NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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Optimisation with Trig Functions (sin/cos/tan)
Lesson

As with other optimisation problems, those that involve the trigonometric functions begin with the step of writing a function that models the essential features of the problem. Subsequent steps include finding the derivative and equating it to zero, thus identifying where the gradient is zero.

Because of the periodic nature of the trigonometric functions, there is often an additional complication of choosing which of many possible solutions is the one required.

We will need the following basic differentiation facts:

$\frac{\mathrm{d}}{\mathrm{d}x}\sin x=\cos x$ddxsinx=cosx
$\frac{\mathrm{d}}{\mathrm{d}x}\cos x=-\sin x$ddxcosx=sinx
$\frac{\mathrm{d}}{\mathrm{d}x}\tan x=\frac{1}{\cos^2x}$ddxtanx=1cos2x

 

Example 1

At a certain latitude on the surface of the earth, the length of daylight time varies throughout the year by $\pm h$±h hours about a mean value of $12$12 hours at each equinox. (For this problem, it will not matter exactly what the value of $h$h is.) Write a trigonometric function that models the hours of daylight over a period of $1$1 year so that the summer solstice occurs at time $T=0$T=0.

Use the model to determine at what time or times during the year the length of daylight is changing most rapidly.

Since the cosine function attains its maximum when the independent variable is $0$0 and again at $2\pi$2π, it matches the requirement that the maximum hours of daylight coincides with $T=0$T=0 and again with $T=365$T=365 days.

The function should have a period of $365$365. To achieve this, we can construct the expression $\cos\frac{2\pi T}{365}$cos2πT365 where $T$T varies from $0$0 to $365$365.

The function should have a mean value of $12$12 and an amplitude of $h$h. So, we write

$\phi(T)=12+h\cos\frac{2\pi T}{365}$ϕ(T)=12+hcos2πT365.

This function gives the length of daylight for each day of the year. By differentiating $\phi(T)$ϕ(T) we obtain the rate of change of $\phi$ϕ. We wish to know when the rate of change is most rapid. Therefore, we need the rate of change of the rate of change. That is, we need the second derivative, which we will equate to zero to find its maximum and minimum values.

 

We differentiate: $\frac{\mathrm{d}\phi}{\mathrm{d}T}=-h\frac{2\pi}{365}\sin\frac{2\pi T}{365}$dϕdT=h2π365sin2πT365.

We differentiate again:
$\frac{\mathrm{d^2}\phi}{\mathrm{d}T^2}=-h\left(\frac{2\pi}{365}\right)^2\cos\frac{2\pi T}{365}$d2ϕdT2=h(2π365)2cos2πT365

Equating the gradient function of the gradient function to zero, we obtain $\cos\frac{2\pi T}{365}=0$cos2πT365=0 which gives $\frac{2\pi T}{365}=\frac{\pi}{2},\frac{3\pi}{2},...$2πT365=π2,3π2,....

The solutions are, to the nearest day, $T=91,274,...$T=91,274,.... The first of these is when the length of daylight is decreasing most rapidly and the second is when the length of daylight is increasing most rapidly. If these values of $T$T are substituted back into the length-of-daylight function, $\phi(T)$ϕ(T), we find that at these times there are 12 hours of daylight. That is, the length of daylight is changing most rapidly at each equinox.

 

Example 2

A tunnel has a right-angle change of direction. The cross section of the tunnel is a square with side 2 metres. Long thin sections of steel beam are to be moved along the tunnel but it is not known how long these can be so that they can be manoeuvred around the right-angle corner.

What is the maximum length of the beam?

In the diagram, the dotted lines represent possible beams. We look for an expression that gives the lengths of these lines in terms of the angles they make with the lower edge of the tunnel diagram.

You could draw some extra diagrams and set up some trigonometric relationships to confirm that the length of a line $L$L can be given in two parts $L_1$L1 and $L_2$L2 separated by the corner, such that $L=L_1+L_2$L=L1+L2, where $\sin\theta=\frac{2}{L_1}$sinθ=2L1 and $\cos\theta=\frac{2}{L_2}$cosθ=2L2.

