Parametric Equations For A Level

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Parametric Equations – Introduction

Parametric equations are a mathematical concept that is introduced to students in year 13 of an A level Maths course. They provide a way to represent curves and functions in terms of one or more parameters, instead of using the traditional x and y variables. By using parametric equations, we can describe the motion of objects, study the behaviour of curves, and solve problems in various fields of mathematics and science.

Introduction to Parametric Equations:
Parametric equations involve expressing the coordinates of points on a curve or surface as functions of an independent parameter, often denoted as t. Instead of the usual Cartesian coordinates (x, y), we have functions x(t) and y(t) that define the position of a point at each value of t.

Parametric Equations for Curves:
Parametric equations allow us to describe complex curves that cannot be easily represented by a single equation. By varying the parameter t, we can trace out different points on the curve. For example, a simple parametric equation for a circle is x(t) = rcos(t) and y(t) = rsin(t), where r is the radius and t varies from 0 to 2π.

Advantages of Parametric Equations:
Parametric equations offer several advantages over traditional equations. They can represent curves with self-intersections, such as the lemniscate. Parametric equations can also describe motion in terms of time, allowing us to model the path of moving objects accurately.

Applications of Parametric Equations:
Parametric equations find applications in various fields. In physics, they help describe the motion of projectiles, planets, and particles. In computer graphics, parametric equations are used to create smooth curves and animations. They also play a crucial role in engineering designs, such as modelling the path of a robotic arm.

Converting between Parametric and Cartesian Equations:
Sometimes, it may be necessary to convert parametric equations into Cartesian form. This involves eliminating the parameter by solving the equations simultaneously. For example, given x(t) = t^2 and y(t) = t + 1, we can eliminate t to obtain the Cartesian equation y = x^(1/2) + 1.

Parametric Equations in Calculus:
Parametric equations are essential in calculus, particularly when dealing with curves and their properties. Calculating derivatives and integrals of parametric equations require applying the chain rule and other techniques specific to parametric functions.

Parametric Equations - Basic Examples

\begin{array}{rl} x=t+2 & y=t^2+3 \\ x-2=t & y=(x-2)^2+3 \end{array}

Parametric Equations Involving Calculus

When y = 0:

\begin{aligned} & 0=t^2-1 \\ & 1=t^2 \\ & \therefore t= \pm 1 \\ & t=1: \quad x=2(1)+1=3 \\ & t=-1: x=2(-1)+1=-1 \end{aligned}

When x = 0: 

\begin{aligned} & 0=2 t+1 \quad \therefore t=-\frac{1}{2} \\ & \therefore y=\left(-\frac{1}{2}\right)^2-1=-\frac{3}{4} \end{aligned}

Crosses axes at the following points:

(-1,0) \quad(3,0)(0,-3 / 4)

Part b) 

\frac{d x}{d t}=2 \quad \frac{d y}{d t}=2 t

 

\frac{d y}{d x}=\frac{d y}{d t} \times \frac{d t}{d x}=2 t \times \frac{1}{2}=t

 

\therefore \frac{d y}{d x}=\frac{x-1}{2}

 

Part a) 

\begin{array}{rlrl} x & =\tan ^2 t & y & =\cos t \\ \frac{d x}{d t} & =2 \tan t \sec ^2 t & \frac{d y}{d t} & =-\sin t \end{array}

 

\begin{aligned} \frac{d y}{d x} & =\frac{-\sin t}{2 \tan t \sec ^2 t} \\ & =-\frac{1}{2} \sin t \cot t \cos ^2 t \\ & =-\frac{1}{2} \sin t \frac{\cos t}{\sin t} \cos ^2 t \\ & =-\frac{1}{2} \cos ^3 t \end{aligned}

Part b) 

\begin{aligned} t=\frac{\pi}{4} \quad \frac{d y}{d x} & =-\frac{1}{2}\left(\cos \frac{\pi}{4}\right)^3 \\ & =-\frac{\sqrt{2}}{8} \end{aligned}

 

\begin{array}{ll} y=-\frac{\sqrt{2}}{8} x+c \quad \text { when } t=\frac{\pi}{4} \quad x=\left(\tan \frac{\pi}{4}\right)^2=1 \\ \frac{\sqrt{2}}{2}=-\frac{\sqrt{2}}{8}+c & y=\cos \frac{\pi}{4}=\frac{\sqrt{2}}{2} \\ c=\frac{5 \sqrt{2}}{8} \quad y=-\frac{\sqrt{2}}{8} x+\frac{5 \sqrt{2}}{8} & \end{array}

Part c) 

1+\tan ^2 t=\sec ^2 t

 

x=\tan ^2 t \quad \begin{aligned} y & =\cos t \\ y^2 & =\cos ^2 t \\ \frac{1}{y^2} & =\sec ^2 t \end{aligned}

 

1+x=\frac{1}{y^2}
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Parametric Equations With Trig

 
 
 
Part a) 
\begin{array}{rlrl} x & =\sin ^2 t & y & =\sin 2 t \\ \frac{d x}{d t} & =2 \sin t \cos t & \frac{d y}{d t}=2 \cos 2 t \\ \frac{d y}{d x} & =\frac{2 \cos 2 t}{\not 2 \sin t \cos t} & \\ & =\frac{\cos 2 t}{\sin t \cos t} & \end{array}
 
Part b) 
\begin{aligned} & t=\frac{\pi}{6} \quad x=\frac{1}{4} \quad y=\frac{\sqrt{3}}{2} \\ & \frac{d y}{d x}=\frac{\cos \left(\frac{\pi}{3}\right)}{\sin \left(\frac{\pi}{6}\right) \cos \left(\frac{\pi}{6}\right)}=\frac{2 \sqrt{3}}{3} \\ & y=-\frac{3}{2 \sqrt{3}} x+c \\ & y=-\frac{\sqrt{3}}{2} x+c \\ & \frac{\sqrt{3}}{2}=-\frac{\sqrt{3}}{8}+c \\ & c=\frac{5 \sqrt{3}}{8} \\ & y=-\frac{\sqrt{3}}{2} x+\frac{5 \sqrt{3}}{8} \end{aligned}
Part c) 
\sin ^2 t+\cos ^2 t=1
 
\begin{aligned} x=\sin ^2 t \quad y & =\sin 2 t \\ y & =2 \sin t \cos t \\ y^2 & =4 \sin ^2 t \cos ^2 t \\ \frac{y^2}{4 \sin ^2 t} & =\cos ^2 t \\ x+\frac{y^2}{4 x} & =1 \end{aligned}

In conclusion, parametric equations provide a powerful tool for describing curves and functions using parameters instead of traditional variables. They offer flexibility in representing complex curves, modelling motion, and solving problems in various fields. Understanding parametric equations is crucial for advancing in mathematics and applying mathematical concepts to real-world situations.

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