Chapter 4.6, Problem 30E

To determine

To find: How fast is the ball's shadow moving along the ground $1/2\text{sec}$ later.

Answer to Problem 30E

The answer: Moving $1500\text{ft}/\text{sec}$ towards the pole.

Explanation of Solution

Given information:

Adaptive Shadow A $50\text{ft}$ high pole is illuminated at the summit. As seen in the diagram below, a ball is dropped from the same height from a place $30ft$ away from the light.

Calculation:

  $\begin{array}{c}s=16t^2\\ =16{\left(\frac{1}{2}\right)}^2\\ =4\text{ft}\\ \frac{x-30}{x}=\frac{50-s}{50}\\ \Rightarrow \frac{x-30}{x}=\frac{46}{50}\\ \Rightarrow 50x-1500=46x\\ \Rightarrow 4x=1500\\ \Rightarrow x=375\text{ft}\end{array}$

Put

  $s=16t^2$ and simplify first, then implicitly differentiate the similar triangle equation.

  $\begin{array}{c}\frac{x-30}{x}=\frac{50-s}{50}\\ \Rightarrow 1-\frac{30}{x}=1-\frac{8}{25}{t}^2\\ \Rightarrow 1-30x^{-1}=1-\frac{8}{25}{t}^2\\ \Rightarrow 0-(-1)30x^{-2}dx=0-\frac{8}{25}(2)tdt\\ \Rightarrow \frac{dx}{dt}=-\frac{16}{25}t\cdot \frac{x^2}{30}\\ \Rightarrow \frac{dx}{dt}=-\frac{8}{375}t{x}^2\end{array}$

Put,

  $\begin{array}{c}t=0.5\\ x=375\end{array}$

Then,

  $\begin{array}{c}$ \frac{dx}{dt}=-\frac{8}{375}(0.5){(375)}^2$=-1500\end{array}$

Therefore, the required moving $1500\text{ft}/\text{sec}$ towards the pole is the ball's shadow moving along the ground $1/2\text{sec}$ later.