Question

A man walks in a straight path away from a streetlight at a rate of 4 feet per second. The man is 6 feet tall and the streetlight is 15 feet tall.
At what rate is the length of his shadow changing?
At what rate is the distance from the top of the streetlight to the tip of his shadow changing when the man is 12 feet from the streetlight?

Answer 1

Answer:

a) 8/3 feet/second

b) 16/3 feet/second

Step-by-step explanation:

Let the length of the shadow be X

And initial distance from the streetlight be D

15/6 = (D + X)/X

15X = 6D + 6X

9X = 6D

X = ⅔D

dX/dD = ⅔

⅔ = dX/dt × dt/dD

dX/dt = ⅔ × 4 = 8/3

Distance from the tip is the hypotenuse 'H'

H² = 15² + (D + X)²

H² = 225 + (D + X)²

Differentiate both sides wrt time

2H × dH/dt = 2(D + X) × (dD/dt + dX/dt)

When D = 12, X = ⅔(12) = 8

D + X = 20

H² = 20² + 15²

H² = 625

H = 25

2(25) × dH/dt = 2(20) × (4 + 8/3)

dH/dt = 16/3