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:
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?
On
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
$\frac{d(x+y}{dt}=\frac{dx}{dt}+\frac{dy}{dt}=0.7+0.3=1 m/s$
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.