A Related Rates Shadow Problem

[middleCmusic]

EDIT: I think I figured it out - sorry for taking up space. I posted my answer below.*

Homework Statement

A light is at the top of a 16-ft pole. A boy 5 ft tall walks away from the pole at a rate of 4 ft/sec.

a) At what rate is the tip of his shadow moving when he is 18 ft from the pole?
b) At what rate is the length of his shadow increasing?

Homework Equations

No relevant equations.

The Attempt at a Solution

So, I know the general idea of how to solve related rates problems and here's what I've gotten so far before I got stuck:

Let x be his distance from the pole and let y be the length of his shadow.
Then by similar triangles, [tex] \frac{16}{x+y}= \frac{5}{y}, [/tex] so we have
[tex]16y=5x+5y,[/tex]
[tex]11y=5x,[/tex]
[tex]y=\frac{5}{11}x.[/tex]

Then what I thought I was supposed to do was [tex]\frac{dy}{dt}=\frac{5}{11}\frac{dx}{dt},[/tex]
but from here I can't see how to apply the fact that he is 18 ft from the pole, since x doesn't appear in the related rates equation. I know that this problem should be easy and I'm probably overcomplicating it, but thanks in advance for your help!

* EDIT: I think I jumped the gun on posting about this one, sorry. Here's what I realized:

Since he is walking away at 4 ft/sec, [tex]\frac{dy}{dt} = \frac{5}{11}*4\frac{ft}{sec} = \frac{20}{11} \frac{ft}{sec}.[/tex] This answers part (b).

Then let [tex]z = x+y,[/tex]
and [tex]\frac{dz}{dt}=\frac{dx}{dt}+\frac{dy}{dt}=4+\frac{20}{11}=\frac{64}{11} \frac{ft}{sec}.[/tex]
This answers part (a).

Is this correct?

[Sorry for wasting space - I tried to delete the thread, but I don't know how.]

 

[dynamicsolo]

The surprise in shadow problems like this is that there IS no dependence on x : the tip of the shadow advances at a constant rate and the shadow lengthens at a constant rate if the person is walking at a constant speed. Because the person-shadow triangle is similar to the lightpole-shadow triangle, the proportion between the two triangles is constant, so all the lengths along the ground will increase (or decrease) uniformly.