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Modelling with trigonometric functions

Modelling with trigonometric functions

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Explainer Video

Tutor: Toby

Summary

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=0.4sin⁡(1440t)°h=0.4 \sin(1440 t) \degreeh=0.4sin(1440t)°

​​

where hhh is the height above sea level in metres and ttt 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 m0.1 \ m0.1 m below sea level, in one cycle.


Note: The angle here is given in degrees.


a) The sine graph will have a maximum at 0.40.40.4 and minimum at −0.4-0.4−0.4. The time for one cycle is found by setting 1440t1440t1440t​ to 360360360, thus is given by 3601440=0.25 min\dfrac {360}{1440} = 0.25 \text{ min}1440360​=0.25 min. The graph is therefore as shown:

Maths; Trigonometry and modelling; KS5 Year 13; Modelling with trigonometric functions


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


b) To find the times when the height is 0.10.10.1 metres beneath sea level, substitute h=−0.1h=-0.1h=−0.1 and solve for ttt​ to give:

0.4sin⁡(1440t)=−0.1sin⁡(1440t)=−0.251440t=sin⁡−1(−0.25)1440t=−14.478,194.478,345.522t=−0.0101,0.1351,0.2399 min\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}0.4sin(1440t)sin(1440t)1440t1440tt​=−0.1=−0.25=sin−1(−0.25)=−14.478,194.478,345.522=−0.0101,0.1351,0.2399 min​​​


Ignore any negative times and convert to seconds:

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


The amount of time the buoy is 0.1 m0.1 \text{ m}0.1 m below sea level is therefore:

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


Example 2

An alternating voltage is given by the equation:

v=6sin⁡2t−10cos⁡2tt≥0v = 6 \sin 2t - 10 \cos2t \qquad t\ge 0v=6sin2t−10cos2tt≥0​​


where vvv is measured in Volts and ttt is time measured in seconds.


a) What is the peak voltage?

b) How long does it take for the voltage for fall below 2V2V2V?


a) First the expression 6sin⁡2t−10cos⁡2t6 \sin 2t - 10 \cos2t6sin2t−10cos2t needs to be put into the form Rsin⁡(x−α)R \sin(x-\alpha)Rsin(x−α):

6sin⁡2t−10cos⁡2t=Rsin⁡(2t−α)6sin⁡2t−10cos⁡2t=R[sin⁡2tcos⁡α−cos⁡2tsin⁡α]6sin⁡2t−10cos⁡2t=Rsin⁡2tcos⁡α−Rcos⁡2tsin⁡α\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}6sin2t−10cos2t6sin2t−10cos2t6sin2t−10cos2t​=Rsin(2t−α)=R[sin2tcosα−cos2tsinα]=Rsin2tcosα−Rcos2tsinα​​​


Equate coefficients to find RRR and α\alphaα:

6=Rcos⁡α10=Rsin⁡αR=62+102=234tan⁡α=106α=1.03 rads\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}610Rtanαα​=Rcosα=Rsinα=62+102​=234​=610​=1.03 rads​​​


The equation for voltage can now be written as:

v=234sin⁡(2t−1.03)v= 2\sqrt{34} \sin (2t- 1.03)v=234​sin(2t−1.03)​​


Therefore, the peak voltage is 234‾\underline{2 \sqrt{34}}234​​ Volts.


b) To find the time taken to fall below 2 V2 \ V2 V, substitute v=2v=2v=2 and solve for ttt:

234sin⁡(2t−1.03)=2sin⁡(2t−1.03)=22342t−1.03=sin⁡−1(2234)2t−1.03=0.172352t=1.20235t=0.601 s\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}234​sin(2t−1.03)sin(2t−1.03)2t−1.032t−1.032tt​=2=234​2​=sin−1(234​2​)=0.17235=1.20235=0.601 s​​​


Therefore, it takes 0.601 s‾\underline{0.601\ s}0.601 s​ for the voltage to fall below 2 V2 \ V2 V.


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FAQs - Frequently Asked Questions

What type of real life situations can be modelled with trigonometric functions?

Trigonometric functions are used to model situations that involve oscillating or cyclic motion.

What is the best way to starting solving a problem involving modelling real life situations using trig functions?

Always start by drawing a sketch of the trig graph, so you can visualise the maximum and minimum values as well as the period of one cycle.

Which trigonometric formulae can be used when modelling real life situations?

You may get problems that involve using identities such as the addition formulae, double angle formulae or any other trig identities.

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