# A calculus problem from high school textbook

A man 150 cm tall, walks away from a source of light situated at the top of a pole 5 m high at the rate of 0.7 m/s. Find the rate at which:

2. the tip of his shadow is moving

when he is 2 m away from pole.

My attempt: I found the rate at which his shadow is lengthening by using the property of similar triangles and by differentiating. However, for the 2nd part, I could not figure out the rate to be found. Can anyone help me out?

Denote the distance of the man from the pole by $$l$$ and the length of the shadow by $$s$$. You are given $$l'=0.7\ m/s$$ and were asked to find $$s'$$ which you already have. What you are looking for is the rate of change of $$l+s$$, i.e., you need to find $$(l+s)'=l'+s'$$. Answer should be ready from there.

Let x= Distance of person from the bottom of the Pole, and y=Distance of shadow tip from person's position
$$\frac{y}{x+y}=\frac{1.5}{5}$$, therefore $$7y=3x$$,
Now rate of shadow lengthening is $$\frac{7dy}{dt}=\frac{3dx}{dt}$$
$$\frac{7dy}{dt}=3 (0.7)$$
$$\frac{dy}{dt}=0.3 m/s$$
Rate at which shadow tip is moving is $$\frac{d(x+y}{dt}=\frac{dx}{dt}+\frac{dy}{dt}=0.7+0.3=1 m/s$$