Chapter 4.6, Problem 60E

a.

To determine

Use Poiseuille’s Law to show that the total resistance of the blood along the path $ABC$ is

  $R=c\left(\frac{a-b\cot \theta }{r_1{}^4}+\frac{b\csc \theta }{r_2{}^4}\right)$

Explanation of Solution

Given information:

The blood vascular system consists of blood vessels (arteries, arterioles, capillaries, and veins) that convey blood from the heart to the organs and back to the heart. This system should work so as to minimize the energy expended by the heart in pumping the blood. In particular, this energy is reduced when the resistance of the blood is lowered. One of Poiseuille’s Laws gives the resistance $R$ of the blood as

Where $L$ is the length of the blood vessel, $r$ is the radius and $C$ is a positive constant determined by the viscosity of the blood.

Formula used:

Trigonometric ratio:

  

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

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 The blood vascular system consists of blood vessels (arteries, arterioles, capillaries, and veins) that convey blood from the heart to the organs and back to the heart. This system should work so as to minimize the energy expended by the heart in pumping the blood. In particular, this energy is reduced when the resistance of the blood is lowered. One of Poiseuille’s Laws gives the resistance $R$ of the blood as

  $R=C\frac{L}{r^4}$

Where $L$ is the length of the blood vessel, $r$ is the radius and $C$ is a positive constant determined by the viscosity of the blood

  

Divide the $ABC$ into two paths, one from $A$ to $B$ and one from $B$ to $C$

Therefore,

  $AB=L_1\text{ and }BC=L_2$

The total resistance of $ABC$ as the sum of the resistance along $AB$ and $BC$

  $R=L_1+L_2$

Now, Use Poiseuille’s Law

  $R=C\frac{L_1}{r_1{}^4}+C\frac{L_2}{r_2{}^4}$

Recall that,

Trigonometric ratio:

  $\begin{array}{l}\csc \theta =\frac{h}{p}\text{ and }\cot \theta =\frac{b}{p}\\ \csc \theta =\frac{L_2}{b}\text{and }\cot \theta =\frac{b}{a-L_1}\\ L_2=b\csc \theta \text{and }{L}_1=a-b\cot \theta \end{array}$

Substitute $L_2=b\csc \theta \text{and }{L}_1=a-b\cot \theta$ to get,

  $R=C\left(\frac{a-b\cot \theta }{r_1{}^4}+\frac{b\csc \theta }{r_2{}^4}\right)$

Conclusion:

Thus the total resistance from $A$ to $C$ is $R=C\left(\frac{a-b\cot \theta }{r_1{}^4}+\frac{b\csc \theta }{r_2{}^4}\right)$

b..

To determine

Prove that the resistance is minimized when $\cos \theta =\frac{r_2{}^4}{r_1{}^4}$ .

Explanation of Solution

Given information:

The blood vascular system consists of blood vessels (arteries, arterioles, capillaries, and veins) that convey blood from the heart to the organs and back to the heart. This system should work so as to minimize the energy expended by the heart in pumping the blood. In particular, this energy is reduced when the resistance of the blood is lowered. One of Poiseuille’s Laws gives the resistance $R$ of the blood as

  $R=C\frac{L}{r^4}$

Where $L$ is the length of the blood vessel, $r$ is the radius and $C$ is a positive constant determined by the viscosity of the blood. (Poiseuille established this law experimentally, but it also follows from Equation $\mathrm{6.7.2},$ ). The figure shows a main blood vessel with radius $r_1$ branching at angle $\theta$ into a smaller vessel with radius $r$

Formula used:

Trigonometric ratio:

  

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

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 the blood vascular system consists of blood vessels (arteries, arterioles, capillaries, and veins) that convey blood from the heart to the organs and back to the heart. This system should work so as to minimize the energy expended by the heart in pumping the blood. In particular, this energy is reduced when the resistance of the blood is lowered. One of Poiseuille’s Laws gives the resistance $R$ of the blood as

  $R=C\frac{L}{r^4}$

Where $L$ is the length of the blood vessel, $r$ is the radius and $C$ is a positive constant determined by the viscosity of the blood. (Poiseuille established this law experimentally, but it also follows from Equation $\mathrm{6.7.2},$ ). The figure shows a main blood vessel with radius $r_1$ branching at angle $\theta$ into a smaller vessel with radius $r$

  

Divide the $ABC$ into two paths, one from $A$ to $B$ and one from $B$ to $C$

Therefore,

  $AB=L_1\text{ and }BC=L_2$

The total resistance of $ABC$ as the sum of the resistance along $AB$ and $BC$

  $R=L_1+L_2$

Now, Use Poiseuille’s Law

  $R=C\frac{L_1}{r_1{}^4}+C\frac{L_2}{r_2{}^4}$

Recall that,

Trigonometric ratio:

  $\begin{array}{l}\csc \theta =\frac{h}{p}\text{ and }\cot \theta =\frac{b}{p}\\ \csc \theta =\frac{L_2}{b}\text{and }\cot \theta =\frac{b}{a-L_1}\\ L_2=b\csc \theta \text{and }{L}_1=a-b\cot \theta \end{array}$

