Mind blowing related rates question

[bcddd214]

A man six feet tall walks at a rate of 5 feet per second away from the light that is 15 feet above the ground. When he is 10 feet from the base of the light at what rate is the tip of his shadow moving? At what rate is the length of his shadow changing?

rate1 = 5ft/s, rate2 = ?, y1 = 15ft, x2 = 10
It is basically a bigger triangle with a smaller one inside.
I am pretty sure this is the formula to use;
AC/(A ?C ? )=BC/(B ?C ? )

but not sure where to go from here.

 

[galactus]

Use similar triangles. They are handy in many related rates problems.

Actually, this is a classic, cliche' related rates problem.

Let x=distance from lamppost to man

Let L=length of shadow.

Thus, the distance between the tip of the shadow and the post is x+L

By similar triangles:

\(\displaystyle \frac{6}{15}=\frac{L}{x+L}\)

\(\displaystyle L=\frac{2}{3}x\)

Now, differentiate to find dL/dt. You are given dx/dt.

For the second part of the question, use x+L. Differentiate. You can use dx/dt and dL/dt from the previous part.