Title: Default for 12.2 Maxima and Minima
Course: HELM Workbooks 2023-2024

Maxima and minima

Consider the curve
\[
y=f(x)\qquad a\le x\le b
\]
shown below.

[Image: Curve from a to b with x0 and x1 minimum and maximum]

By inspection we see that there is no \(y\)-value greater than that at \(x=a\) (i.e. \(f(a)\)) and there is no value smaller than that at \(x=b\) (i.e. \(f(b)\)). However, the points on the curve at \(x_0\) and \(x_1\) merit comment. It is clear that in the near neighbourhood of \(x_0\) all the \(y\)-values are greater than the \(y\)-value at \(x_0\) and, similarly, in the near neighbourhood of \(x_1\) all the \(y\)-values are less than the \(y\)-value at \(x_1\).

We say \(f(x)\) has a global maximum at \(x=a\) and a global minimum at \(x=b\) but also has a local minimum at \(x=x_0\) and a local maximum at \(x=x_1\).

Our primary purpose in this section is to see how we might locate the position of the local maxima and the local minima for a smooth function \(f(x)\).

A stationary point on a curve is one at which the derivative has a zero value. In the figure below we have sketched a curve with a maximum and a curve with a minimum.

[Image: two parabolas with maximum and minimum values at x0]

By drawing tangent lines to these curves in the near neighbourhood of the local maximum and the local minimum it is obvious that at these points the tangent line is parallel to the \(x\)-axis so that
\[
\dfrac{\mathrm{d}f}{\mathrm{d}x}\bigg|_{x_0}=0
\]

A generalisation of the third law of indices states:

Key Point

Points on the curve \(y=f(x)\) at which\(\dfrac{\mathrm{d}f}{\mathrm{d}x}= 0\) are called stationary points of the function.

However, be careful! A stationary point is not necessarily a local maximum or minimum of the function but may be an exceptional point called a point of inflection, illustrated below..

[Image: inflexion at x0]

Example

Sketch the curve \(y=(x-2)^2+2\) and locate the stationary points on the curve.

Solution

Here \(f(x) = (x−2)^2+ 2\) so \(\dfrac{\mathrm{d}f}{\mathrm{d}x}= 2(x−2)\). At a stationary point \(\dfrac{\mathrm{d}f}{\mathrm{d}x}= 0\) so we have \(2(x−2) = 0\) so \(x= 2\). We conclude that this function has just one stationary point located at \(x= 2\) (where \(y= 2\)). By sketching the curve \(y=f(x)\) it is clear that this stationary point is a local minimum.

[Image: minimum turning point at 2,2]

Exercise

Locate the position of the stationary points of \(\displaystyle f(x)=[value]\).

1. First find \(\displaystyle \frac{\mathrm{d}f}{\mathrm{d}x}=\)
2. Now locate the stationary points by solving \(\displaystyle \frac{\mathrm{d}f}{\mathrm{d}x}=0\). (Give your answer as a set of points, e.g. {[x1,y1],[x2,y2]} note the appearance of square brackets).
Feedback: 1. We have \(\displaystyle f(x) = [value]\). Differentiating once we obtain: \[\frac{\mathrm{d}f}{\mathrm{d}x}=[value].\] 2. To find the stationary points we solve \(\displaystyle \frac{\mathrm{d}f}{\mathrm{d}x}=0\) as suggested. We obtain, \[\begin{align}[value]=0\\\Rightarrow\; x=[value] \;\mbox{ or }\; x= [value] \end{align}\] When \(x=[value]\), \(\displaystyle f(x) = [value]\) and when \(x= [value]\), \(\displaystyle f(x) = [value]\), so the stationary points are \( \left([value],[value]\right) \) and \(\left([value],[value]\right)\). You can confirm your deductions by sketching the curve.

Exercise

Sketch the curve \(y= \cos{(2x)}\) where \(0.1\leq x\leq3\pi4\) and on it locate the position of the global maximum, global minimum and any local maxima or minima.
Feedback: [Image]

Distinguishing between local maxima and minima

We might ask if it is possible to predict when a stationary point is a local maximum, a local minimum or a point of inflection without the necessity of drawing the curve. To do this we highlight the general characteristics of curves in the neighbourhood of local maxima and minima.

For example: at a local maximum (located at \(x_0\) say) describes the situation:

[Image: parabola with x0 maximum]

To the left of the maximum \(\dfrac{\mathrm{d}f}{\mathrm{d}x}>0\)

To the right of the maximum \(\dfrac{\mathrm{d}f}{\mathrm{d}x}<0\)

If we draw a graph of the derivative \(\dfrac{\mathrm{d}f}{\mathrm{d}x}\) against \(x\) then, near a local maximum, it must take one of two basic shapes described:

[Image: inflexions at x0 with angle between y equals 0 as alpha, second graph alpha is 180 degrees]

In case (a) \( \dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\mathrm{d}f}{\mathrm{d}x}\right)\bigg|_{x_0}\equiv \tan\alpha<0\) whilst in case (b) \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\mathrm{d}f}{\mathrm{d}x}\right)\bigg|_{x_0}=0\).

