A street light is at the top of a 15 foot tall pole. A 6 foot tall woman 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 50 feet from the base of the pole?

1 Answer
May 10, 2018

d'(t_0)=20/3=6,bar6 ft/s

Explanation:

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Using Thales Proportionality theorem for the triangles AhatOB, AhatZH

The triangles are similar because they have hatO=90°, hatZ=90° and BhatAO in common.

We have (AZ)/(AO)=(HZ)/(OB) <=>

ω/(ω+x)=6/15 <=>

15ω=6(ω+x) <=>

15ω=6ω+6x <=>

9ω=6x <=>

3ω=2x <=>

ω=(2x)/3

Let OA=d then

d=ω+x=x+(2x)/3=(5x)/3

  • d(t)=(5x(t))/3

  • d'(t)=(5x'(t))/3

For t=t_0 , x'(t_0)=4 ft/s

Therefore, d'(t_0)=(5x'(t_0))/3 <=>

d'(t_0)=20/3=6,bar6 ft/s