Modelling with trigonometric functions

​​In a nutshell

You can use trigonometric functions to model real life situations. Trigonometric functions are used to model situations that involve oscillating or cyclic motion. Problems may involve use of trig identities.

Modelling with trigonometric functions

When modelling a real life situation, a sketch of the model can help. Take care to use the correct units for the angle given in the question.

Example 1

The movement of a buoy on the surface of water can be modelled by the equation

​$$h=0.4 \sin(1440 t) \degree$$

​​

where $$h$$ is the height above sea level in metres and $$t$$ is the time in minutes.

​

a) Draw a sketch of the graph to find the maximum height of the buoy above sea level and the time taken, in seconds, from one maximum height to another.

b) Find the amount of time, in seconds, that the buoy is $$0.1 \ m$$ below sea level, in one cycle.

Note: The angle here is given in degrees.

a) The sine graph will have a maximum at $$0.4$$ and minimum at $$-0.4$$. The time for one cycle is found by setting $$1440t$$​ to $$360$$, thus is given by $$\dfrac {360}{1440} = 0.25 \text{ min}$$. The graph is therefore as shown:

Therefore, the maximum height above sea level is $$\underline{0.4 \text{ m}}$$ and the time taken from one maximum to another is $$0.25 \text{ min} = \underline{15 \text{ s}}$$.

b) To find the times when the height is $$0.1$$ metres beneath sea level, substitute $$h=-0.1$$ and solve for $$t$$​ to give:

$$\begin{aligned}0.4\sin(1440t) &=-0.1 \\\sin(1440t) &= -0.25 \\1440t &= \sin^{-1}(-0.25) \\1440t&= -14.478, 194.478, 345.522 \\t &= -0.0101, 0.1351, 0.2399 \text{ min}\end{aligned}$$​​

Ignore any negative times and convert to seconds:

$$t = 8.103 \text{ s}, 14.397 \text{ s}$$​​

The amount of time the buoy is $$0.1 \text{ m}$$ below sea level is therefore:

Time = $$14.397 - 8.103 \ \underline{= 6.29 \text{ s}} \ (3 \ s.f.)$$​

Example 2

An alternating voltage is given by the equation:

$$v = 6 \sin 2t - 10 \cos2t \qquad t\ge 0$$​​

where $$v$$ is measured in Volts and $$t$$ is time measured in seconds.

a) What is the peak voltage?

b) How long does it take for the voltage for fall below $$2V$$?

a) First the expression $$6 \sin 2t - 10 \cos2t$$ needs to be put into the form $$R \sin(x-\alpha)$$:

$$\begin{aligned}6 \sin 2t - 10 \cos2t &= R \sin(2t-\alpha) \\6 \sin 2t - 10 \cos2t &= R[\sin 2t \cos \alpha - \cos 2t \sin \alpha] \\6 \sin 2t - 10 \cos2t&= R\sin 2t \cos \alpha - R\cos 2t \sin \alpha \\\end{aligned}$$​​

Equate coefficients to find $$R$$ and $$\alpha$$:

$$\begin{aligned}6 &= R \cos \alpha \\10 &=R \sin \alpha \\\\R &= \sqrt{6^2+10^2} = 2 \sqrt{34} \\\\\tan \alpha&= \dfrac {10}{6} \\\alpha&= 1.03 \ rads\end{aligned}$$​​

The equation for voltage can now be written as:

$$v= 2\sqrt{34} \sin (2t- 1.03)$$​​

Therefore, the peak voltage is $$\underline{2 \sqrt{34}}$$ Volts.

b) To find the time taken to fall below $$2 \ V$$, substitute $$v=2$$ and solve for $$t$$:

$$\begin{aligned}2\sqrt{34} \sin (2t- 1.03) &= 2 \\\sin (2t- 1.03) &= \dfrac {2}{2\sqrt{34}} \\2t- 1.03 &= \sin^{-1} \Big( \dfrac {2}{2\sqrt{34}}\Big) \\2t- 1.03 &= 0.17235 \\2t&= 1.20235 \\t&= 0.601 \ s\end{aligned}$$​​

Therefore, it takes $$\underline{0.601\ s}$$ for the voltage to fall below $$2 \ V$$.