Chapter 5.6, Problem 33E

To determine

To find: the rate at which the shadow’s length is decreasing.

Answer to Problem 33E

The rate at which the shadow’s length is decreasing is $-7.4844\text{ in}\text{./min}\text{.}$

Explanation of Solution

Given information :

From the given diagram in the question, we can see that

  $\begin{array}{l}\frac{d\theta }{dt}={0.27}^{\circ }\text{/min}\text{.}\\ \frac{d\theta }{dt}={0.27}^{\circ }\times \frac{\pi }{180}\text{ radians/min}\text{. }\left[{\text{Since, 1}}^{\circ }=\frac{\pi }{180}\text{ radians}\right]\\ \frac{d\theta }{dt}=0.0015\times 3.14\text{ radians/min}\text{. }\left[\text{Since, }\pi =3.14\right]\\ \frac{d\theta }{dt}=0.00471\text{ radians/min}\text{.}\end{array}$

All variables are differentiable functions of t .

Calculation:

We have to calculate the rate at which the shadow’s length is decreasing.

Let x represent the length of the shadow.

At the moment, $x=60\text{ ft}\text{.}$

Use trigonometry to get an equation,

  $\begin{array}{l}\text{tan}\theta =\frac{\text{perpendicular}}{\text{base}}=\frac{80}{60}\\ \theta ={\text{tan}}^{-1}\left(\frac{80}{60}\right)\\ \theta ={\text{tan}}^{-1}\left(1.33\right)\\ \theta \approx {53.13}^{\circ }\end{array}$

Again, use trigonometry,

  $\begin{array}{l}\text{tan}\theta =\frac{80}{x}\\ x=\frac{80}{\text{tan}\theta }\\ x=80\text{cot}\theta \left[\text{Since, }\frac{1}{\text{tan}\theta }=\text{cot}\theta \right]\end{array}$

Differentiate with respect to t ,

  $\begin{array}{l}\frac{dx}{dt}=80\left[-{\text{cosec}}^2{53.13}^{\circ }\right]\left(0.00471\right)\\ \frac{dx}{dt}=80\left[\frac{1}{{\sin }^2{53.13}^{\circ }}\right]\left(0.00471\right)\left[\text{Since, cosec}\theta =\frac{1}{\sin \theta }\right]\\ \frac{dx}{dt}=\frac{-0.3768}{{\left(0.79\right)}^2}\left[\text{Since, }\sin 53.13=0.79\right]\\ \frac{dx}{dt}=\frac{-0.3768}{0.6241}\\ \frac{dx}{dt}=-0.6237\text{ ft}\text{./min}\text{.}\end{array}$

We have to find the answer in inches.

Therefore,

  $\begin{array}{l}\frac{dx}{dt}=-0.6237\times 12\left[\text{Since, 1ft}\text{.}=\text{12 inches}\right]\\ \frac{dx}{dt}=-7.4844\text{ in}\text{./min}\text{.}\end{array}$

Hence, the rate at which the shadow’s length is decreasing is $-7.4844\text{ in}\text{./min}\text{.}$