We reach the conclusion that at a stationary point which is a maximum the value of the second derivative \(\dfrac{\mathrm{d}^2f}{\mathrm{d} x^2}\) is either negative or zero.

Near a local minimum the above graphs are inverted. Below shows a local minimum.

[Image: parabola with minimum x0]

To the left of the minimum \(\dfrac{\mathrm{d}f}{\mathrm{d}x}<0\)

To the right of the minimum \(\dfrac{\mathrm{d}f}{\mathrm{d}x}>0\)

The figure below shows the two possible graphs of the derivative:

[Image: two graphs with inflexion at x0 and angle between x is 0 is beta]

Here, for case (a) \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\mathrm{d}f}{\mathrm{d}x}\right)\bigg|_{x_0}=\tan\beta >0\) whilst in (b) \( \dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\mathrm{d}f}{\mathrm{d}x}\right)\bigg|_{x_0}=0.\).

In this case we conclude that at a stationary point which is a minimum the value of the second derivative \(\dfrac{\mathrm{d} ^2f}{\mathrm{d} x^2}\) is either positive or zero. For the third possibility for a stationary point - a point of inflection - the graph of \(f(x)\) against \(x\) and of \(\dfrac{\mathrm{d}f}{\mathrm{d}x}\) against \(x\) take one of two forms as shown below.

[Image: 4 graphs describing the when df by dx is positive and negative.]

For either of these cases \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\mathrm{d}f}{\mathrm{d}x}\right)\bigg|_{x_0}=0\)

The sketches and analysis of the shape of a curve \(y=f(x)\) in the near neighbourhood of stationary points allow us to make the following important deduction:

A generalisation of the third law of indices states:

Key Point

If \(x_0\) locates a stationary point of \(f(x)\), so that \(\displaystyle \frac{\mathrm{d}f}{\mathrm{d}x}\|_{x_0}= 0\), then the stationary point

\(\quad\) is a local minimum if \(\dfrac{\mathrm{d}^2f}{\mathrm{d}x^2}\bigg|_{x_0}>0\)

\(\quad\) is a local maximum if \(\dfrac{\mathrm{d}^2f}{\mathrm{d}x^2}\bigg|_{x_0}<0\)

\(\quad\) is inconclusive if \(\dfrac{\mathrm{d}^2f}{\mathrm{d}x^2}\bigg|_{x_0}=0\).

Example

Find the stationary points of the function \(f(x)=x^3-6x\).

Are these stationary points local maxima or local minima?

Solution

\(\dfrac{\mathrm{d}f}{\mathrm{d}x}= 3x^2−6\). At a stationary point \(\dfrac{\mathrm{d}f}{\mathrm{d}x}= 0\) so \(3x^2−6 = 0\), implying \(x=\pm\sqrt{2}\).Thus \(f(x)\) has stationary points at \(x=\sqrt{2}\)and\(x=−\sqrt{2}\). To decide if these are maxima or minima we examine the value of the second derivative of \(f(x)\)at the stationary points.

\(\dfrac{\mathrm{d}^2f}{\mathrm{d}x^2}= 6x\) so \(\dfrac{\mathrm{d}^2f}{\mathrm{d}x^2}\bigg|_{x=\sqrt{2}}= 6\sqrt{2}>0\). Hence \(x=\sqrt{2}\) locates a local minimum.

Similarly \(\dfrac{\mathrm{d}^2f}{\mathrm{d}x^2}\bigg|_{x=−\sqrt{2}}=−6\sqrt{2}<0\). Hence \(x=−\sqrt{2}\) locates a local maximum.

A sketch of the curve agrees with this analysis:

[Image: sin wave with maximum and minimum at minus square root 2 and square root 2.]

Exercise

For the function \(f(x) = [value]\), \([value] \leq x \leq [value] \), find the positions of any local minima or maxima and distinguish between them.

1. Calculate the first derivative, \(\dfrac{\mathrm{d}f}{\mathrm{d}x}=\)
2. The stationary points are located at: \(x = \) (write your answer as a list, e.g. [x1,x2,x3])
3. Now calculate the second derivative \(\displaystyle \dfrac{\mathrm{d}^2f}{\mathrm{d}x^2}=\)
4. Finally, evaluate the second derivative at each stationary point and draw appropriate conclusions:
\(\quad\) At \(x=\) , \(\dfrac{\mathrm{d}^2f}{\mathrm{d}x^2}=\) . Hence, this is a

\(\quad\) At \(x=\) , \(\dfrac{\mathrm{d}^2f}{\mathrm{d}x^2}=\) . Hence, this is a

