A street light is on top of a 14 feet pole. A person who is 6ft tall walks away from the pole at a rate of?

Question

6ft per second. At what speed is the length of the person's shadow growing?

Answer 1

Let
x = distance of person from light pole
s = length of shadow

Given
14 = height of street light
6 = height person
dx/dt = 6 ft/sec

Find ds/dt.

By similar triangles we have:

s/6 = (x + s)/14
14s = 6x + 6s
8s = 6x
s = (6/8)x = (3/4)x
ds/dx = 3/4

ds/dt = (ds/dx)(dx/dt) = 6(3/4) = 18/4 = 9/2 ft/sec

Answer 2

Let a be the distance of the person from the pole and b the length of their shadow. By similar triangles, b / (a + b) = 6/14. Mash this equation around a bit and you get b = (3/4)a. So if a is increasing at 6ft / sec the shadow is growing at (3/4)*6 = 4ft 6in / sec.

Answer 3

ok, permit a be distance from tip of shadow to person, and b be distance from person to gentle. So, utilising similarity of triangles, 6/a=14/(a+b), or 14a=6a+6b, or 8a=6b, so a=6b/8. So, da/db=6/8. Now da/dt = (da/db)(db/dt)=(6/8)*5= 30/8. SO, its 30/8.