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 1.2 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:

(dy/dt) = - 0.45 m/s (the minus sign indicates that the height of the shadow decreases as the man moves towards the wall).

Step-by-step explanation:

An image of the problem is presented in the attached file to this solution. The image shows the exact moment when the 2m tall man is 4m away from the wall.

First of, when the man is 4m away from the wall, he is 8m away from the spotlight, that is, x = 8m

Using the concept of similar triangles, we can obtain the value of y at this point.

(x/2) = [(x+4)/y]

x = 8

(8/2) = (12/y)

y = 3m when x = 8m

From the image,

But generally, for x and y at any point,

(x/2) = (12/y)

xy = 24

Differentiating with respect to t

(d/dt)(xy) = (d/dt)12

y (dx/dt) + x (dy/dt) = 0

x (dy/dt) = - y (dx/dt)

At x = 8m, y = 3m, (dx/dt) = 1.2 m/s

8 (dy/dt) = - 3 (1.2)

(dy/dt) = - 3.6/8 = - 0.45 m/s

The minus sign indicates that the height of the shadow decreases as the man moves towards the wall.