Question13. A street light is mounted at the top of a 15 -ft-tall pole. A man 6 ft tall walks away from the pole with a speed of $5 \mathrm{ft} / \mathrm{s}$ along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?

Studdy Solution
Substitute the known values and solve for $\frac{ds}{dt}$:
Given $\frac{dx}{dt} = 5 \, \text{ft/s}$, substitute this into the differentiated equation:
$\frac{ds}{dt} = \frac{2}{3} \times 5$
$\frac{ds}{dt} = \frac{10}{3} \, \text{ft/s}$
The rate at which the tip of the shadow is moving is the sum of the man's speed and the rate of change of the shadow's length:
$\frac{d(x+s)}{dt} = \frac{dx}{dt} + \frac{ds}{dt} = 5 + \frac{10}{3}$
$\frac{d(x+s)}{dt} = \frac{15}{3} + \frac{10}{3} = \frac{25}{3} \, \text{ft/s}$
The tip of the shadow is moving at a rate of:
$$\boxed{\frac{25}{3} \, \text{ft/s}}$$