MA 231
Supplemental Problems on the Chain Rule
df
df dx df dy
=
+
= fxx + fy y dt dx dt dy dt
∆f = f (a + ∆x, b + ∆y) − f (a, b) ≈ fx(a, b)∆x + fy (a, b)∆y.
Note that problem 1c is changed from what was given in class:
1. Find
dz
if
dt
(a) z = f (x, y) = x2 y, x = t2 , y = t2
(b) z = f (x, y) = xexy , x = et , y = t2
y
(c) z = f (x, y) = , x = 3t + 1, y = t2
x
2. Suppose n, w measure the distance north and west from the corner of a tulip
field. f (n, w) = w1/2 n3/2 + n2 gives the height of the tulips when you are n feet
to the north and w feet to the west.
(a) You are tiptoeing through the tulips and notice that when you are 1 foot
north and 3 feet west that you are moving at 4 ft/sec northward and −2
ft/sec westward. How fast is the height of the tulips changing?
(b) Suppose you move from (1, 3) to (3, 2). Estimate the change in the height
of the tulips.
3. As a function of temperature T and nutrient load N , the population P of bacteria
in a lake is given by P (N, T ) = N e−T /10 .
(a) Suppose that when T = 10 and N = 40 you observe that the temperature is
rising at 3 degrees per hour and the nutrient load is dropping at 5 kilograms
per hour. How fast is the bacterial population changing?
(b) Suppose that at T = 10 and N = 40 you know that the bacterial population
is 33.7 (in appropriate units). Esimate the bacterial population when T = 9,
N = 37.
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Answers
The answers are given in expanded form to make it easier to check.
1a. z = 2(t2 )t2 (2t) + (t2 )2 2t
1b. z = (exy + xyexy )x + x2 exy y = (ee t + et t2 ee t )et + e2t ee t 2t
t 2
1c. z = −
t 2
t 2
y 1 t2
1
x
+
=
−
1+
y
2t
2
2
x
x
(t + 1)
(t + 1)
2. fn = 32 w1/2 n1/2 + 2n, fw = 12 w−1/2 n3/2 .
2a. z = ( 32 31/2 + 2)4 + ( 12 3−1/2 )(−2)
2b. ∆z = f (3, 2) − f (1, 3) ≈ ( 32 31/2 + 2)2 + ( 12 3−1/2 )(−1)
N −T /10
e
3. PN = e−T /10 , PT = − 10
3a. P = PN N + PT T = e−1 (−5) + (−4e−1 )(3)
3b. ∆P = P (37, 9)−33.7 = P (37, 9)−P (40, 10) ≈ PN (50, 10)∆N +PT (40, 10)∆T
= e−1 (−3) + (−4e−1 )(−1).
So P (37, 9) = 33.7 + e−1 (−3) + (−4e−1 )(−1).
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