# Thread: Another related rate model

1. ## Another related rate model

I posted a question almost the same as this yesterday.. and I understand how to do them. I followed the way the people who answered my post solved the problem, but it's not working for this question.

A 1.7 m tall man walks away from a 9-m high streetlight at 0.8 m/s. How fast is the end of his shadow moving when he is 20 m from the lamppost.

My textbook says the answer is 0.986 m/s, but I keep on getting 0.186 m/s.

PLEASE HELP, i have a test tommorrow, and I know a question like this is going to be on it

2. That's because you found the rate at which the shadow length is changing.

The tip of the shadow is x+y feet from the light. Therefore, the rate it is

changing is dx/dt+dy/dt. You have 0.8 + 0.186 = 0.986

3. Hello, jmailloux!

You did it again . . . you made $s$ the wrong part.

A 1.7-m tall man walks away from a 9-m high streetlight at 0.8 m/s.
How fast is the end of his shadow moving when he is 20 m from the lamppost?
Code:
      *
|     *
|           *
9 |           |     *
|        1.7|           *
|     x     |       s-x       *
----*-----------+-----------------------*----
: - - - - - - - - s - - - - - - - - :

This time, we make $s$ be the entire distance,
. . from the lampost to the end of his shadow.

From the similar right triangles, we have: . $\frac{s - x}{1.7} \:=\:\frac{s}{9}$

Then: . $9s - 9x\:=\:1.7s\quad\Rightarrow\quad s\:=\:\frac{9}{7.3}x$

Differentiate with respect to time: . $\frac{ds}{dt}\:=\:\frac{9}{7.3}\left(\frac{dx}{dt} \right)$

Since $\frac{dx}{dt} = 0.8\text{ m/s}:\;\frac{dx}{dt} \:=\:\frac{9}{7.3}(0.8) \:\approx\:0.986\text{ m/s}$