\(\quad\) At \(x=\) , \(\dfrac{\mathrm{d}^2f}{\mathrm{d}x^2}=\) . Hence, this is a
Feedback: 1. For \(f(x)=[value]\) we have \[\frac{\mathrm{d}f}{\mathrm{d}x}=[value].\] 2. Hence stationary points are at values of \(x\) in the range specified for which \([value]\), i.e. at \([value] = \pi\) or \([value] = 2\pi\) or \([value] = 3\pi\) (making sure \(x\) is within the range \([value] \leq x \leq [value]\)). Then, stationary points are located at \(x=\displaystyle [value]\), \(x= \displaystyle [value]\) and \(x=\displaystyle [value]\). 3. For the second derivative we have \[\frac{\mathrm{d}^2f}{\mathrm{d}x^2} = [value].\] 4. So, we have \[\begin{align} &\frac{\mathrm{d}^2f}{\mathrm{d}x^2}\Bigr|_{x= [value]} &=& [value]\\ &\frac{\mathrm{d}^2f}{\mathrm{d}x^2}\Bigr|_{x= [value]} &=& [value]\\ &\frac{\mathrm{d}^2f}{\mathrm{d}x^2}\Bigr|_{x= [value]} &=& [value] \end{align}\] Hence, \(x=\displaystyle [value]\) is a [value], \(x= \displaystyle [value]\) is a [value], and \(x=\displaystyle [value]\) is a [value]. You can check against a sketch: [Image: Graph of a cos wave showing three stationary points between 0.1 and some upper bound.]

Exercise

Determine the local maxima and/or minima of the function \(\displaystyle f(x)=[value]\).

1. \(\dfrac{\mathrm{d}f}{\mathrm{d}x}=\)
2. Thus \(\dfrac{\mathrm{d}f}{\mathrm{d}x}=0\) when \(x=\) (give your answer as a list, e.g. [x1,x2])
3. Now obtain the value of the second derivatives at the stationary points
\(\quad\)At \(x=\) , \(\dfrac{\mathrm{d}^2f}{\mathrm{d}x^2}=\) . Hence, this point is a
\(\quad\)At \(x=\) , \(\dfrac{\mathrm{d}^2f}{\mathrm{d}x^2}=\) . Hence, this point is a
Feedback: 1. For \(f(x)=\displaystyle [value]\) we have \[\frac{\mathrm{d}f}{\mathrm{d}x}=[value]=[value].\] 2. Hence stationary points are located at \(x=\displaystyle [value]\) and \(x=\displaystyle [value]\). 3. For the second derivative we have \[\frac{\mathrm{d}^2f}{\mathrm{d}x^2} = [value]\]so, we have \[\begin{align} &\frac{\mathrm{d}^2f}{\mathrm{d}x^2}\Bigr|_{x=[value]} &=& [value]\\ &\frac{\mathrm{d}^2f}{\mathrm{d}x^2}\Bigr|_{x=[value]} &=& [value] \end{align}\] Hence, for \(x= \displaystyle [value]\) the value is inconclusive but \(x=\displaystyle [value]\) is a local minimum.

Conclusion following previous exercise

Using this analysis we cannot decide whether the stationary point at \(x=0\) is a local maximum, minimum or a point of inflection. However, just to the left of \(x=0\) the value of \(\dfrac{\mathrm{d}f}{\mathrm{d}x}\) (which equals \(x^2(4x-1)\)) is negative whilst just to the right of \(x=0\) the value of \(\dfrac{\mathrm{d}f}{\mathrm{d}x}\) is negative again. Hence the stationary point at \(x=0\) is a point of inflection. This is confirmed by sketching the curve below.

[Image: inflection at zero zero and minimum at x equals 1 over 4 and y equals minus zero point zero zero one three.]

Exercise

A materials store is to be constructed next to a \(3\) metre high stone wall (shown as \(OA\) in the cross section in the diagram). The roof \((AB)\) and front \((BC)\) are to be constructed from corrugated metal sheeting. Only [value] metre length sheets are available. Each of them is to be cut into two parts such that one part is used for the roof and the other is used for the front. Find the dimensions \(x, y\) of the store that result in the maximum cross-sectional area. Hence determine the maximum cross-sectional area.

[Image: stone wall to left 3 metres, top left A top right B y right length, x bottom length, O bottom left and C bottom right.]

1. Give the cross sectional area of the store, \(A=\)
2. Give the equation that relates \(AB\) and \(BC\): (use AB and BC as variables)
3. Now, rewrite it as a polynomial in terms of \(x\) and \(y\):
4. Calculate the values of \(x\) and \(y\) that maximise the cross sectional area (give a precise answer):
\(\quad x=\)
\(\quad y=\)
5. Finally, give the maximum value of the cross sectional area (give a precise answer): \(\max(A) =\)

