#### 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?

#### Answer

The required rate at which the length of his shadow on the building decreases is 0.6 m/s. that is dydt=−0.6 m/s when he is 4 m form 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=1.6 m/s .

**Formula used:**

(1) Chain rule:dydx=dydu⋅dudx

(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=1.6 m/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=12−4=8 m

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 [Q dydx=dydu⋅dudx]=24ddt(x− 1)dxdt=24(−1)(x−1−1)dxdt

On further simplification the rate at which his shadow changes on the wall is obtained as shown below.

dydy=−24(1x2)dxdt=(−24x2) dxdt m/s

Substitutes 4 for x and 1.6 for dxdt in dydt.

dydt=(−24(82))(1.6)=(−2464)(1.6)=−38.464=−0.6 m/s

Therefore, the required rate at which the length of his shadow on the building decreases is 0.6 m/s. that is dydt=−0.6 m/s when he is 4 m form the building.