Derivative Word Problem - Trig Needed?

[Char. Limit]

Homework Statement

Now this is a problem that my sister had on her Calculus homework, and I can't seem to figure it out. I believe a similar triangles argument is necessary, but I'm not sure, and trig always was my weak spot. The problem is as follows:

A person 2 meters tall walks directly away from a streetlight that is 8 meters above the ground. If the person is walking at a constant rate and the person's shadow is lengthening at the rate of 4/9 m/s, at what rate, in m/s, is the person walking?

Homework Equations

The Attempt at a Solution

Well, we drew a nice triangle like so:

We know that dx/dt = 4/9 and that dX/dt (the x-axis on the larger triangle) is constant. But I can't seem to complete the problem...

 

[Dick]

It isn't true that dX/dt is constant. It's getting longer too. What is true is that the ratio of the height and the length of the smaller and larger triangles are equal. Those are the similar triangles. Write that expression down.

 

[Char. Limit]

What, so like this?

\frac{8}{x+X} = \frac{2}{x}

And then use dx/dt=4/9 to get an expression for X in terms of t?

 

[Dick]

What, so like this?

\frac{8}{x+X} = \frac{2}{x}

And then use dx/dt=4/9 to get an expression for X in terms of t?

Solve that equation for X then differentiate both sides to get an expression for dX/dt in terms of dx/dt.

 

[Char. Limit]

Thanks! I can't believe I missed that.