Chapter 4.6, Problem 62E

a.

To determine

To calculate: Two light sources of identical strength are placed $10\text{m}$ apart an object is to be placed at a point $P$ on a line $l$ parallel to the line joining the light sources and at a distance $d$ meters from it (see the figure). We want to locate $P$ on $l$ so that the intensity of illumination is minimized. We need to use the fact that the intensity of illumination for a single source is directly proportional to the strength of the source and inversely proportional to the source of the distance from the source.

Find an expression for the intensity $I\left(x\right)$ at the point $P$

Answer to Problem 62E

  $T_1=\frac{D}{c_1},T_2=\frac{2h\sec \theta }{c_1}+\frac{D-2h\tan \theta }{c_2},T_3=\frac{\sqrt{4h^2+D^2}}{c_1}$

Explanation of Solution

Given information:

Two light sources of identical strength are placed $10\text{m}$ apart an object is to be placed at a point $P$ on a line $l$ parallel to the line joining the light sources and at a distance $d$ meters from it (see the figure). We want to locate $P$ on $l$ so that the intensity of illumination is minimized. We need to use the fact that the intensity of illumination for a single source is directly proportional to the strength of the source and inversely proportional to the source of the distance from the source.

Formula used:

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

Trigonometric ratio:

  $\csc \theta =\frac{h}{p}\text{ and }\cot \theta =\frac{b}{p}$

Speed: the distance covered per unit time is called speed. Speed is directly proportional to distance and inversely proportional to time.

Speed =distance/time, Time =distance/speed, Distance =speed x time

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Calculation:

As per the given problem

Draw the diagram of two light sources of identical strength are placed $10\text{m}$ apart an object is to be placed at a point $P$ on a line $l$ parallel to the line joining the light sources and at a distance $d$ meters from it (see the figure). We want to locate $P$ on $l$ so that the intensity of illumination is minimized. We need to use the fact that the intensity of illumination for a single source is directly proportional to the strength of the source and inversely proportional to the source of the distance from the source.

  

  $I\left(x\right)\alpha \frac{\text{strangth of source}}{{\left(\text{distance from source}\right)}^2}$

Adding the intensities from the left and right light bulbs to get,

  $\begin{array}{l}I\left(x\right)=\frac{k}{x^2+d^2}+\frac{k}{{\left(10-x\right)}^2+d^2}\\ =\frac{k}{x^2+d^2}+\frac{k}{x^2+d^2-20x+100}\end{array}$

Distance the first sound travels is $D$ and at speed $c_1$

Recall that,

Speed: the distance covered per unit time is called speed. Speed is directly proportional to distance and inversely proportional to time.

Speed =distance/time, Time =distance/speed, Distance =speed x time

  $\begin{array}{l}{c}_1=\frac{D}{T_1}\\ \Rightarrow T_1=\frac{D}{c_1}\end{array}$

Now, the sum of the distance the first sound travels is $2PR$ at speed $c_1$ because it has not reached the second layer of the stone and $RS$ at speed $c_2$ because it is now traveling on the second layer of the stone

Recall that,

Speed: the distance covered per unit time is called speed. Speed is directly proportional to distance and inversely proportional to time.

Speed =distance/time, Time =distance/speed, Distance =speed x time

  $T_2=\frac{2PR}{c_1}+\frac{SQ}{c_2}$

Recall that,

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

And

Trigonometric ratio:

  $\tan \theta =\frac{p}{b}\text{ and }\cot \theta =\frac{b}{p}$

  $\begin{array}{l}PR=\sqrt{h^2+h^2{\tan }^2\theta }\\ =\sqrt{h^2\left(1+{\tan }^2\theta \right)}\\ =\sqrt{h^2{\sec }^2\theta }\\ =h\sec \theta \end{array}$

The maximum horizontal distance is $D$

So,

  $\begin{array}{l}RS=D-P{R}_x-S{Q}_x\\ \text{}\text{}\text{}\text{}=D-2h\tan \theta \end{array}$

Now substitute the values to get,

  $T_2=\frac{2h\sec \theta }{c_1}+\frac{D-2h\tan \theta }{c_2}$

The third sound travels two times the distance of $PO$ at speed $c_1$

Recall that,

Speed: the distance covered per unit time is called speed. Speed is directly proportional to distance and inversely proportional to time.

