Chapter 4, Problem 65RE

(a)

To determine

To find:

The height in order to maximize the intensity of illumination.

Answer to Problem 65RE

The light pole should be $28.284feet$ in order to maximize the intensity of illumination.

Explanation of Solution

Given:

The equation is

  $l=\frac{k\cos \theta }{d^2}$

Concept used:

If the quantities $x$ and $y$ are related by an equation

  $y\alpha \frac{1}{x}$

For some constant $k\ne 0$

That is y varies inversely proportional as $x$ or y is directly proportional to $x$

The constant $k$ is called constant proportionality .

Calculation:

The original equation given in order utilize variable that can be easily solved the given diagram

  $\begin{array}{l}{a}^2+b^2=c^2\\ h^2+{40}^2=d^2\\ d=\sqrt{h^2+1600}\\ \cos \theta =\frac{h}{d}\\ \cos \theta =\frac{h}{\sqrt{h^2+1600}}\end{array}$

Now after substituting $d$ and $\cos \theta$

  $\begin{array}{l}l=\frac{kh}{{\left(h^2+1600\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ l=kh{\left(h^2+1600\right)}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}$

Differentiating with respect to $x$

  $\frac{dl}{dx}=-3k{h}^2{\left(h^2+1600\right)}^{-\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+k{\left(h^2+1600\right)}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Using the zero product property of the equation

  $\begin{array}{l}0=-3k{h}^2{\left(h^2+1600\right)}^{-\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+k{\left(h^2+1600\right)}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ h=\pm \sqrt{800}\\ h=28.284\end{array}$

The light pole should be $28.284feet$ in order to maximize the intensity of illumination.

(b)

To determine

To find:

The rate of change of the intensity of the light equation.

Answer to Problem 65RE

The value for the rate of the intensity of the light at the point on the woman’s back is $-2.139e^{-6}k$

Explanation of Solution

Given:

The equation is

  $l=\frac{k\cos \theta }{d^2}$

Concept used:

If the quantities $x$ and $y$ are related by an equation

  $y\alpha \frac{1}{x}$

For some constant $k\ne 0$

That is y varies inversely proportional as $x$ or y is directly proportional to $x$

The constant $k$ is called constant proportionality .

Calculation:

The original equation given in order utilize variable that can be easily solved the given diagram

  $\begin{array}{l}{a}^2+b^2=c^2\\ h^2+{40}^2=d^2\\ d=\sqrt{h^2+1600}\\ \cos \theta =\frac{h}{d}\\ \cos \theta =\frac{h}{\sqrt{h^2+1600}}\end{array}$

Now after substituting $d$ and $\cos \theta$

  $\begin{array}{l}l=\frac{kh}{{\left(h^2+1600\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\\ l=kh{\left(h^2+1600\right)}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}$

Using the zero product property of the equation

  $\begin{array}{l}0=-3k{h}^2{\left(h^2+1600\right)}^{-\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+k{\left(h^2+1600\right)}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ h=\pm \sqrt{800}\\ h=28.284\end{array}$

The light pole should be $28.284feet$ in order to maximize the intensity of illumination.

The equation is

  $\begin{array}{l}j=h-4\\ j=24.284\end{array}$

  $l=kj{\left(j^2+r^2\right)}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Differentiating with respect to $t$

  $\begin{array}{l}\frac{dl}{dt}=-\frac{3}{2}kj{\left(j^2+1600\right)}^{-\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\mathrm{2.40.4}\\ \frac{dl}{dt}=-2.139e^{-6}k\end{array}$

The value for the rate of the intensity of the light at the point on the woman’s back is $-2.139e^{-6}k$ .