Page 78 - How To Solve Word Problems In Calculus
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√ dx
Substituting x = 50, y = 10 97, and = 20, we obtain
dt
√ dy
(−40)(20) = 10 97
dt
dy 800 80
=− √ =−√
dt 10 97 97
The negative sign indicates that the runner’s distance to second
√
base is decreasing at the rate of 80/ 97 (approximately 8.1) ft/sec.
5.
x z
y
dx dz
Given: = 5 Find: (a) 2 minutes later
dt dt
dy dA
=−5 (b) 2 minutes later
dt dt
2
2
(a) z = x + y 2
dz dx dy
2z = 2x + 2y
dt dt dt
dz dx dy
z = x + y
dt dt dt
After 2 minutes, x = 70 + 10 = 80 cm and y = 70 − 10 =
√
2
2
60 cm. At this instant, z = 80 + 60 = 100 cm.
dz
100 = 80 · 5 + 60 · (−5)
dt
= 400 − 300
65