Speed =distance/time, Time =distance/speed, Distance =speed x time

  $T_3=\frac{2PO}{c_1}$

Recall that,

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

And

Trigonometric ratio:

  $\tan \theta =\frac{p}{b}\text{ and }\cot \theta =\frac{b}{p}$

  $\begin{array}{l}PO=\sqrt{h^2+{\left(\frac{D}{2}\right)}^2}\\ =\sqrt{h^2+{\frac{D}{4}}^4}\end{array}$

Substitute the value to get,

  $\begin{array}{l}{T}_3=\frac{2\sqrt{h^2+{\frac{D}{4}}^4}}{c_1}\\ T_3{}^2=\frac{4h^2+D^2}{c_1{}^2}\\ T_3=\frac{\sqrt{4h^2+D^2}}{c_1}\end{array}$

Conclusion:

Thus $T_1=\frac{D}{c_1},T_2=\frac{2h\sec \theta }{c_1}+\frac{D-2h\tan \theta }{c_2},T_3=\frac{\sqrt{4h^2+D^2}}{c_1}$

b.

To determine

To calculate: Two light sources of identical strength are placed $10\text{m}$ apart an object is to be placed at a point $P$ on a line $l$ parallel to the line joining the light sources and at a distance $d$ meters from it (see the figure). We want to locate $P$ on $l$ so that the intensity of illumination is minimized. We need to use the fact that the intensity of illumination for a single source is directly proportional to the strength of the source and inversely proportional to the source of the distance from the source

If $d=5\text{m}$ , use graphs of $I\left(x\right)$ and $I^{\prime }\left(x\right)$ to show that the intensity is minimized when $x=5\text{m}$ , that is, when $P$ is at the midpoint of $l$

Answer to Problem 62E

  $I\left(x\right)$ is minimum at $x=5$

Explanation of Solution

Given information:

Two light sources of identical strength are placed $10\text{m}$ apart an object is to be placed at a point $P$ on a line $l$ parallel to the line joining the light sources and at a distance $d$ meters from it (see the figure). We want to locate $P$ on $l$ so that the intensity of illumination is minimized. We need to use the fact that the intensity of illumination for a single source is directly proportional to the strength of the source and inversely proportional to the source of the distance from the source.

Formula used:

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

Trigonometric ratio:

  $\csc \theta =\frac{h}{p}\text{ and }\cot \theta =\frac{b}{p}$

Speed: the distance covered per unit time is called speed. Speed is directly proportional to distance and inversely proportional to time.

Speed =distance/time, Time =distance/speed, Distance =speed x time

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Calculation:

As per the question:

Draw the diagram of two light sources of identical strength are placed $10\text{m}$ apart an object is to be placed at a point $P$ on a line $l$ parallel to the line joining the light sources and at a distance $d$ meters from it (see the figure). We want to locate $P$ on $l$ so that the intensity of illumination is minimized. We need to use the fact that the intensity of illumination for a single source is directly proportional to the strength of the source and inversely proportional to the source of the distance from the source.