(The following information pages are applications, if you want to move to the next exercise simply move to 6 in the menu on the left)
Feedback: 1. Note that the store has the shape of a trapezium. So the cross-sectional area (\(A\)) of the store is given by the formula: Area \(=\) average length of parallel sides \(\times\) distance between parallel sides: \[ A=\frac{1}{2}(y+3)x \qquad\qquad (1) \] 2. The lengths \(x\) and \(y\) are related through the fact that \(AB+BC= [value]\), where \(BC=y\) and \(AB=\sqrt{x^2+ (3-y)^2}\). 3. Hence\(\sqrt{x^2+ (3-y)^2}+y= [value]\). This equation can be rearranged in the following way: \[ \sqrt{x^2+ (3-y)^2}= [value]-y \iff x^2+ (3-y)^2= ([value]-y)^2\quad \mbox{ i.e. }\;\; [value]= [value] \] which implies that \[[value].\qquad\qquad (2)\] 4. It is necessary to eliminate either \(x\) or \(y\) from (1) and (2) to obtain an equation in a single variable. Using \(y\) instead of \(x\) as the variable will avoid having square roots appearing in the expression for the cross-sectional area. Hence \[ [value].\qquad \qquad (3) \] Substituting for \(y\) from Equation (3) into Equation (1) gives \[ A=[value]=[value] \] To find turning points, we evaluate \(\dfrac{\mathrm{d}A}{\mathrm{d}x}\) from above to get \[ \dfrac{\mathrm{d}A}{\mathrm{d}x}=[value] \] Solving the equation \(\dfrac{\mathrm{d}A}{\mathrm{d}x}= 0\) gives \(\displaystyle [value] = 0 \;\implies\; [value]\). Hence \(x=\pm [value]=\pm [value]\). Only \(x>0\) is of interest, so \(x=[value] = [value]\) gives the required turning point. Check: Differentiating \(\dfrac{\mathrm{d}A}{\mathrm{d}x}\) and using the positive \(x\) solution gives \[ \frac{\mathrm{d}^2A}{\mathrm{d}x^2}=[value]=[value] <0 \] Since the second derivative is negative then the cross-sectional area is a maximum. This is the only turning point identified for \(A >0\) and it is identified as a maximum. To find the corresponding value of \(y\), substitute \(x= [value]=[value]\) into \([value]\) to get \( [value]= [value]\). So the values of \(x\) and \(y\) that yield the maximum cross-sectional area are \([value]\,m\) and \([value]\,m\) respectively. 5. To find the maximum cross-sectional area, substitute for \(x=[value]\) into \(\dfrac{\mathrm{d}A}{\mathrm{d}x}\) to get \(A=[value] = [value]\). So the maximum cross-sectional area of the store is \([value]\,m^2\)to \(2\) d.p.

Exercise

Equivalent resistance in an electrical circuit

Current distributes itself in the wires of an electrical circuit so as to minimise the total power consumption i.e. the rate at which heat is produced. The power \((p)\) dissipated in an electrical circuit is given by the product of voltage \((v)\) and current \((i)\) flowing in the circuit, i.e. \(p = vi.\) The voltage across a resistor is the product of current and resistance \((r).\) This means that the power dissipated in a resistor is given by \(p = i^2r.\)

Suppose that an electrical circuit contains three resistors \(r_1, r_2, r_3\) and \(i_1\) represents the current flowing through both \(r_1\) and \(r_2\), and that \((i - i_1)\) represents the current flowing through \(r_3\) (see diagram):

[Image: Electrical circuit diagram with three resistances R1, R2, R3. R1 and R2 (on same branch) in parallel with R3.]

1. Write down an expression for the power dissipated in the circuit: \(p=\) (for subscripts write e.g. x1 for \(x_1\))
2. Show that the power dissipated is a minimum when \(\ \ i_1=\dfrac{r_3}{r_1+r_2+r_3}\ i\).

\(\quad\)Differentiating with respect to , we get:

\(\qquad p' =\) and

\(\qquad p'' =\)

\(\quad\) Hence, for \(i_1=\dfrac{r_3}{r_1+r_2+r_3}\ i\) the power dissipated is a minimum since ...

\(\quad\)(select one to complete the sentence)

3. If \(R\) is the equivalent resistance of the circuit, i.e. of \(r_1,\ r_2\) and \(r_3\), for minimum power dissipation and the corresponding voltage \(V\) across the circuit is given by \(V = iR = i_1(r_1 + r_2)\), find an expression for \(R\).

\(\quad R=\)

Note: In this problem \(R_1\) and \(R_2\) could be replaced by a single resistor. However, treating them as separate allows the possibility of considering more general situations (variable resistors or temperature dependent resistors).
Feedback: 1. The power dissipated in the circuit may be written as \[p=[value].\] 2. Differentiating with respect to \(i_1\) we obtain \[\dfrac{\mathrm{d}p}{\mathrm{d}i_1}=[value].\]This is zero when \[[value].\]To check if this represents a minimum we differentiate a second time to find \[\frac{\mathrm{d}^2p}{\mathrm{d}i_1^2}=[value]\]which is positive for any value of \(i_1\). So the previous result represents a minimum. 3. Finally, to find \(R\), substitute for \(i_1\) in \(iR=i_1(r_1+r_2)\) to obtain \[[value]\] which means \[R = [value].\]

Example

Water wheel efficiency

Introduction

A water wheel is constructed with symmetrical curved vanes of angle of curvature \(\theta\). Assuming that friction can be taken as negligible, the efficiency, \(\eta\), i.e. the ratio of output power to input power, is calculated as \[ \eta=\frac{2(V-v)(1+\cos\theta )v}{V^2}\]where \(V\) is the velocity of the jet of water as it strikes the vane, \(v\) is the velocity of the vane in the direction of the jet and \(\theta\) is constant. Find the ratio, \(v/V\), which gives maximum efficiency and find the maximum efficiency.

