# Homework Help: Related Rates (Derivatives)

1. Jul 10, 2011

### erix335

1. The problem statement, all variables and given/known data

A man 6 feet tall walks away from a streetlight that is 18 feet tall. If the length of his shadow is changing at a rate of 3 feet per second when he is 25 feet away from the base of the light, how fast is he walking away from the light at this moment?

2. Relevant equations

Trying to figure it out.

3. The attempt at a solution

Diagram drawn to no avail.

Last edited by a moderator: Jul 11, 2011
2. Jul 11, 2011

### lanedance

hi erix335

consider the triangle made by the shadow tip and the street light

and then the triangle made by the shadow tip and the man

you should be able to see they are similar triangles

use this to try and write the length of the shadow, s, as a function of the distance, d, the man is from the pole s(d)=?.

once you have that you differentiate to find the rate of change

3. Jul 11, 2011

### erix335

2s(ds/dt)= (25-3t)^2 + ?(18-6t)^2
2(30.8 hyp of triangle) = 2(25-3t)3+2*18-6t)6
61.4=(50-6t)3+(36-12t)6
=366-90t
-304.6/-90=t
t=304.6/90 or 3.38....which seems to be a right answer but the second part of my equation had no real thought behind it.

I seem to be stuck in the initial equation, and the 3.38 is the value of t, which means I would have to put it under 25, giving me a bit of a ridiculous answer.

4. Jul 11, 2011

### lanedance

sorry i can't understand what you have done, can you explain it?

5. Jul 11, 2011

### erix335

took the derivative of the speed and hypotenuse equation...or so I thought, then filled in for T and put it under the distance (25)

6. Jul 11, 2011

### lanedance

what was the hypotenuse equation?

i would start by finding s(x), as a function of the distance the man is from the pole
s = shadow length
x = distance of man from pole (sorry, had to change d=x for noatation in the derivative, otherwise it was too hard to read)

then using the chain rule
$$\frac{ds}{dt}= \frac{ds}{dx}\frac{dx}{dt}$$

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