How would i find the speed of her shadow?

Question

A street light is at the top of a 12 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 4 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 40 ft from the base of the pole?

Answer 1

construct 2 similar right triangles

angles the same, sides proportional

heights are 12 and 6

bases are x+40 and x

12/6= (40+x)/x=2

2x=40+x

x=40

bases are 80 and 40

if she walks 4 ft/sec, her shadow's tip moves twice as fast = 2x4 = 8 feet per second

hypotenuses are sqr(6400+144)= sqr6544 and sqr(1600+36)=sqr1636

h^2-a^2+b^2, a'=0, b'=4, b=40, a=6

take derivatives

2hh'=2aa'+2bb'

hh'=40(4)=160

h'=160/sqr1636=3.956 feet per second rate of change of the sunlight speed over her head

at time zero, the tip of the shadow is zero as there is no shadow when she's at the base of the light pole.

10 seconds later, at 4 ft/s she's 40 feet away and the tip of the shadow is 80 feet from the base of the light pole. the tip moved 80 feet in 10 seconds or at an average rate of 80/10 = 8 ft/sec

in 10 more seconds, she's now 80 feet away, and the tip of the shadow is further out, moving the same 8 ft/sec

base = x, height still her 6 feet

larger right triangle has height 12 feet, base = 80+x

12/6 = (80+x)/x = 2

2x = 80+x

x = 80

base is 80 tip of the shadow is 160 feet away

having moved 160 feet in 20 seconds = 160/20 = 8 ft/sec

from 10 to 20 seconds, the shadow's tip moved 80 feet. 80/(20-10) = 8 ft/sec

the shadow's tip moves twice as fast as she moves: 2x4 = 8 feet per second