Mathematical statement of the problem

We need to express the efficiency in terms of a single variable so that we can find the maximum value. \[ \text{Efficiency} =\frac{2(V-v)(1+\cos\theta )v}{V^2}=2\left(1-\frac{v}{V}\right)\frac{v}{V}(1+\cos\theta ).\]Let \(\eta=\) Efficiency and \(x = \dfrac{v}{V}\) then \(\eta = 2x(1 - x)(1 + \cos\theta ).\) We must find the value of \(x\) which maximises \(\eta\) and we must find the maximum value of \(\eta\). To do this we differentiate \(\eta\) with respect to \(x\) and solve \(\dfrac{\mathrm{d}\eta}{\mathrm{d}x}=0\) in order to find the stationary points.

Mathematical analysis

Now \(\eta = 2x(1 - x)(1 + \cos\theta ) = (2x- 2x^2)(1 + \cos\theta )\).

So \(\dfrac{\mathrm{d}\eta}{\mathrm{d}x} =(2-4x)(1+\cos\theta )\).

Now \(\dfrac{\mathrm{d}\eta}{\mathrm{d}x} = 0 \Rightarrow 2-4x = 0\Rightarrow x = \dfrac{1}{2}\) and the value of \(\eta\) when \(x = \dfrac{1}{2}\) is \[ \eta= 2\left(\frac{1}{2}\right)\left(1-\frac{1}{2}\right)(1+\cos\theta )=\frac{1}{2}(1+\cos\theta ).\]This is clearly a maximum not a minimum, but to check we calculate \(\dfrac{\mathrm{d}^2\eta}{\mathrm{d}x^2}=-4(1+\cos\theta)\) which is negative which provides confirmation.

Interpretation

Maximum efficiency occurs when \(\dfrac{v}{V}=\dfrac{1}{2}\) and the maximum efficiency is given by \[ \eta =\frac{1}{2}(1+\cos\theta ).\]

Example

Refraction

The problem

A light ray is travelling in a medium \((A)\) at speed \(c_A\). The ray encounters an interface with a medium \((B)\) where the velocity of light is \(c_B\). Between two fixed points \(P\) in media \(A\) and \(Q\) in media \(B\), find the path through the interface point \(O\) that minimizes the time of light travel (see figure below). Express the result in terms of the angles of incidence and refraction at the interface and the velocities of light in the two media.

[Image: Geometry of light rays at an interface.]

The solution

The light ray path is shown as \(POQ\) in the above figure where \(O\) is a point with variable horizontal position \(x\). The points \(P\) and \(Q\) are fixed and their positions are determined by the constants \(a,\ b,\ d\) indicated in the figure. The total path length can be decomposed as \(PO + OQ\) so the total time of travel \(T(x)\) is given by \[ T (x)= PO/c_A + OQ/c_B.\qquad\qquad (1)\]Expressing the distances \(PO\) and \(OQ\) in terms of the fixed coordinates \(a,\ b,\ d,\) and in terms of the unknown position \(x\), Equation (1) becomes \[ T(x)= \frac{\sqrt{a^2+x^2}}{c_A}+\frac{\sqrt{b^2+(d-x)^2}}{c_B}\qquad\qquad (2)\]It is assumed that the minimum of the travel time is given by the stationary point of \(T(x)\) such that \[ \frac{\mathrm{d}T}{\mathrm{d}x}=0.\qquad \qquad (3)\]Using the chain rule to compute (3) given (2) leads to \[ \frac{1}{2}\ \frac{2x}{c_A\sqrt{a^2+x^2}}+\frac{1}{2}\ \frac{2x-2d}{c_B\sqrt{b^2+(d-x)^2}}=0.\]After simplification and rearrangement \[ \frac{x}{c_A\sqrt{a^2+x^2}}=\frac{d-x}{c_B\sqrt{b^2+(d-x)^2}}.\]Using the definitions \(\sin\theta_A=\dfrac{x}{\sqrt{a^2+x^2}}\) and \(\sin\theta_B=\dfrac{d-x}{\sqrt{b^2+(d-x)^2}}\) this can be written as \[ \frac{\sin\theta_A}{c_A}=\frac{\sin\theta_B}{c_B}.\qquad\qquad (4)\]Note that \(\theta_A\) and\(\theta_B\) are the incidence angles measured from the interface normal as shown in the figure. Equation (4) can be expressed as \[ \frac{\sin\theta_A}{\sin\theta_B}=\frac{c_A}{c_B}\]which is the well-known law of refraction for geometrical optics and applies to many other kinds of waves. The ratio \(\dfrac{c_A}{c_B}\) is a constant called the refractive index of medium \((B)\) with respect to medium \((A)\).

Example

Fluid power transmission

Introduction

Power transmitted through fluid-filled pipes is the basis of hydraulic braking systems and other hydraulic control systems. Suppose that power associated with a piston motion at one end of a pipeline is transmitted by a fluid of density \(\rho\) moving with positive velocity \(V\) along a cylindrical pipeline of constant cross-sectional area \(A\). Assuming that the loss of power is mainly attributable to friction and that the friction coefficient \(f\) can be taken to be a constant, then the power transmitted, \(P\) is given by \[ P=\rho g A(hV -cV^3)\], where \(g\) is the acceleration due to gravity and \(h\) is the head which is the height of the fluid above some reference level (\(=\) the potential energy per unit weight of the fluid). The constant \(c=\dfrac{4fl}{2gd}\) where \(l\) is the length of the pipe and \(d\) is the diameter of the pipe. The power transmission efficiency is the ratio of power output to power input.

