Chapter 4.6, Problem 59E

a.

To determine

To calculate: The point $C$ should the bird fly in order to minimize the total energy expended in returning to its nesting area In general, if it takes 1.4 times as much energy to fly over water as it does over land.

Answer to Problem 59E

The to minimize energy the birds fly to the end point about $\approx 5.1\text{km}$ from $\text{B}$

Explanation of Solution

Given information:

A bird with these tendencies is released from an island that is 5 km from the nearest point $B$ on a straight shoreline, flies to a point $C$ on the shoreline, and then flies along the shoreline to its nesting area $D$ . Assume that the bird instinctively chooses a path that will minimize its energy expenditure. Points $B$ and $D$ are 13 km apart

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:

  

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

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 a bird with these tendencies is released from an island that is 5 km from the nearest point $B$ on a straight shoreline, flies to a point $C$ on the shoreline, and then flies along the shoreline to its nesting area $D$ . Assume that the bird instinctively chooses a path that will minimize its energy expenditure. Points $B$ and $D$ are 13 km apart

  

Let $x$ the distance from $B\text{ to C}$

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:

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

  $\begin{array}{l}{\left(AC\right)}^2={\left(AB\right)}^2+{\left(BC\right)}^2\\ =5^2+x^2\\ AC=\sqrt{x^2+25}\end{array}$

The distance from $C$ to $D$ is $13-x$

Therefore, the total distance is

  $\begin{array}{l}=AC+CD\\ =\sqrt{x^2+25}D+13-x\end{array}$

Let $k$ is the energy per km it takes to fly over land. Then

The total energy is

  $E\left(x\right)=1.4k\times \sqrt{x^2+25}D+k\left(13-x\right)$

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 $x$

  $\begin{array}{l}{E}^{\prime }\left(x\right)=1.4k\times \frac{2x}{2\sqrt{x^2+25}}+k\left(-1\right)\\ =\frac{1.4kx}{\sqrt{x^2+25}}-k\\ =\frac{k\left(1.4x-\sqrt{x^2+25}\right)}{\sqrt{x^2+25}}\end{array}$

Solve for $E^{\prime }\left(x\right)=0$ , and simplified

  $\begin{array}{l}\frac{k\left(1.4x-\sqrt{x^2+25}\right)}{\sqrt{x^2+25}}=0\\ \Rightarrow 1.4x-\sqrt{x^2+25}=0\\ \Rightarrow \sqrt{x^2+25}=1.4x\end{array}$

Take square root on both sides,

  $\begin{array}{l}{x}^2+25=1.96x^2\\ \Rightarrow 0.96x^2=25\\ \Rightarrow x=\sqrt{\frac{25}{0.96}}\\ \approx 5.1\text{km}\end{array}$

Evaluate the $E^{\prime }\left(x\right)$ at the critical point and end point

At $x=0$

  $E\left(0\right)=20k$

At $x=5.1$

  $E\left(5.1\right)\approx 17.9k$

At $x=13$

  $E\left(13\right)=19.5k$

Conclusion:

Thus to minimize energy the birds fly to the end point about $\approx 5.1\text{km}$ from $\text{B}$

b.

To determine

To calculate: The ratio $W/L$ corresponding to the minimum expenditure of energy where

  $W$ and $L$ denote the energy (in joules) per kilometer flown over water and land, respectively.

Answer to Problem 59E

The ratio of $\frac{W}{L}$ minimize the energy if the birds aims for the point that is $x$ km from $B$

Explanation of Solution

Given information:

A bird with these tendencies is released from an island that is 5 km from the nearest point $B$ on a straight shoreline, flies to a point $C$ on the shoreline, and then flies along the shoreline to its nesting area $D$ . Assume that the bird instinctively chooses a path that will minimize its energy expenditure. Points $B$ and $D$ are 13 km apart

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:

  

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

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 a bird with these tendencies is released from an island that is 5 km from the nearest point $B$ on a straight shoreline, flies to a point $C$ on the shoreline, and then flies along the shoreline to its nesting area $D$ . Assume that the bird instinctively chooses a path that will minimize its energy expenditure. Points $B$ and $D$ are 13 km apart

  

Let $x$ the distance from $B\text{ and C}$

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:

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

  $\begin{array}{l}{\left(AC\right)}^2={\left(AB\right)}^2+{\left(BC\right)}^2\\ =5^2+x^2\\ AC=\sqrt{x^2+25}\end{array}$

The distance from $C$ to $D$ is $13-x$

Therefore, the total distance is

  $\begin{array}{l}=AC+CD\\ =\sqrt{x^2+25}D+13-x\end{array}$

If $\frac{W}{L}$ is large, the bird would fly to a point $C$ that is closer to $B$ than a $D$ to minimize the energy used flying over water.

If $\frac{W}{L}$ is small, the bird would fly to a point $C$ that is closer to $D$ than a $B$ to minimize the distance of the flight.

