Chapter 5.6, Problem 29E

To determine

To find: the rate at which the length of the person’s shadow is changing when he is $10\text{ ft}\text{.}$ from the base of the light.

Answer to Problem 29E

The rate at which the length of the person’s shadow is changing is $-3\text{ ft}\text{./sec}\text{.}$

Explanation of Solution

Given information :

Streetlight

  

From the given data in the problem, we draw a diagram. The diagram looks like two triangles. When any two triangles are similar, the ratio between any two matching sides, two sides opposite an angle of the same degree is equal to the ratio of any two other matching sides. Therefore, two matching sides is

  $\frac{16}{6}=\frac{x+z}{z}$

  $x=10\text{ ft}\text{.}$

  $\frac{dx}{dt}=-5\text{ ft}\text{./sec}\text{.}$ It is negative because it moved towards a streetlight.

All variables are differentiable functions of t .

Calculation:

We have to calculate the rate at which the length of the person’s shadow is changing when he is $10\text{ ft}\text{.}$ from the base of the light.

So,

  $\begin{array}{l}\frac{16}{6}=\frac{x+z}{z}\\ 16z=6\left(x+z\right)\\ 16z=6x+6z\\ 16z-6z=6x\\ 10z=6x\\ z=\frac{6}{10}x\left[\begin{array}{l}\text{Divide both numerator }\\ \text{and denominator by 2}\text{.}\end{array}\right]\\ z=\frac{3}{5}x\end{array}$

Differentiate with respect to t .

  $\frac{dz}{dt}=\frac{3}{5}\cdot \frac{dx}{dt}$

Putting the value of $\frac{dx}{dt}=-5$ in above equation,

  $\begin{array}{l}\frac{dz}{dt}=\frac{3}{5}\left(-5\right)\\ \frac{dz}{dt}=-3\text{ feet/sec}\text{.}\end{array}$

Hence, the rate at which the length of the person’s shadow is changing is $-3\text{ ft}\text{./sec}\text{.}$