Word Problem - Relate Rates

**Gogo_Z** · October 30th 2008, 04:14 PM

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.

Help please and thank you

**skeeter** · October 30th 2008, 04:45 PM

let x = guy's distance from the post
s = shadow length

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.

**Gogo_Z** · October 30th 2008, 04:52 PM

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.

**skeeter** · October 30th 2008, 05:03 PM

you're right ... my mistake.

**Gogo_Z** · October 30th 2008, 05:10 PM

Why do you add them for b) ?

**skeeter** · October 30th 2008, 05:15 PM

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.

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

**Gogo_Z** · October 30th 2008, 05:17 PM

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

Thread: Word Problem - Relate Rates

LinkBack

Thread Tools

Display

Word Problem - Relate Rates

Similar Math Help Forum Discussions

Relate Rates - Vertical Gain

Rates Word Problem

rates of change word problem

Word Problem - Rates

Word Problem-Related Rates

Search Tags