# Homework Help: Related Rates light shadow

1. Oct 28, 2007

### momogiri

Question
A street light is mounted at the top of a 15 foot tall pole. A man 6 ft tall walks away from the pole with a speed of 6 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?

Attempt
Well actually, I've drawn it out and stuff, but I was wondering if the tip of shadow would be moving in the same speed as the man..
I'm a bit confused at how to tackle this problem.. :/

2. Oct 28, 2007

### Gib Z

Draw up some triangles, and remember to use Pythagoras.

3. Oct 28, 2007

### momogiri

Yeah, I drew up my triangle within a triangle, it's attached, so..

The change in speed of the shadow is not the same as the speed of the man, right?
And 40 ft refers to the distance between the pole and the man right? Not between the shadow tip and the pole?

#### Attached Files:

• ###### tri.JPG
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4. Oct 28, 2007

### HallsofIvy

Yes, the problem says "when he is 40 ft from the pole".

But you don't need Pythagoras since the length of the hypotenuse is not relevant. Use "similar triangles".

Last edited by a moderator: Oct 28, 2007
5. Oct 28, 2007

### Gib Z

I was thinking of something else :( Sorry

6. Oct 28, 2007

### momogiri

Ok, so I tried to do it with similar triangles..

With my image that I attached earlier, I had the similar trigangles
y/6
and
(x+y)/15

(y being the distance of the shadow and the man
x being distance from man and pole)

So $$\frac{y}{6} = \frac{x+y}{15}$$
and I solved for x
$$15y = 6x+6y$$
$$9y = 6x$$
$$x = \frac{3}{2}y$$
and took derivatives of both sides wrt t
$$dx/dt = \frac{3}{2}dy/dt$$
And since the change in x is 6ft/sec
I just plugged in the numbers and found dy/dt to be 4
But it's wrong.

It's probably because I didn't incorporate 40ft as x somewhere, but I'm looking at what I did, and I just don't know what I should do :/

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