calculus - related rates

Question

A man of height 1.9 meters walks away from a 5-meter lamppost at a speed of 1.7 m/s. How fast is the tip of his shadow moving? Use 3 decimal places and correct units.

Answer 1

This is a very interesting problem that incorporates calculus with geometry! First, let's let m'(t) be the velocity of the man:

m'(t) = 1.7

We can integrate that to get the function of the position of the man:

m(t) = 1.7t

Now, using similar triangles, we can write the function for the position of the shadow (s(t)) with respect to m(t):

s(t) / 5 = (s(t) - m(t)) / 1.9

Now, we just need to simply it:

1.9 * s(t) = 5 * s(t) - 5 * m(t)

3.1 * s(t) = 8.5t

s(t) = 2.74193548t

And now we differentiate to get the velocity of the shadow (s'(t)):

s'(t) = 2.74193548

So the shadow is moving at 2.742 m/s.

This is a great example of how math builds on itself and you still need to remember things that you may have learned from years earlier.