Chapter 4.1, Problem 14E

To determine

To find: the rate of shadow moving when distance is 40 ft.

Answer to Problem 14E

The tip of the shadow is moving with the speed of $\frac{25}{3}$ ft/s.

Explanation of Solution

Given information:

Given that the street light is mounted at the top of 15 ft tall pole and a man 6ft tall walks away from pole @ 5ft/s straight.

Calculation:

Let $x$ be the distance from yhe light to the men, and let $l$ be the length of his shadow. Furthur, let $s$ is the distance from pole to the light of the shadow.

The height of the pole is $15$ feet and the height of the man is 6 feet.

Speed is $\frac{dx}{dt}=5$ ft/s and the moment of interest is $x=40$ ft.

That is $\frac{ds}{dt}$

To get an equation involving rates of change, differentiate both sides of the equations from part $\left(d\right)$ with respect to $t$ . When doing this treat the changing quantities as functions of time, and use the chain rule. From equation $\left(1\right)$

  $\begin{array}{l}\frac{d}{dt}\left(\frac{x+1}{15}\right)=\frac{d}{dt}\left(\frac{1}{6}\right)\\ \frac{1}{15}\frac{dx}{dt}+\frac{1}{15}\frac{dl}{dt}=\frac{1}{6}\frac{dl}{dt}\\ \frac{1}{15}\frac{dx}{dt}=\frac{1}{10}\frac{dl}{dt}\\ \frac{dl}{dt}=\frac{2}{3}\frac{dx}{dt}\end{array}$

From equation $\left(2\right)$

  $\begin{array}{c}\frac{d}{dt}\left(s\right)=\frac{d}{dt}\left(x+1\right)\\ \frac{ds}{dt}=\frac{dx}{dt}+\frac{dl}{dt}\\ =\frac{dx}{dt}+\frac{2}{3}\frac{dx}{dt}\\ \frac{ds}{dt}=\frac{5}{3}\frac{dx}{dt}\end{array}$

Now substitute $\frac{dx}{dt}=5$ ft/s in $\frac{ds}{dt}=\frac{5}{3}\frac{dx}{dt}$

  $\frac{ds}{dt}=\frac{5}{3}\left(5\right)=\frac{25}{3}ft/s$

Therefore, the tip of the shadow is moving with the speed of $\frac{25}{3}$ ft/s