# Related rates problem with a twist

1. Apr 29, 2007

### sebasalekhine7

I have this problem, and I have attepmted a classical approach without much success.

A man 5 ft tall runs at a rate of 8ft/sec towards a source of light that arises vertically at a point A. The height of the light source H, is given by the formula h(t)=t^3 +1, in feet, where time t is measured in seconds. At what rate is the length of the man's shadow decreasing at time=10seconds?

I have tried establishing the geometrical relationships between the height h(t) and the distance between the point A and the tip of the man's shadow.
say, if we call "s" the length of the shadow, and 'x' the distance between the man and point A, we have that h/5=x/s but then I do not have any other information about the problem. What's next??

2. Apr 30, 2007

### HallsofIvy

Take another look at your picture! At any given instant, we have two triangles: one with a vertical side of length h (the light), the other with a vertical side of length 5 (the man). The horizontal sides of those triangles are s+ x and s respectively. (Notice that there is NO triangle with side length x: there is no triangle with vertex at the man.)

Using "similar triangles", you have s/5= (x+s)/h. Not to convert that from a "static" equation to one involving rates of change, differentiate both sides with respect to t. You are given that dx/dt= -8. Knowing that h(t)= t3+ 1, you can calculate dh/dt, as a function of t, and so find ds/dt as a function of t.