Thus, $L(\theta)=\frac{2}{\sin\theta}+\frac{2}{\cos\theta}$L(θ)=2sinθ+2cosθ

If we imagine one of the red dotted lines being moved around the corner, we see that it will rotate around the corner when its length is at a minimum. Therefore, we differentiate the function $L(\theta)$L(θ) and look for the $\theta$θ that gives the minimum.

$L'(\theta)=\frac{-2}{\sin^2\theta}\cos\theta+\frac{2}{\cos^2\theta}\sin\theta$L(θ)=2sin2θcosθ+2cos2θsinθ.

Equating this to zero gives $-\sin^3\theta+\cos^3\theta=0$sin3θ+cos3θ=0. That is, $\left(\cos\theta-\sin\theta\right)\left(\cos^2\theta+\cos\theta\sin\theta+\sin^2\theta\right)=0$(cosθsinθ)(cos2θ+cosθsinθ+sin2θ)=0.

Thus, the only real solution occurs when $\cos\theta=\sin\theta$cosθ=sinθ. That is, when $\theta=\frac{\pi}{4}$θ=π4 or $45^\circ$45°, as we might have guessed. Then using $L(\theta)=\frac{2}{\sin\theta}+\frac{2}{\cos\theta}$L(θ)=2sinθ+2cosθ, we calculate

$L\left(\frac{\pi}{4}\right)=4\sqrt{2}$L(π4)=42.

 

Worked Examples

Question 1

We want to find the maximum value of the area of the rectangle shown below, where the measurements are in centimetres and $0\le\theta\le\frac{\pi}{2}$0θπ2.

  1. Order the following steps you would take to find the maximum value of the area.

    $A$A Find the derivative of the area with respect to the angle.
    $B$B Test each possible maximum value.
    $C$C Find any possible values of the angle which might maximise the area.
    $D$D Find an expression for the area in terms of the angle.
    $E$E Calculate the maximum value of the area.
    Step 1 $\editable{}$
    Step 2 $\editable{}$
    Step 3 $\editable{}$
    Step 4 $\editable{}$
    Step 5 $\editable{}$
  2. Find an expression for $x$x in terms of $\theta$θ.

  3. Find an expression for $y$y in terms of $\theta$θ.

  4. Hence find an expression for the area, $A$A, in terms of $\theta$θ.

  5. By differentiating $A$A with respect to $\theta$θ, and then using a CAS calculator, find the possible values of $\theta$θ which could maximise the area.

  6. What can we conclude about $A$A when $\theta=\frac{\pi}{4}$θ=π4?

    Since $A'$A is positive when $\theta<\frac{\pi}{4}$θ<π4 and positive when $\theta>\frac{\pi}{4}$θ>π4, $\theta=\frac{\pi}{4}$θ=π4 gives the minimum value of $A$A.

    A

    Since $A'$A is negative when $\theta<\frac{\pi}{4}$θ<π4 and negative when $\theta>\frac{\pi}{4}$θ>π4, $\theta=\frac{\pi}{4}$θ=π4 gives the maximum value of $A$A.

    B

    Since $A'$A is negative when $\theta<\frac{\pi}{4}$θ<π4 and positive when $\theta>\frac{\pi}{4}$θ>π4, $\theta=\frac{\pi}{4}$θ=π4 gives the minimum value of $A$A.

    C

    Since $A'$A is positive when $\theta<\frac{\pi}{4}$θ<π4 and negative when $\theta>\frac{\pi}{4}$θ>π4, $\theta=\frac{\pi}{4}$θ=π4 gives the maximum value of $A$A.

    D

    Since $A'$A is positive when $\theta<\frac{\pi}{4}$θ<π4 and positive when $\theta>\frac{\pi}{4}$θ>π4, $\theta=\frac{\pi}{4}$θ=π4 gives the minimum value of $A$A.