Problem in words

Assuming that the head of the fluid, \(h\), is a constant find the value of the fluid velocity, \(V\), which gives a maximum value for the output power \(P\). Given that the input power is \(P_i= \rho gAVh\), find the maximum power transmission efficiency obtainable.

Mathematical statement of the problem

We are given that \(P=\rho gA(hV-cV^3)\) and we want to find its maximum value and hence maximum efficiency.

To find stationary points for \(P\) we solve \(\dfrac{ \mathrm{d} P}{ \mathrm{d} V}= 0.\)

To classify the stationary points we can differentiate again to find the value of \(\dfrac{ \mathrm{d}^2P}{ \mathrm{d} V^2}\) at each stationary point and if this is negative then we have found a local maximum point. The maximum efficiency is given by the ratio \(P/P_i\) at this value of \(V\) and where \(P_i = \rho gAVh.\) Finally we should check that this is the only maximum in the range of \(P\) that is of interest.

Mathematical analysis

\[ P=\rho gA(hV-cV^3)\]\[ \dfrac{ \mathrm{d} P}{ \mathrm{d} V}=\rho gA(h-3cV^2)\]\[ \dfrac{ \mathrm{d} P}{ \mathrm{d} V}=0\;\; \mbox{ gives }\;\; \rho gA(h-3cV^2)=0\]\[ \Rightarrow\quad V^2=\dfrac{h}{3c}\quad \Rightarrow\quad V=\pm\ \sqrt{\dfrac{h}{3c}}\] and as \(V\) is positive \[\Rightarrow\quad V=\sqrt{\dfrac{h}{3c}}.\]

To show this is a maximum we differentiate \(\dfrac{ \mathrm{d} P}{ \mathrm{d} V}\) again giving \(\dfrac{ \mathrm{d}^2P}{ \mathrm{d} V^2} =\rho gA(-6cV)\). Clearly this is negative, or zero if \(V=0\). Thus \(V =\sqrt{\dfrac{h}{3c}}\) gives a local maximum value for \(P\).

We note that \(P=0\) when \(E=\rho gA(hV-cV^3)=0\), i.e. when \(hV-cV^3=0\), so \(V=0\) or \(V=\sqrt{\dfrac{h}{C}}\). So the maximum at \(V=\sqrt{\dfrac{h}{3C}}\) is the only max in this range between \(0\) and \(V=\sqrt{\dfrac{h}{C}}\).

The efficiency \(E\), is given by (input power/output power), so here \[E=\frac{\rho gA(hV-cV^3)}{\rho gAVh}=1-\dfrac{cV^2}{h}\]

At \(V =\sqrt{\dfrac{h}{3c}}\) then \(V^2=\dfrac{h}{3c}\) and therefore \(E = 1-\dfrac{c\dfrac{h}{3c}}{c}=1-\dfrac{1}{3}=\dfrac{2}{3}\) or \(66\frac{2}{3}\%\).

Interpretation

Maximum power transmitted through the fluid when the velocity \(V=\sqrt{\dfrac{h}{3c}}\) and the maximum efficiency is \(66\frac{2}{3}\)%. Note that this result is independent of the friction and the maximum efficiency is independent of the velocity and (static) pressure in the pipe.

[Image: Graphs of transmitted power as a function of fluid velocity for two values of the head.]

The figure above shows the maxima in the power transmission for two different values of the head in an oil filled pipe (oil density \(1100 \text{kg m}^{-3}\)) of inner diameter \(0.01 \text{m}\) and coefficient of friction \(0.01\) and pipe length \(1 \text{m}\).

Example

Crank used to drive a piston

Introduction

A crank is used to drive a piston as in the figure below.

[Image: Diagram of a crank used to drive a piston.]

Problem

The angular velocity of the crankshaft is the rate of change of the angle \(\theta,\ \omega= \mathrm{d}\theta/\mathrm{d}t.\) The piston moves horizontally with velocity \(v_p\) and acceleration \(a_p\); \(r\) is the length of the crank and \(l\) is the length of the connecting rod. The crankpin performs circular motion with a velocity of \(v_c\) and centripetal acceleration of \(\omega^2r\). The acceleration \(a_p\) of the piston varies with \(\theta\) and is related by \[ a_p =\omega^2 r\left(\cos\theta+\frac{r\cos2\theta}{l}\right)\]

Find the maximum and minimum values of the acceleration \(a_p\) when \(r = 150\) mm and \(l = 375\) mm.

Mathematical statement of the problem

We need to find the stationary values of \(a_p =\omega^2 r\left(\cos\theta+\dfrac{r\cos2\theta}{l}\right)\) when \(r = 150\) mm and \(l = 375\) mm. We do this by solving \(\dfrac{\mathrm{d}a_p}{\mathrm{d}\theta}=0\) and then analysing the stationary points to decide whether they are a maximum, minimum or point of inflexion.