  

  $I\left(x\right)\alpha \frac{\text{strangth of source}}{{\left(\text{distance from source}\right)}^2}$

Adding the intensities from the left and right light bulbs to get,

  $\begin{array}{l}I\left(x\right)=\frac{k}{x^2+d^2}+\frac{k}{{\left(10-x\right)}^2+d^2}\\ =\frac{k}{x^2+d^2}+\frac{k}{x^2+d^2-20x+100}\end{array}$

Substitute $d=5$ and $x=5$ to get

  $\begin{array}{l}I\left(x\right)=\frac{k}{x^2+d^2}+\frac{k}{{\left(10-x\right)}^2+d^2}\\ =\frac{k}{x^2+5^2}+\frac{k}{5^2+5^2}\\ =k\left(\frac{1}{x^2+25}+\frac{1}{50}\right)\mathrm{......}\left(1\right)\end{array}$

Draw the graph of intensity $I\left(x\right)$

  

Therefore $I\left(x\right)$ has a minimum value at $x=5\text{in }\left(0,10\right)$

Recall that,

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Differentiate with respect to $x$

  $I^{\prime }\left(x\right)=k\left(\frac{-2x}{{\left(x^2+25\right)}^2}+\frac{-2\left(x-10\right)}{{\left(x-10\right)}^2+25}\right)\left(2\right)$

Draw the graph of intensity $I^{\prime }\left(x\right)$

  

Graph of $I^{\prime }\left(x\right)$ is zero at $x=5$ , more over $I^{\prime }\left(x\right)$ is negative before $x=5$ and is positive after $x=5$ , indicates that $I\left(x\right)$ is minimum at $x=5$

Conclusion:

Thus $I\left(x\right)$ is minimum at $x=5$

c.

To determine

To calculate: Two light sources of identical strength are placed $10\text{m}$ apart an object is to be placed at a point $P$ on a line $l$ parallel to the line joining the light sources and at a distance $d$ meters from it (see the figure). We want to locate $P$ on $l$ so that the intensity of illumination is minimized. We need to use the fact that the intensity of illumination for a single source is directly proportional to the strength of the source and inversely proportional to the source of the distance from the source

a) If $d=10\text{m}$, show that the intensity (perhaps surprisingly ) is not minimized at the midpoint.

Answer to Problem 62E

  $I\left(x\right)$ has minimized at $x=0$ and $x=10$ . And the midpoint is the most brightly lit point.

Explanation of Solution

Given information:

Two light sources of identical strength are placed $10\text{m}$ apart an object is to be placed at a point $P$ on a line $l$ parallel to the line joining the light sources and at a distance $d$ meters from it (see the figure). We want to locate $P$ on $l$ so that the intensity of illumination is minimized. We need to use the fact that the intensity of illumination for a single source is directly proportional to the strength of the source and inversely proportional to the source of the distance from the source.

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

Trigonometric ratio:

  $\csc \theta =\frac{h}{p}\text{ and }\cot \theta =\frac{b}{p}$

Speed: the distance covered per unit time is called speed. Speed is directly proportional to distance and inversely proportional to time.

Speed =distance/time, Time =distance/speed, Distance =speed x time

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Calculation:

As per the question:

Draw the diagram of two light sources of identical strength are placed $10\text{m}$ apart an object is to be placed at a point $P$ on a line $l$ parallel to the line joining the light sources and at a distance $d$ meters from it (see the figure). We want to locate $P$ on $l$ so that the intensity of illumination is minimized. We need to use the fact that the intensity of illumination for a single source is directly proportional to the strength of the source and inversely proportional to the source of the distance from the source.

  

  $I\left(x\right)\alpha \frac{\text{strangth of source}}{{\left(\text{distance from source}\right)}^2}$

Adding the intensities from the left and right light bulbs to get,

  $\begin{array}{l}I\left(x\right)=\frac{k}{x^2+d^2}+\frac{k}{{\left(10-x\right)}^2+d^2}\\ =\frac{k}{x^2+d^2}+\frac{k}{x^2+d^2-20x+100}\end{array}$

Substitute $d=10$ in equation $\left(1\right)\text{and }\left(2\right)$

  $\begin{array}{l}I\left(x\right)=\frac{k}{x^2+{10}^2}+\frac{k}{{\left(10-x\right)}^2+{10}^2}\\ =\frac{1}{x^2+100}+\frac{1}{x^2-20x+200}\end{array}$

Draw the graph of $I\left(x\right)$

  

And

  $I^{\prime }\left(x\right)=\frac{-2x}{{\left(x^2+25\right)}^2}+\frac{-2\left(x-10\right)}{x^2-20x+200}$

Draw the graph of $I^{\prime }\left(x\right)$

  

From the graph for $d=10$ , $I\left(x\right)$ has minimized at $x=0$ and $x=10$ . And the midpoint is the most brightly lit point.

