
2
(15 points) 6. Let φ(x, y, z) be the electric potential due to a point charge in three
dimensions, that is,
k
φ(x, y, z) = ,
r
p
where r = x2 + y 2 + z 2 and k > 0 is a physical constant. Find the
level sets of φ and its negative gradient F~ = −∇φ.
Be sure to write your final answer for F~ = −∇φ in terms of ~r = (x, y, z)
and indicate what famous law of nature this is.
Answer. The level sets are where kr = c for some constant c ∈ R;
thus, r = kc , so the level sets are spheres.
~ = −∇φ =
E
k
(x2
+ y2
3
+ z2) 2
(x, y, z) =
k
~r.
||~r||3
(10 points) 7. A mountain climber wishes to challenge herself as much as possible in
a climb up Mt. Mathematics. The altitude in feet on the mountain is
given by A(x, y) = −x2 − y 2 + 7200. If the climber is standing at the
base at the point (60, 60), what direction should she climb to make
the steepest ascent?
Answer We can find the direction of greatest change for the function as this will correspond the the steepest accent. Taking the gradient of A(x, y) and plugging in (60, 60) gives the direction vector
< −120, −120 >. Of course any vector in this same direction is acceptable such as < −1, −1 >
3