Mathematical analysis

\[ a_p =\omega^2r\left(\cos\theta+\dfrac{r\cos2\theta}{l}\right)\;\; \mbox{ so }\;\; \dfrac{\mathrm{d}a_p}{\mathrm{d}\theta}=\omega^2r\left(-\sin\theta-\dfrac{2r\sin2\theta}{l}\right)\]. To find the maximum and minimum acceleration we need to solve \[\dfrac{\mathrm{d}a_p}{\mathrm{d}\theta}=0\Leftrightarrow \omega^2 r\left(-\sin\theta-\dfrac{2r\sin2\theta}{l}\right)=0.\]\[\sin\theta+\dfrac{2r}{l}\sin2\theta=0\Leftrightarrow \sin\theta+\dfrac{4r}{l}\sin\theta\cos\theta=0\]\[\Leftrightarrow\quad \sin\theta\left(1+\dfrac{4r}{l}\cos\theta\right)=0\]\[\Leftrightarrow\quad \sin\theta=0\;\; \mbox{or}\;\; \cos\theta=-\dfrac{l}{4r}\;\; \mbox{ and as } r=150 \mbox{ mm }\;\; \mbox{ and }\;\; l=375 \mbox{ mm }\]\[\Leftrightarrow\quad \sin\theta=0\;\; \mbox{ or } \;\;\cos\theta=-\dfrac{5}{8}\].

CASE 1: \(\sin\theta = 0\)

If \(\sin\theta=0\) then \(\theta=0\) or \(\theta=\pi\). If \(\theta=0\) then \(\cos\theta=\cos2\theta=1\) so \[ a_p =\omega^2r\left(\cos\theta+\dfrac{r\cos2\theta}{l}\right)=\omega^2r\left(1+\dfrac{r}{l}\right) =\omega^2r\left(1+\dfrac{2}{5}\right)=\dfrac{7}{5}\omega^2r\]If \(\theta=\pi\) then \(\cos\theta=-1,\ \cos2\theta=1\) so \[ a_p=\omega^2r\left(\cos\theta+\dfrac{r\cos2\theta}{l}\right)= \omega^2r\left(-1+\dfrac{r}{l}\right)=\omega^2r\left(-1+\dfrac{2}{5}\right)=-\dfrac{3}{5}\omega^2r\]In order to classify the stationary points, we differentiate \(\dfrac{\mathrm{d}a_p}{\mathrm{d}\theta}\) with respect to \(\theta\) to find the second derivative: \[ \dfrac{\mathrm{d}^2a_p}{\mathrm{d}\theta^2}=\omega^2r\left(-\cos\theta-\dfrac{4r\cos2\theta}{l}\right) =-\omega^2r\left(\cos\theta+\dfrac{4r\cos2\theta}{l}\right)\]

At \(\theta=0\) we get \(\dfrac{\mathrm{d}^2a_p}{\mathrm{d}\theta^2}=-\omega^2r\left(1+\dfrac{4r}{l}\right)\) which is negative.

So \(\theta=0\) gives a maximum value and \(a_p =\dfrac{7}{5}\omega^2r\) is the value at the maximum.

At \(\theta=\pi\) we get \(\dfrac{\mathrm{d}^2a_p}{\mathrm{d}\theta^2}=-\omega^2r\left(-1+\dfrac{4}{l}\right) =-\omega^2r\left(\dfrac{3}{5}\right)\) which is negative.

So \(\theta=\pi\) gives a maximum value and \(a_p=-\dfrac{3}{5}\omega^2r\)

CASE 2: \(\cos\theta = -\dfrac{ 5}{ 8}\)

If \(\cos\theta =-\dfrac{5}{8}\) then \(\cos2\theta = 2\cos^2\theta -1 = 2\left(\dfrac{5}{8}\right)^2 -1\) so \(\cos2\theta =-\dfrac{7}{32}\). \[ a_p = \omega^2r\left(\cos\theta+\dfrac{r\cos2\theta}{l}\right)= \omega^2r\left(-\dfrac{5}{8}+-\dfrac{7}{32}\times \dfrac{2}{5}\right)=\dfrac{57}{80}\omega^2r.\]At \(\cos\theta=-\dfrac{5}{8}\) we get \(\dfrac{\mathrm{d}^2a_p}{\mathrm{d}\theta^2}= \omega^2r\left(-\cos\theta-\dfrac{4r\cos2\theta}{l}\right)=\omega^2r\left(\dfrac{5}{8}+\dfrac{4r}{l}\ \dfrac{7}{32}\right)\) which is positive.

So \(\cos\theta= -\dfrac{5}{8}\) gives a minimum value and \(a_p=-\dfrac{57}{80}\omega^2r\)

Thus the values of \(a_p\) at the stationary points are: \[\dfrac{7}{5}\omega^2r\;\mbox{ (maximum), }\;\; -\dfrac{3}{5}\omega^2r\;\mbox{ (maximum) and }\;\; -\dfrac{57}{80}\omega^2r\;\mbox{ (minimum). }\]So the overall maximum value is \(1.4\omega^2r = 0.21\omega^2\) and the minimum value is\(-0.7125 \omega^2r = -0.106875\omega^2\) where we have substituted \(r = 150\) mm (\(=\) 0.15 m) and \(l =375\) mm (\(= 0.375\) m).

