Question

A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.7 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?

Answer 1

Answer:

0.6375 m/s

Explanation:

Let x be the distance of the man from the building

from the figure attached

initially the value of x=12

Given:

$\frac{dx}{dt}=-1.7m/s$

where the negative sign depicts that the distance of the man from the building is decreasing.

Now, Let The length of the shadow be = y

we have to calculate $\frac{dy}{dt}$ when x=4

from the similar triangles

we have,

$\frac{2}{12-x}=\frac{y}{12}$    

or

$y=\frac{24}{12-x}$

Differentiating with respect to time 't' we get

$\frac{dy}{dt}=-\frac{24}{12-x}^2\frac{-dx}{dt}$

or

$\frac{dy}{dt}=\frac{24}{12-x}^2\frac{dx}{dt}$

Now for x = 4, and $\frac{dx}{dt}=-1.7m/s$  we have,

$\frac{dy}{dt}=\frac{24}{12-4}^2\times (-1.7)$

or

$\frac{dy}{dt}=-0.6375m/s$

here, the negative sign depicts the decrease in length and in the question it is asked the decreasing rate  thus, the answer is 0.6375m/s