# Application of Calculus problem

• Feb 22nd 2010, 11:33 AM
kycon90
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???
• Feb 22nd 2010, 12:47 PM
Soroban
Hello, kycon90!

Did you make a sketch?

Quote:

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