Page 55 - How To Solve Word Problems In Calculus

100
3 2
Since its volume is 100 in , x y = 100 and y = . It follows that
x 2
100
2
S(x) = 2x + 4x ·
x 2
2
S(x) = 2x + 400x −1
Differentiating,

S (x) = 4x − 400x −2
400
= 4x −
x 2
The rate of change is determined by letting x = 5.

400

S (5) = 20 − = 4
25
2. (a) The rate of change of f is the value of its derivative at x = 16.
√
f (x) = x = x 1/2
1 1 1
−1/2
f (x) = x = = √
2 2x 1/2 2 x
1

f (16) =
8
f (x + h) − f (x)
(b) The average rate of change is the value where
h
x = 16 and h = 9.
f (x + h) − f (x) f (25) − f (16) 5 − 4 1
= = =
h 9 9 9
√
Alternatively, if y = x, the rate of change is y/ x.
When x = 16, y = 4 and when x = 25, y = 5.

y 5 − 4 1
= =
x 25 − 16 9
3. (a) When t = 0, s = 25 and when t = 2, s = 101
s 101 − 25
v av = = = 38 ft/min
t 2 − 0

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