#### To determine

**To find:** The quantities in the given problem.

#### Answer

The quantities in the given problem is dxdt=5 ft/s.

#### Explanation

**Given:**

Given: a man 6 ft tall walks away from street light mounted on a 15-ft-tall pole at a rate of 5 ft/s. If we let t be time (in s) and x be the distance from the pole to the man ( in ft ), then we are given that dxdt=5 ft/s.

#### To determine

**To find:** The unknown.

#### Explanation

**Formula used:**

Properties of similar triangles

Unknown: the rate at which the tip of his shadow is moving when he is 40 ft from the pole. If we let y be the distance from the man to the tip of his shadow (in ft. ), then we want to find ddt(x+y) when x=40 ft .

#### To determine

**To draw:** The picture of the position of the man and his shadow from the pole.

#### Explanation

Draw the picture as,

From Figure 1, it is observed that position of the man and his shadow from the pole.

**Notation**:

- x - Distance to the man from the pole.
- y - Distance to the man from the tip of his shadow.
- (x+y) - Distance to the tip of his shadow from pole.

#### To determine

**To write:** The equation that relates the quantities.

#### Answer

The Equation is y =23x.

#### Explanation

Equation which relates the quantities: by using the properties of similar triangles in the triangle △CAB and △CDE

x+yy=15615y =6x+6y9y =6xy =23x

#### To determine

**To find:** How fast is the tip of his shadow moving when he is 40 ft from the pole.

#### Answer

The required rate at which the tip of his shadow moves is ddt(x+y)=253 ft/s.

#### Explanation

The tip of his shadow moving at a rate of

ddt(x+y)=ddt(x+23x) [Q y=23x]=ddt(53x)=53⋅dxdt

Substitutes 5 for dxdt in ddt(x+y)

ddt(x+y)=53(5)=253 ft/s

Therefore, the required rate at which the tip of his shadow moves is ddt(x+y)=253 ft/s.