Interpretation

The maximum acceleration occurs when \(\theta=0\) and \(a_p=0.21\omega^2\).

The minimum acceleration occurs when \(\cos\theta =-\dfrac{5}{8}\) and \(a_p = -0.106875\omega^2\).

Exercise

Locate the stationary points of the following functions and distinguish among them as maxima, minima and points of inflection.

(Write null for the values of \(x\), \(y\) and \(\dfrac{\mathrm{d}^2f}{\mathrm{d}x^2}\) if no stationary point exists.)

1. \(f(x)=[value] \quad\) [Remember \(\dfrac{\mathrm{d}}{\mathrm{d}x}(\ln |x|) = \dfrac{1}{x}\)]

\(\quad \dfrac{\mathrm{d}f}{\mathrm{d}x} =\)

Number of stationary points \(=\)
Stationary point located at \(x=\) , \(y=\) where \(\dfrac{\mathrm{d}^2f}{\mathrm{d}x^2}=\) , is a
2. \(f(x)=[value]\)

\(\quad \dfrac{\mathrm{d}f}{\mathrm{d}x} =\)

Number of stationary points \(=\)
Stationary point located at \(x=\) , \(y=\) where \(\dfrac{\mathrm{d}^2f}{\mathrm{d}x^2}=\) , is a
3. \(f(x)=\displaystyle [value]\qquad -1<x<2\)

\(\quad \dfrac{\mathrm{d}f}{\mathrm{d}x} =\)

Number of stationary points \(=\)
Stationary point located at \(x=\) , \(y=\) where \(\dfrac{\mathrm{d}^2f}{\mathrm{d}x^2}=\) , is a
Feedback: Worked solution 1. \(\displaystyle \frac{\mathrm{d}f}{\mathrm{d}x} = [value]\) When \(\dfrac{\mathrm{d}f}{\mathrm{d}x}=0 \implies [value]\), so there is a single stationary point located at \(x=[value]\). Differentiating a second time and evaluating at the stationary point we get \[\frac{\mathrm{d}^f}{\mathrm{d}x^2}\Bigr|_{x=[value]} = [value]\Bigr|_{x=[value]} = [value].\]Hence, \(([value],[value])\) is a [value]. 2. \(\displaystyle \frac{\mathrm{d}f}{\mathrm{d}x} = [value]\) When \(\dfrac{\mathrm{d}f}{\mathrm{d}x}=0 \implies [value]\), so there is a single stationary point located at \(x=[value]\). Differentiating a second time and evaluating at the stationary point we get \[\frac{\mathrm{d}^f}{\mathrm{d}x^2}\Bigr|_{x=[value]} = [value]\Bigr|_{x=[value]} = [value].\]However, \(\dfrac{\mathrm{d}f}{\mathrm{d}x} > 0\) on either side of \(x = [value]\) so \(([value], [value])\) is a [value]. 3. \(\displaystyle \frac{\mathrm{d}f}{\mathrm{d}x} = \displaystyle [value]\) \(\dfrac{\mathrm{d}f}{\mathrm{d}x}=0\) has no real roots (since \(b^2<4ac\)), so \(f(x)\) has no stationary points. Nevertheless \(f(x)\) does have a point of inflection at \(x = 1, y = 0\) as a sketch shows, although at that point \(\frac{\mathrm{d}y}{\mathrm{d}x} \neq 0\).

Exercise

Write omega for \(\omega\), pi for \(\pi\) and atan for \(\tan^{-1}\).

A perturbation in the temperature of a stream leaving a chemical reactor follows a decaying sinusoidal variation, according to \[ T(t)=5\text{exp}(-at)\sin(\omega t)\]where \(a\) and \(\omega\) are positive constants.

1. Sketch the variation of temperature with time.
2. By examining the behaviour of \(\dfrac{\mathrm{d}T}{\mathrm{d}t}\), show that the maximum temperatures occur at times of \(\left(\tan^{-1}\left(\dfrac{\omega}{a}\right)+2\pi n\right)/\omega\).

\(\quad \dfrac{\mathrm{d}T}{\mathrm{d}t} = \)

\(\quad\)At stationary points, \(\quad\omega t = \) \(+\)

\(\quad\)Maximum temperatures are located at \(t\) such that \(\dfrac{\mathrm{d}^2T}{\mathrm{d}t^2}\) which is true

\(\quad\)Hence, maximum temperatures occur at times \(t=\left(\tan^{-1}\left(\dfrac{\omega}{a}\right)+2\pi n\right)/\omega\), where \(2n=\) is an integer.
Feedback: Worked solution 1. The sketch: [Image: Oscillatory decay of temperature with time.] 2. \(\dfrac{\mathrm{d}T}{\mathrm{d}t}=0\) implies that \([value]\), so \(\tan(\omega t) > 0\) and \[\omega t = [value] + k\pi,\quad k\mbox{ an integer }\]Examination of \(\dfrac{\mathrm{d}^2T}{\mathrm{d}t^2}\) reveals that only even values of \(k\) give \(\dfrac{\mathrm{d}^2T}{\mathrm{d}t^2}<0\) for a maximum. So setting \(k=2n\) gives the required answer.