Word Problem-Related Rates

• Dec 9th 2007, 12:14 PM
doctorgk
Word Problem-Related Rates
You are standing under a 20 foot lamppost. Your 5 foot friend is 10 feet away from you and sprints away at a speed of 20 feet per second.
(a) how fast is the tip of her shadow moving
(b) at that same instant, how fast is the length of her shadow changing.

Drawing if it helps

b
.-
. ----
. -----
. -------
. ---------
. -----------
. -------------
.....................-
a --------------c

ab is the light post

• Dec 9th 2007, 03:39 PM
TwistedOne151
Similar triangles
Let x be the distance from the lamppost to your friend, and y the distance from the lamppost to the tip of her shadow. Then y-x is the length of her shadow, and by similar triangles, $\frac{y}{20}=\frac{y-x}{5}$

B
|**-
|****-
|******-
|********|
|********|*-
|********|***-
|********|*****-
A----------C--------P
\----x-----/\-(y-x)-/

Solving for y:
$\frac{y}{20}=\frac{y-x}{5}$

$5y=20(y-x)$
$20x=15y$
$y=\tfrac{4}{3}x$

For (a), we want $\frac{dy}{dt}$. We use the chain rule $\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$, along with y in terms of x above and the fact that $\frac{dx}{dt}=20$ from the problem.

For (b), we remember that the shadow length is y-x, and so we want $\frac{d}{dt}(y-x)=\frac{dy}{dt}-\frac{dx}{dt}$, which should follow easily from the answer to (a).

--Kevin C.