Conclusion:

Thus $I\left(x\right)$ has minimized at $x=0$ and $x=10$ . And the midpoint is the most brightly lit point.

d.

To determine

To calculate: Two light sources of identical strength are placed $10\text{m}$ apart an object is to be placed at a point $P$ on a line $l$ parallel to the line joining the light sources and at a distance $d$ meters from it (see the figure). We want to locate $P$ on $l$ so that the intensity of illumination is minimized. We need to use the fact that the intensity of illumination for a single source is directly proportional to the strength of the source and inversely proportional to the source of the distance from the source

Somewhere between $d=5\text{m}\text{and }d=10\text{m}$ there is transitional value of $d$ at which the point of minimal illumination abruptly changes. Estimate this value of $d$ graphical methods. Then find the exact value of $d$ .

Answer to Problem 62E

  $d\approx 7.07$

Explanation of Solution

Given information:

Two light sources of identical strength are placed $10\text{m}$ apart an object is to be placed at a point $P$ on a line $l$ parallel to the line joining the light sources and at a distance $d$ meters from it (see the figure). We want to locate $P$ on $l$ so that the intensity of illumination is minimized. We need to use the fact that the intensity of illumination for a single source is directly proportional to the strength of the source and inversely proportional to the source of the distance from the source.

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

Trigonometric ratio:

  $\csc \theta =\frac{h}{p}\text{ and }\cot \theta =\frac{b}{p}$

Speed: the distance covered per unit time is called speed. Speed is directly proportional to distance and inversely proportional to time.

Speed =distance/time, Time =distance/speed, Distance =speed x time

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.

Calculation:

As per the question:

Draw the diagram of two light sources of identical strength are placed $10\text{m}$ apart an object is to be placed at a point $P$ on a line $l$ parallel to the line joining the light sources and at a distance $d$ meters from it (see the figure). We want to locate $P$ on $l$ so that the intensity of illumination is minimized. We need to use the fact that the intensity of illumination for a single source is directly proportional to the strength of the source and inversely proportional to the source of the distance from the source.

  

  $I\left(x\right)\alpha \frac{\text{strangth of source}}{{\left(\text{distance from source}\right)}^2}$

Adding the intensities from the left and right light bulbs to get,

  $\begin{array}{l}I\left(x\right)=\frac{k}{x^2+d^2}+\frac{k}{{\left(10-x\right)}^2+d^2}\\ =\frac{k}{x^2+d^2}+\frac{k}{x^2+d^2-20x+100}\end{array}$

If $d=6$ get minimum at mid point. But if $d=8$ get the minima at end points.

So there is some $d$ between $6\text{an}d8$ , which have minima at midpoint and endpoint simultaneously. That is $I\left(0\right)=I\left(5\right)$

From the graph $d\approx 7$

  

Now,

  $\begin{array}{l}\frac{1}{d^2}+\frac{1}{100+d^2}=\frac{2}{d^2+25}\\ \left(25+d^2\right)\left(100+d^2\right)+d^2\left(25+d^2\right)=2d^2(100+d^2)\\ \Rightarrow 2500=50d^2\\ \Rightarrow d^2=50\end{array}$

Take square root on both sides to get,

  $\begin{array}{l}d=5\sqrt{2}\\ \approx 7.07\end{array}$

Conclusion:

Thus $d\approx 7.07$