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Parametric equations

Modelling with parametric equations

Modelling with parametric equations

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Further kinematics

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Constant acceleration formulae: SUVAT

Vertical motion under gravity

Modelling in mechanics

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Quantities and units in mechanics

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Parametric differentiation

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Binomial expansion II

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Parametric equations

Parametric equations

Using trigonometric identities

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Points of intersection of parametric curves

Modelling with parametric equations

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y = |f(x)| and y = f(|x|)

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Proof by exhaustion

Explainer Video

Tutor: Alice

Summary

Modelling with parametric equations  

​​In a nutshell

Parametric equations can be used to model many real life situations. There are many applications and some may relate to mechanics problems, including projectile motion.



Real life models

Mechanics problems can be modelled using parametric equations. For example, the equations for xxx and yyy can represent the displacement in different directions and the parameter can represent time.


Example

The motion on a skateboard ramp can be modelled by the parametric equations for −π2≤t≤π2-\frac {\pi} {2} \le t \le \frac {\pi} {2}−2π​≤t≤2π​ as:


x=0.5t−4sin⁡ty=1−3cos⁡t\begin{aligned}x&=0.5t - 4\sin t \\y&=1-3 \cos t\end{aligned}xy​=0.5t−4sint=1−3cost​​​


A sketch of the curve is shown. Find 

a) The length of the straight line ABABAB.

b) The depth of the ramp.

Maths; Parametric equations; KS5 Year 13; Modelling with parametric equations


First, find the coordinates of some points on the curve. Use t=−π2t=- \dfrac{\pi}{2}t=−2π​ and π2 \dfrac{\pi}{2}2π​ as these values are given as the limits for the range of ttt. The curve also intersects the yyy-axis, so find where x=0x=0x=0. Use a table of values: 


​ttt​​
​−π2-\dfrac {\pi}{2}−2π​​​
​000​​
​π2\dfrac{\pi}{2}2π​​​
​xxx​​
​4−π44- \dfrac{\pi}{4}4−4π​​​
​000​​
​π4−4\dfrac {\pi}{4} - 44π​−4​​
​yyy​​
​111​​
​−2-2−2​​
​111​​


​​

a) From the table, it can be seen that point AAA has coordinates (π4−4,1)\Big( \frac{\pi}{4}- 4, 1\Big)(4π​−4,1) and BBB has coordinates (4−π4,1)\Big( 4-\frac{\pi}{4}, 1\Big)(4−4π​,1). Therefore the distance ABABAB is the difference in the xxx coordinates:

AB=(4−π4)−(π4−4)AB=8−π2‾\begin{aligned}AB &= \Big( 4-\dfrac{\pi}{4}\Big) - \Big(\dfrac{\pi}{4}- 4 \Big) \\\\AB&=\underline{8 - \dfrac{\pi}{2}}\\\end{aligned}ABAB​=(4−4π​)−(4π​−4)=8−2π​​​​​


b) The depth of the ramp is given by the difference in yyy-coordinates. Points AAA and BBB have yyy-coordinate 111 and the lowest point is on the yyy-axis, where x=0x=0x=0. At this point, y=−2y=-2y=−2. 

Therefore, the depth is 1−(−2)=3 m‾1-(-2) = \underline{3 \text{ m}}1−(−2)=3 m​.




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FAQs - Frequently Asked Questions

Can mechanics problems be modelled using parametric equations?

Mechanics problems can be modelled using parametric equations. For example, the equations for x and y can represent the displacement and the parameter can represent time.​

What's the best way to start solving a problem involving modelling a real life problem with parametric equations?

The best way to solve the problem is by drawing a sketch of the curve and labelling the main points. These can be the points of intersection of the curve with the coordinate axes or the ends of the curve, given by the limits for t. Use a table of values to help.

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