# Rate of how fast a shadow grows as you walk

• Jul 10th 2012, 06:11 AM
daigo
Rate of how fast a shadow grows as you walk
http://i.imgur.com/hWTgs.png

Quote:

There is a lamp post 15 feet tall casting a shadow 'B' ft. long of a person that is 6 feet tall standing 'A' ft. from the lamp post. If the person moves away from the lamp post at 5 feet per second, how fast does the shadow lengthen?
So here I thought it might be something to formulate a triangle, or actually two:

http://i.imgur.com/qFMoJ.png

So the problem is asking me to find the derivative of the shadow's length with respect to time: $\displaystyle \frac{dB}{dt}$

I think the triangles are similar:

$\displaystyle \frac{15}{A+B} = \frac{6}{B} \\ \\ \cdots \\ B = \frac{2}{3}A$

So we have B and I will attempt to take the derivative of this:

$\displaystyle \frac{dB}{dt} = [\frac{2}{3}A]'$

Constant/Chain rules:

$\displaystyle \frac{dB}{dt} = \frac{2}{3} \cdot [A'] \\ = \frac{2}{3} \cdot \frac{dA}{dt}$

Now how do I find the derivative of 'A'? It must have something to do with the movement velocity because it was given information and I haven't used it yet. Unless I am being tricked and it is extraneous/unnecessary information.
• Jul 10th 2012, 08:44 AM
ebaines
Re: Rate of how fast a shadow grows as you walk
You're almost there! The term dA/dt is the rate of change of length A per unit time, which is precisely the velocity of point A. You were told the velocity of A is 5 ft/s, so there you go:

dB/dt = 2/3 dA/dt = 2/3 (5 ft/s) = 10/3 ft/s.
• Jul 10th 2012, 08:49 AM
tom@ballooncalculus
Re: Rate of how fast a shadow grows as you walk
Quote:

Originally Posted by daigo
Now how do I find the derivative of 'A'? It must have something to do with the movement velocity because it was given information and I haven't used it yet.

Yes... what is the meaning of A' ...?

The derivative of A wrt t...

the rate of change of distance wrt time...

what we normally call...