# Thread: Related Rate models question

1. ## Related Rate models question

I'm getting the biggest headache trying to solve this question.. and its probably the easiest question ever but I just can't get the right answer and its driving me nuts. And i have a test on monday on this stuff.. so any help would be appreciated beyond belief.

A man 2 m tall walks away from a lamppost whose light is 5 m above the ground. If he walks at a speed of 1.5 m/s, at what rate is his shadow increasing when he is 10 m from the lamppost.

The answer is supposed to be 1 m/s, but I keep on getting 2.5 m/s, and I have no clue what I'm doing wrong.

Thanks for the help!!!

2. Let x=shadow length and y=distance bewtween man and lamppost.

Similar triangles:

$\frac{x}{2}=\frac{x+y}{5}$

$x=\frac{2y}{3}$

Differentiate:

$\frac{dx}{dt}=\frac{2}{3}\frac{dy}{dt}$

We know that dy/dt=3/2

We have:

$\frac{dx}{dt}=\frac{2}{3}\cdot\frac{3}{2}=1$

3. Hello, jmailloux!

A man 2 m tall walks away from a lamppost whose light is 5 m above the ground.
If he walks at a speed of 1.5 m/s, at what rate is his shadow increasing
when he is 10 m from the lamppost.

The answer is supposed to be 1 m/s, but I keep on getting 2.5 m/s,
and I have no clue what I'm doing wrong. . I know!

You made a very common error.
I know ... I still do it from time to time.
Code:
      *
|   *
|       *
|           *
5|           |   *
|          2|       *
|           |           *
* - - - - - + - - - - - - - *
:     x     :       s       :

I bet you let $s$ equal the whole bottom of the diagram.
That would give us the rate at which the tip of his shadow is moving.
. . But they want the rate at which the length of his shadow is changing.

From the similar right triangles, we have: . $\frac{s}{2}\:=\:\frac{x+s}{5}$

Then: . $5s\:=\:2x + 2s\quad\Rightarrow\quad 3s \,= \,2x\quad\Rightarrow\quad s \,= \,\frac{2}{3}x$

Differentiate with respect to time: . $\frac{ds}{dt}\:=\:\frac{2}{3}\cdot\frac{dx}{dt}$

The man's speed is 1.5 m/sec: . $\frac{dx}{dt} = 1.5$

Therefore: . $\frac{ds}{dt}\:=\:\frac{2}{3}(1.5) \:=\:1\text{ m/s}$

4. thank you soo much

it makes a lot more sense after you explained it, and i did make s equal to the whole bottom of the diagram.

anyways.. thanks for the help!