# Thread: Related Rates/Differential problem

1. ## Related Rates/Differential problem

hi guys can you please help me with this problem:

A street light is mounted at the top of a 15ft-tall pole. A man 6ft tall walks away from the pole with a speed of 5ft/s along a straight path. How fast is the tip of his shadow moving when he is 40ft from the pole?

is this picture I made the right idea?

I don't quite see the relevance of the 6ft since the line reaches from the top anyway. I believe I use tan =opposite/adjacent

2. Originally Posted by RezMan
hi guys can you please help me with this problem:

A street light is mounted at the top of a 15ft-tall pole. A man 6ft tall walks away from the pole with a speed of 5ft/s along a straight path. How fast is the tip of his shadow moving when he is 40ft from the pole?

is this picture I made the right idea?

I don't quite see the relevance of the 6ft since the line reaches from the top anyway. I believe I use tan =opposite/adjacent
Your diagram is incorrect the man should be parallel to the pole. Then let x be the distance from the pole to the man and y the distance from the man to the tip of his shadow. Then use similar triangles to relate x and y.

3. but why exactly would I even need to find x and y? What exactly would be in charge of the rate the base(the 40ft line) grows? Is it not the top angle?

4. You still have an extra line in your picture. See the attached pdf

Man.pdf

From this we can use similar triangles to get

$\frac{x+y}{15}=\frac{y}{6} \iff 2x+2y=5y \iff y =\frac{2}{3}x$

Do you think you can finish from here?

5. $x=40-\frac{2}{3}x$?
and then we use that to tan=6/y?

6. Originally Posted by RezMan
$x=40-\frac{2}{3}x$?
and then we use that to tan=6/y?
First why are you trying to use the tangent function? Second you are making the classic error in related rate problems you are substituting known values too soon (before you have taken the time derivative.)

x is the distance from the pole to the man and y is the distance from the man to the tip of his shadow.

So you need to ask yourself what quantities do we need and how are the related?

The only known information is that

$\frac{dx}{dt}\bigg|_{x=40 \text{ ft}}=5 \frac{\text{ft}}{\text{s}}$

See if you can finish from here.

7. now I get it! I should be looking for the rate of the hypotenuse increase/or actually the whole base. so find that and multiply by rate. Thanks man!