# Word Problem - Relate Rates

• Oct 30th 2008, 04:14 PM
Gogo_Z
Word Problem - Relate Rates
A 6ft tall guy walks directly away from a street lamp that is 18 ft above the ground at a rate of 4ft/s. Determine the following at the instant he is 24 ft from the base of the lamp.
a) the rate of increase in the length of his shadow.
b) the speed of the end of her shadow

I get 2ft/s for a) but i don't know if that's right, and for b) i don't know what to do.

• Oct 30th 2008, 04:45 PM
skeeter
let x = guy's distance from the post

use proportions from the two similar triangles ...

$\displaystyle \frac{18}{x+s} = \frac{6}{s}$

simplify the algebra ...

$\displaystyle 2s = x$

take the time derivative ...

$\displaystyle 2\frac{ds}{dt} = \frac{dx}{dt}$

since $\displaystyle \frac{dx}{dt} = 4$ , $\displaystyle \frac{ds}{dt} = 8$

speed of the end of the shadow is $\displaystyle \frac{dx}{dt} + \frac{ds}{dt}$

note that his distance from the post doesn't matter.
• Oct 30th 2008, 04:52 PM
Gogo_Z
Quote:

Originally Posted by skeeter
take the time derivative ...

$\displaystyle 2\frac{ds}{dt} = \frac{dx}{dt}$

since $\displaystyle \frac{dx}{dt} = 4$ , $\displaystyle \frac{ds}{dt} = 8$

speed of the end of the shadow is $\displaystyle \frac{dx}{dt} + \frac{ds}{dt}$

note that his distance from the post doesn't matter.

$\displaystyle 2\frac{ds}{dt} = \frac{dx}{dt}$ if $\displaystyle \frac{dx}{dt}$ is 4ft/s then shouldn't it be divided by 2 from the left side to get 2ft/s, I just don't understand where the 8 comes from.
• Oct 30th 2008, 05:03 PM
skeeter
you're right ... my mistake.
• Oct 30th 2008, 05:10 PM
Gogo_Z
Why do you add them for b) ?
• Oct 30th 2008, 05:15 PM
skeeter
the speed of the end of the shadow is the rate that the entire distance from the post to the end of the shadow, $\displaystyle (x + s)$, changes.

$\displaystyle \frac{d}{dt}(x+s) = \frac{dx}{dt} + \frac{ds}{dt}$
• Oct 30th 2008, 05:17 PM
Gogo_Z
so 6ft/s should be final answer, thanks for the help man