2 General parametric form

We will assume that any curve in the $xy$ plane may be written in parametric form:

$\underset{/}{\underset{\⏟}{x=g\left(t\right)y=h\left(t\right)}}\underset{\backslash }{\underset{\⏟}{t_0\le t\le t_1}}$

parametric equations of the curve  parameter range

in which $g\left(t\right),h\left(t\right)$ are given functions of $t$ and the parameter $t$ ranges over the values $t_0\to t_1$ . As we give values to $t$ within this range then corresponding values of $x,y$ are calculated from $x=g\left(t\right)$ , $y=h\left(t\right)$ which can then be plotted on an $xy$ plane.

In HELM booklet  12.3, we discovered how to obtain the derivative $\frac{dy}{dx}$ from a knowledge of the parametric derivatives $\frac{dy}{dt}$ and $\frac{dx}{dt}$ . We found

$\frac{dy}{dx}=\frac{dy}{dt}÷\frac{dx}{dt}\text{and }\frac{d^{2}y}{d{x}^2}=\left(\frac{dx}{dt}\frac{d^{2}y}{d{t}^2}-\frac{dy}{dt}\frac{d^{2}x}{d{t}^2}\right)÷{\left(\frac{dx}{dt}\right)}^3$

Note that derivatives with respect to the parameter $t$ are often denoted by a dot:

$\frac{dx}{dt}\equiv ẋ\frac{dy}{dt}\equiv ẏ\frac{d^{2}x}{d{t}^2}\equiv ẍ\text{etc}$

so that

$\frac{dy}{dx}=\frac{ẏ}{ẋ}\text{and}\frac{d^{2}y}{d{x}^2}=\frac{ẋÿ-ẏẍ}{{ẋ}^3}$

Knowledge of the derivative is sometimes useful in curve sketching.

Example 8

Sketch the curve $x=t^3+3t^2+2ty=3-2t-t^2-3\le t\le 1.$

Solution

$x=t^3+3t^2+2t=t\left(t+2\right)\left(t+1\right)$ $y=3-2t-t^2=-\left(t+3\right)\left(t-1\right)$

so that $x=0$ when $t=0,-1,-2$ and $y=0$ when $t=-3,1$ . We calculate the values of $x,y$ at various values of $t$ :

$\frac{dx}{dt}=3t^2+6t+2,\frac{dy}{dt}=-2-2t\therefore \frac{dy}{dx}=\frac{-2\left(1+t\right)}{3t^2+6t+2}$

so there is a turning point when $t=-1$ . The reader is urged to calculate $\frac{d^{2}y}{d{x}^2}$ and to show that this is negative when $t=-1$ (i.e. at $x=0,y=4$ ) indicating a maximum when. (The reader should check that vertical tangents occur at $t=-0.43$ and $t=-1.47$ , to 2 d.p.)

We can now make a reasonable sketch of the curve: