# Math Help - Word Problem - Relate Rates

1. ## 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.

2. let x = guy's distance from the post

use proportions from the two similar triangles ...

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

simplify the algebra ...

$2s = x$

take the time derivative ...

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

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

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

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

3. Originally Posted by skeeter
take the time derivative ...

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

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

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

note that his distance from the post doesn't matter.
$2\frac{ds}{dt} = \frac{dx}{dt}$ if $\frac{dx}{dt}$ 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.

4. you're right ... my mistake.

5. Why do you add them for b) ?

6. the speed of the end of the shadow is the rate that the entire distance from the post to the end of the shadow, $(x + s)$, changes.

$\frac{d}{dt}(x+s) = \frac{dx}{dt} + \frac{ds}{dt}$

7. so 6ft/s should be final answer, thanks for the help man