Substitute $L_2=b\csc \theta \text{and }{L}_1=a-b\cot \theta$ to get,

  $R\left(\theta \right)=C\left(\frac{a-b\cot \theta }{r_1{}^4}+\frac{b\csc \theta }{r_2{}^4}\right)$

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 $\theta$

  $\begin{array}{l}{R}^{\prime }\left(\theta \right)=C\left\{\frac{-b}{r_1{}^4}\left(-{\csc }^2\theta \right)+\frac{b}{r_2{}^4}\left(-\csc \theta \cot \theta \right)\right\}\\ =\frac{Cb}{{\sin }^2\theta }\left(\frac{1}{r_1{}^4}-\frac{\cos \theta }{r_2{}^4}\right)\end{array}$

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

  $\begin{array}{l}\frac{Cb}{{\sin }^2\theta }\left(\frac{1}{r_1{}^4}-\frac{\cos \theta }{r_2{}^4}\right)=0\\ \Rightarrow \frac{1}{r_1{}^4}-\frac{\cos \theta }{r_2{}^4}=0\\ \Rightarrow \frac{1}{r_1{}^4}=\frac{\cos \theta }{r_2{}^4}\\ \Rightarrow \cos \theta =\frac{r_2{}^4}{r_1{}^4}\end{array}$

The critical number obtained the positive value which means that the function is minimum at this point.

Conclusion:

Thus the function is minimum at $\cos \theta =\frac{r_2{}^4}{r_1{}^4}$ .

c.

To determine

Find the optimal branching angle (correct to the nearest degree) when the radius of the smaller blood vessel is two-thirds of the largest vessel.

Answer to Problem 60E

The optimal branching angle is ${79}^o$

Explanation of Solution

Given information:

Ornithologist has determined that some species of birds tend to avoid flights over large bodies of water during daylight hours. It is believed that more energy is required to fly over water than over land because air generally rises over land and falls over water during the day. 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:

Trigonometric ratio:

  

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

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 the blood vascular system consists of blood vessels (arteries, arterioles, capillaries, and veins) that convey blood from the heart to the organs and back to the heart. This system should work so as to minimize the energy expended by the heart in pumping the blood. In particular, this energy is reduced when the resistance of the blood is lowered. One of Poiseuille’s Laws gives the resistance $R$ of the blood as

  $R=C\frac{L}{r^4}$

Where $L$ is the length of the blood vessel, $r$ is the radius and $C$ is a positive constant determined by the viscosity of the blood. (Poiseuille established this law experimentally, but it also follows from Equation $\mathrm{6.7.2},$ ). The figure shows a main blood vessel with radius $r_1$ branching at angle $\theta$ into a smaller vessel with radius $r$

  

Divide the $ABC$ into two paths, one from $A$ to $B$ and one from $B$ to $C$

Therefore,

  $AB=L_1\text{ and }BC=L_2$

The total resistance of $ABC$ as the sum of the resistance along $AB$ and $BC$

  $R=L_1+L_2$

Now, Use Poiseuille’s Law

  $R=C\frac{L_1}{r_1{}^4}+C\frac{L_2}{r_2{}^4}$

Recall that,

Trigonometric ratio:

  $\begin{array}{l}\csc \theta =\frac{h}{p}\text{ and }\cot \theta =\frac{b}{p}\\ \csc \theta =\frac{L_2}{b}\text{and }\cot \theta =\frac{b}{a-L_1}\\ L_2=b\csc \theta \text{and }{L}_1=a-b\cot \theta \end{array}$

Substitute $L_2=b\csc \theta \text{and }{L}_1=a-b\cot \theta$ to get,

  $R\left(\theta \right)=C\left(\frac{a-b\cot \theta }{r_1{}^4}+\frac{b\csc \theta }{r_2{}^4}\right)$

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 $\theta$

  $\begin{array}{l}{R}^{\prime }\left(\theta \right)=C\left\{\frac{-b}{r_1{}^4}\left(-{\csc }^2\theta \right)+\frac{b}{r_2{}^4}\left(-\csc \theta \cot \theta \right)\right\}\\ =\frac{Cb}{{\sin }^2\theta }\left(\frac{1}{r_1{}^4}-\frac{\cos \theta }{r_2{}^4}\right)\end{array}$

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

  $\begin{array}{l}\frac{Cb}{{\sin }^2\theta }\left(\frac{1}{r_1{}^4}-\frac{\cos \theta }{r_2{}^4}\right)=0\\ \Rightarrow \frac{1}{r_1{}^4}-\frac{\cos \theta }{r_2{}^4}=0\\ \Rightarrow \frac{1}{r_1{}^4}=\frac{\cos \theta }{r_2{}^4}\\ \Rightarrow \cos \theta =\frac{r_2{}^4}{r_1{}^4}\end{array}$

Given, if $r_2=\frac{2}{3}{r}_1$

Then the optimal branching angle is

  $\begin{array}{l}\cos \theta =\frac{{\left(\frac{2}{3}{r}_1\right)}^4}{r_1{}^4}\\ ={\left(\frac{2}{3}\right)}^4\\ =\frac{16}{81}\\ \theta ={\cos }^{-1}\left(\frac{16}{81}\right)\\ \approx {79}^o\end{array}$

Conclusion:

Thus the optimal branching angle is ${79}^o$