HW14-HW15
Qinfeng Li
September 25
Problem 1. (HW 14, # 3) The temperature at a point (x, y) is T (x, y), measured in degrees
√ Celsius. A bug crawls so that its position after t seconds is
given by x = 1 + t, y = 2 + 31 t, where x and y are measured in centimeters.
The temperature function satisfies Tx (2, 3) = 4 and Ty (2, 3) = 3. How fast is
the temperature rising on the bug’s path after 3 seconds? (Round your answer
to two decimal places.)
√
√
Solution. After 3 seconds, x = 1 + t = 1 + 3 = 2, y = 2 + 13 t = 2 + 13 (3) =
dy
1
1
√1
√1
3, dx
dt = 2 1+t = 2 1+3 = 4 , and dt = 3 . Then
dT
dx
dy
1
1
= Tx (2, 3)
+ Ty (2, 3)
= 4( ) + 3( ) = 2.
dt
dt
dt
4
3
Thus the temperature is rising at a rate of 2◦ C/s.
Problem 2. (HW 14, # 9) Find the directional derivative of the function at
the given point in the direction of the vector v.
g(r, s) = tan−1 (rs), (1, 2), v = 5i + 10j.
Du g(1, 2) =
1
s
r
Solution. gr = 1+(rs)
2 ·s = 1+r 2 s2 , and similarly gs = 1+r 2 s2 , hence ∇g(1, 2) =
2
1
√ 1
(5i + 10j) =
5 i + 5 j. The unit vector in the direction of v is u =
52 +102
√1 i + √2 j, so
5
5
Du g(1, 2) = ∇g(1, 2) · u =
2
1
1
2
4
i+ j· √ i+ √ j = √ .
5
5
5
5
5 5
Problem 3. (HW 15, # 5) Find all points at which the direction of fastest
change of the function f (x, y) = x2 + y 2 − 2x − 4y is i + j. (Enter your answer
as an equation.)
Solution. The direction of fastest change is ∇f (x, y) = (2x − 2)i + (2y − 4)j,
so we need to find all points (x, y) where ∇f (x, y) is parallel to i + j. By
comparing ∇f and i + j, we conclude 2x − 2 = 2y − 4 is the necessary and
sufficient condition. Hence at points on the line y = x + 1, f (x, y) has the
fastest change.
1
Problem 4. (HW 15, # 6) The temperature T in a metal ball is inversely
proportional to the distance from the center of the ball, which we take to be the
origin. The temperature at the point (1, 2, 2) is 120◦ . (a) Find the rate of change
of T at (1, 2, 2) in the direction toward the point (1, −1, 1).
(b) Show that at any point in the ball the direction of greatest increase in temperature is given by a vector that points towards the origin.
k
where k is a constant
x2 +y 2 +z 2
k
√
= 120, hence k = 360.
12 +22 +32
Solution. By hypothesis, we can assume T = √
we are going to find. Since T (1, 2, 2) = 120,
Therefore,
3
∇T (x, y, z) = −360(x2 + y 2 + z 2 )− 2 (x, y, z).
(a) u =
(1,−1,1)
√
,
3
Du T (1, 2, 2) = ∇T (1, 2, 2) · u = −
1
40
40
(1, 2, 2) · √ (1, −1, 1) = − √ .
3
3
3 3
(b) We know that along the gradient vector ∇T , the function increases most
rapidly. From the expression of ∇T , the gradient direction is really parallel to
−(x, y, z), and this is exactly the negative position vector, which points toward
the origin, hence the direction of greatest increase in temperature is given by a
vector points toward the origin.
2