    A

    Since $A'$A is negative when $\theta<\frac{\pi}{4}$θ<π4 and negative when $\theta>\frac{\pi}{4}$θ>π4, $\theta=\frac{\pi}{4}$θ=π4 gives the maximum value of $A$A.

    B

    Since $A'$A is negative when $\theta<\frac{\pi}{4}$θ<π4 and positive when $\theta>\frac{\pi}{4}$θ>π4, $\theta=\frac{\pi}{4}$θ=π4 gives the minimum value of $A$A.

    C

    Since $A'$A is positive when $\theta<\frac{\pi}{4}$θ<π4 and negative when $\theta>\frac{\pi}{4}$θ>π4, $\theta=\frac{\pi}{4}$θ=π4 gives the maximum value of $A$A.

    D
  7. Hence find the maximum value of the area of the rectangle.

Question 2

A trough for holding water is formed by taking a piece of sheet metal $210$210 cm wide and folding the $70$70 cm on either end up as shown below.

We want to find the angle $\theta$θ which will maximise the amount of water that the trough can hold.

  1. Write an expression for $y$y in terms of $\theta$θ.

  2. Write an expression for $x$x in terms of $\theta$θ.

  3. Hence, write an expression for the area of the cross section of the trough, $A$A, in terms of $\theta$θ.

  4. By differentiating $A$A with respect to $\theta$θ, and then using a CAS calculator, find the possible values of $\theta$θ which could maximise the area of the cross-section of the trough and hence maximise the amount of water it can hold.

    Give your answer as an exact value.

  5. What can we conclude about $A$A when $\theta=\frac{\pi}{3}$θ=π3?

    Since $A'$A is negative when $\theta<\frac{\pi}{3}$θ<π3 and negative when $\theta>\frac{\pi}{3}$θ>π3, $\theta=\frac{\pi}{3}$θ=π3 gives the maximum value of $A$A.

    A

    Since $A'$A is positive when $\theta<\frac{\pi}{3}$θ<π3 and positive when $\theta>\frac{\pi}{3}$θ>π3, $\theta=\frac{\pi}{3}$θ=π3 gives the minimum value of $A$A.

    B

    Since $A'$A is negative when $\theta<\frac{\pi}{3}$θ<π3 and positive when $\theta>\frac{\pi}{3}$θ>π3, $\theta=\frac{\pi}{3}$θ=π3 gives the minimum value of $A$A.

    C

    Since $A'$A is positive when $\theta<\frac{\pi}{3}$θ<π3 and negative when $\theta>\frac{\pi}{3}$θ>π3, $\theta=\frac{\pi}{3}$θ=π3 gives the maximum value of $A$A.

    D

    Since $A'$A is negative when $\theta<\frac{\pi}{3}$θ<π3 and negative when $\theta>\frac{\pi}{3}$θ>π3, $\theta=\frac{\pi}{3}$θ=π3 gives the maximum value of $A$A.

    A

    Since $A'$A is positive when $\theta<\frac{\pi}{3}$θ<π3 and positive when $\theta>\frac{\pi}{3}$θ>π3, $\theta=\frac{\pi}{3}$θ=π3 gives the minimum value of $A$A.

    B

    Since $A'$A is negative when $\theta<\frac{\pi}{3}$θ<π3 and positive when $\theta>\frac{\pi}{3}$θ>π3, $\theta=\frac{\pi}{3}$θ=π3 gives the minimum value of $A$A.

    C

    Since $A'$A is positive when $\theta<\frac{\pi}{3}$θ<π3 and negative when $\theta>\frac{\pi}{3}$θ>π3, $\theta=\frac{\pi}{3}$θ=π3 gives the maximum value of $A$A.

    D
  6. Hence, find the maximum value of the area of the cross section of the trough.

    Give your answer as an exact value.

 

Outcomes

M8-11

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

91578

Apply differentiation methods in solving problems

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