  $E\left(x\right)=W\times \sqrt{x^2+25}D+L\left(13-x\right)$

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 $x$

  $E^{\prime }\left(x\right)=\frac{Wx}{\sqrt{x^2+25}}-L$

Solve for $E^{\prime }\left(x\right)=0$ , and simplified

  $\begin{array}{l}\frac{Wx}{\sqrt{x^2+25}}-L=0\\ \Rightarrow \frac{W}{L}=\frac{\sqrt{x^2+25}}{x}\end{array}$

Conclusion:

Thus the ratio of $\frac{W}{L}$ minimize the energy if the birds aims for the point that is $x$ km from $B$

c.

To determine

To calculate: the value of $W/L$ be in order for the bird to fly directly to its nesting area $D$ ? And the value of $W/L$ be for the bird to fly to $B$ and then along the shore to $D$ ?

Answer to Problem 59E

  $\frac{W}{L}$ have no value for which the birds should go directly to $B$

Explanation of Solution

Given information:

A bird with these tendencies is released from an island that is 5 km from the nearest point $B$ on a straight shoreline, flies to a point $C$ on the shoreline, and then flies along the shoreline to its nesting area $D$ . Assume that the bird instinctively chooses a path that will minimize its energy expenditure. Points $B$ and $D$ are 13 km apart

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:

  

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

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 a bird with these tendencies is released from an island that is 5 km from the nearest point $B$ on a straight shoreline, flies to a point $C$ on the shoreline, and then flies along the shoreline to its nesting area $D$ . Assume that the bird instinctively chooses a path that will minimize its energy expenditure. Points $B$ and $D$ are 13 km apart

  

Let $x$ the distance from $B\text{ and C}$

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:

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

  $\begin{array}{l}{\left(AC\right)}^2={\left(AB\right)}^2+{\left(BC\right)}^2\\ =5^2+x^2\\ AC=\sqrt{x^2+25}\end{array}$

The distance from $C$ to $D$ is $13-x$

Therefore, the total distance is

  $\begin{array}{l}=AC+CD\\ =\sqrt{x^2+25}D+13-x\end{array}$

If $\frac{W}{L}$ is large, the bird would fly to a point $C$ that is closer to $B$ than a $D$ to minimize the energy used flying over water.

If $\frac{W}{L}$ is small, the bird would fly to a point $C$ that is closer to $D$ than a $B$ to minimize the distance of the flight.

  $E\left(x\right)=W\times \sqrt{x^2+25}D+L\left(13-x\right)$

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 $x$

  $E^{\prime }\left(x\right)=\frac{Wx}{\sqrt{x^2+25}}-L$

Solve for $E^{\prime }\left(x\right)=0$ , and simplified

  $\begin{array}{l}\frac{Wx}{\sqrt{x^2+25}}-L=0\\ \Rightarrow \frac{W}{L}=\frac{\sqrt{x^2+25}}{x}\end{array}$

Go directly to $D$ means $x=13$ ,

Therefore,

  $\begin{array}{l}\Rightarrow \frac{W}{L}=\frac{\sqrt{{13}^2+25}}{13}\\ \approx 1.07\end{array}$

Conclusion:

Thus $\frac{W}{L}$ have no value for which the birds should go directly to $B$

d.

To determine

To calculate: The energy how many times does it take a bird to fly over water than over land if the ornithologist observe that the birds of a certain species reach the shore at a point 4 km from $B$ ?

Answer to Problem 59E

  $1.6$ times more energy it take a bird to fly over water than over land.

Explanation of Solution

Given information:

A bird with these tendencies is released from an island that is 5 km from the nearest point $B$ on a straight shoreline, flies to a point $C$ on the shoreline, and then flies along the shoreline to its nesting area $D$ . Assume that the bird instinctively chooses a path that will minimize its energy expenditure. Points $B$ and $D$ are 13 km apart

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:

  

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

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 a bird with these tendencies is released from an island that is 5 km from the nearest point $B$ on a straight shoreline, flies to a point $C$ on the shoreline, and then flies along the shoreline to its nesting area $D$ . Assume that the bird instinctively chooses a path that will minimize its energy expenditure. Points $B$ and $D$ are 13 km apart

  

Let $x$ the distance from $B\text{ and C}$

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:

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

  $\begin{array}{l}{\left(AC\right)}^2={\left(AB\right)}^2+{\left(BC\right)}^2\\ =5^2+x^2\\ AC=\sqrt{x^2+25}\end{array}$

The distance from $C$ to $D$ is $13-x$

Therefore, the total distance is

  $\begin{array}{l}=AC+CD\\ =\sqrt{x^2+25}D+13-x\end{array}$

Let $k$ is the energy per km it takes to fly over land. Then

The total energy is

  $E\left(x\right)=1.4k\times \sqrt{x^2+25}D+k\left(13-x\right)$

If the birds choose the path that minimize used energy, Then

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 $x$

  $\begin{array}{l}\frac{1.4kx}{\sqrt{x^2+25}}-k=0\\ \Rightarrow 1.4kx=k\sqrt{x^2+25}\end{array}$

Substitute $1.4kx=c,x=4\text{ and }k=1,$ to get,

  $\begin{array}{l}\frac{1.4kx}{\sqrt{x^2+25}}-k=0\\ \Rightarrow c\times 4=1\times \sqrt{x^2+25}\\ \Rightarrow c\approx 1.6\end{array}$

Conclusion:

Thus $1.6$ times more energy it take a bird to fly over water than over land.