   Chapter 3.9, Problem 18E

Chapter
Section
Textbook Problem

A spotlight on the ground shines on a wall 12m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.6 m/s. how fast is the length of his shadow on the building decreasing when he is 4 m from the building?

To determine

To find: How fast is the length of man’s shadow on the building decreasing when he is 4 meter away from the building?

Explanation

Given:

A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spot light towards the building at a speed of 1.6 m/s.

Let t be time (in sec) and x be the horizontal distance between man and spotlight (in m), then we are given that dxdt=1.6m/s .

Formula used:

(1) Chain rule:dydx=dydududx

(2) Properties of similar triangles.

Calculation:

Let A be the position of the spot light on the ground and B be the position of the wall, and C be the position of the man on the ground.

Let x be the distance between the man and spotlight and y is the height of his shadow on the wall as shown in the figure-1 given below.

Since the distance between the man and spotlight change with time therefore the height of his shadow on the wall also changes with the time, the variables  x and y is function of the time variable t.

It is given that the man walks from spotlight towards the building at a speed of dxdt=1.6m/s.

Obtain dydx  when he is 4 m from the building.

Since 4 m is the distance from man to building, the distance from the spotlight to man is.

x=124=8m

Apply the properties of the similar triangles in the above Figure 1.

y2=12xy=24x

Differentiate with respect to the time variable t.

dydt=ddt(24x)=24ddt(1x)dxdt

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