A street light is at the top of a 12 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 5 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 45 ft from the base of the pole?

In t sec the lady would be at a distance of (speed)(time) = (5 ft/sec)(t sec) = 5t ft.

As the height of the pole is double the height of the lady, from similar triangles, the tip of the shadow would be at a distance x = 2 (5t) ft = 10t.

The distance equation is x = 10t ft.

So, the speed is x' = 10 ft/sec.

The speed is a constant, like the speed of the lady.

To solve this related rates problem, we can use similar triangles and the concept of related rates.

Let ( x ) be the distance between the woman and the base of the pole, and let ( y ) be the length of the woman's shadow.

From the given information, we have:

Height of the pole, ( h ) = 12 ft

Height of the woman, ( a ) = 6 ft

Speed of the woman, ( \frac{dx}{dt} ) = 5 ft/sec

We need to find ( \frac{dy}{dt} ) when ( x = 45 ) ft.

Using similar triangles, we can write:

[ \frac{y}{a} = \frac{x + y}{h} ]

Differentiating both sides with respect to time ( t ), we get:

[ \frac{1}{a} \frac{dy}{dt} = \frac{1}{h} \left( \frac{dx}{dt} + \frac{dy}{dt} \right) ]

Substituting the given values:

[ \frac{1}{6} \frac{dy}{dt} = \frac{1}{12} \left( 5 + \frac{dy}{dt} \right) ]

Multiplying through by 12 to clear the fractions:

[ 2 \frac{dy}{dt} = 5 + \frac{dy}{dt} ]

Combining like terms:

[ \frac{dy}{dt} = 5 \text{ ft/sec} ]

So, the tip of her shadow is moving at a speed of 5 ft/sec when she is 45 ft from the base of the pole.