# Related rates question using similar triangles

So I have this related rates problem here:

A man $$6$$ feet tall is standing still in a gymnasium which has a ceiling that is $$30$$ feet high. Ten feet in front of him a bright light starts to fall from the ceiling and the distance it falls in $$t$$ seconds is $$16t^2$$ feet. After one second, what is the rate of change in the length of his shadow created by the falling light?

Can someone help me set this up? It's similar to the shadow questions I used to do but the guy is still this time, he isn't walking. I'm stuck on how to represent the falling light.

I drew a diagram trying to represent the problem. However, I feel like I have too many variables to relate and I can't figure how to write them in terms of another.

• It's the distance from the man to the base of the ceiling. EDIT, I think my distance is wrong. Commented Dec 2, 2016 at 3:52
• See my edit. Now I guess I use similar triangles. y/[10 +x]=6/x I guess Commented Dec 2, 2016 at 3:56
• That's the general idea, except the height of the lamp is $30 - 16 t^2$.
– dxiv
Commented Dec 2, 2016 at 3:58
• Ohh right. That makes a lot of sense. I would then sub for y, differentiate and solve for dx/dt I guess. Commented Dec 2, 2016 at 4:01

Your idea about similar triangles was exactly right. Having an accurate diagram makes things a lot easier! Solving for $$x$$, we have (assuming $$y=30-16t^2$$): \begin{align} \frac{y}{10+x}&=\frac{6}{x} \\[0.25ex] yx&=60+6x \\[0.5ex] x(y-6)&=60 \\ x&=\frac{60}{y-6} \\ x&=\frac{60}{24-16t^2} \\ x&=\frac{15}{6-4t^2} \end{align}
All we need to do now is find $$\frac{dx}{dt}$$ and evaluate it at $$t=1$$, which isn't terribly difficult. \begin{align} \frac{dx}{dt}&=\frac{-15(-8t)}{(6-4t^2)^2} \\ \frac{dx}{dt}&=\frac{120t}{2^2(3-2t^2)^2} \\ \frac{dx}{dt}&=\frac{30t}{(3-2t^2)^2} \\ \left.\frac{dx}{dt}\right|_{t=1}&=\frac{30}{(3-2)^2} \\[0.75ex] &=30\:\mathrm{ft/s} \end{align}