# Thread: Application of Calculus problem

1. ## Application of Calculus problem

A street light is at the top of a 14 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 5 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 30 ft from the base of the pole?

This is a problem where they give you what I believe is a dv/dt of 5 ft/s, but I don't know what I have to do to find the speed of the tip of her shadow. I tried drawing triangles, but I don't know exactly what I need to do, and I can't think of any equation that would help me find this, y = mx+b since it's a straight line???

2. Hello, kycon90!

Did you make a sketch?

A streetlight is at the top of a 14 ft tall pole.
A woman 6 ft tall walks away from the pole with a speed of 5 ft/sec.
How fast is the tip of her shadow moving when she is 30 ft from the base of the pole?
Code:
    A o
|   * - C
|       o
|       |   *
14|       |       *
|      6|           *
|       |               *
|       |                   *
B o-------o-----------------------o E
: - x - D - - -  s-x  - - - - - :
: - - - - - - s - - - - - - - - :

The lightpole is: $\displaystyle AB = 14\text{ ft}$

The woman is: $\displaystyle CD = 6\text{ ft}$
Her distance from the pole is: $\displaystyle BD = x\text{ ft},\;\;\frac{dx}{dt} = 5 \text{ ft/sec}$

The tip of her shadow $\displaystyle (E)$ is $\displaystyle s$ ft from the pole: .$\displaystyle s \,=\,BE$
. . Hence: .$\displaystyle DE \,=\,s-x$

Since $\displaystyle \Delta ABE \sim \Delta CDE$, we have: .$\displaystyle \frac{14}{s} \:=\:\frac{6}{s-x} \quad\Rightarrow\quad s \:=\:\frac{7}{4}x$

Differentiate with respect to time: .$\displaystyle \frac{ds}{dt} \:=\:\frac{7}{4}\,\frac{dx}{dt}$

Therefore: .$\displaystyle \frac{ds}{dt} \:=\:\frac{7}{4}(5) \:=\:8.75\text{ ft/sec}$