At what rate is the length of his shadow on the building decreasing when he is 4m from the building?
Illustration
Given:
height of the man = 2m
distance of the spotlight to the wall = 12m
horizontal velocity of the man = 1.6m/s
Required: the rate how fast is the length of the man's shadow on the building decreasing when he is 4m from the building.
Solution:
By applying similar triangles we have,
y12=212−xy=2412−x=24(12−x)−1
By taking the derivative with respect to time,
dydt=dydx(dxdt)=24⋅dydx(12−x)−1dxdtdydt=24(−1)(12−x)−2⋅(−1)dxdtdydt=24(12−x)−2dxdt
We know that dxdt=1.6 and x=4m so
dydt=24(12−4)−2(1.6)dydt=0.6m/s
The length of the shadow is decreasing at a rate of 0.6m/s when the man is 4m from the building.
Friday, December 30, 2016
Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 16
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