Question

A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks along the x-axis from the spotlight toward the building at a speed of 1.8 m/s, which is taken as the given dx/dt, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?

Answer 1

Answer:

0.675 m/s

Step-by-step explanation:

Let height of shadow= y,CD=x

Height of man=2 m

Speed of man= $\frac{dx}{dt}=1. 8 m/s$

$\triangle ABD\sim\triangle ECD$

Therefore, $\frac{AB}{EC}=\frac{BD}{CD}$

$\frac{y}{2}=\frac{12}{x}$

$xy=24$

Differentiate w.r.t t

$x\frac{dy}{dt}+y\frac{dx}{dt}=0$

$x\frac{dy}{dt}=-y\frac{dx}{dt}$

$\frac{dy}{dt}=-\frac{y}{x}\frac{dx}{dt}$

When the man is 4 m from  the building

Then, we have x=12-4=8 m

$\frac{dx}{dt}=1.8 m/s$

Substitute the values in above equation then, we get

$8y=24$

$y=\frac{24}{8}=3$

Substitute the values then we get

$\frac{dy}{dt}=-\frac{3}{8}\times 1.8=-0.675 m/s$

Hence, the length of his shadow on the building decreasing at the rate 0.675 m/s.