1 Parametric differentiation

In this subsection we consider the parametric approach to describing a curve:

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

$∕\setminus$

$\text{parametric equations}\text{parametric range}$

As various values of $t$ are chosen within the parameter range the corresponding values of $x,y$ are calculated from the parametric equations. When these points are plotted on an $xy$ plane they trace out a curve. The Cartesian equation of this curve is obtained by eliminating the parameter $t$ from the parametric equations. For example, consider the curve:

$x=2costy=2sint0\le t\le 2\pi .$

We can eliminate the $t$ variable in an obvious way - square each parametric equation and then add:

$x^2+y^2=4{cos}^{2}t+4{sin}^{2}t=4\therefore x^2+y^2=4$

which we recognise as the standard equation of a circle with centre at $\left(0,0\right)$ with radius 2.

In a similar fashion the parametric equations

$x=2ty=4t^2-\infty <t<\infty$

describes a parabola . This follows since, eliminating the parameter $t$ :

$t=\frac{x}{2}\therefore y=4\left(\frac{x^2}{4}\right)\text{so}y=x^2$

which we recognise as the standard equation of a parabola.

The question we wish to address in this Section is ‘how do we obtain the derivative $\frac{dy}{dx}$ if a curve is given in parametric form?’ To answer this we note the key result in this area:

We note that this result allows the determination of $\frac{dy}{dx}$ without the need to find $y$ as an explicit function of $x$ .

Example 13

Determine the equation of the tangent line to the semicircle with parametric equations

$x=costy=sint0\le t\le \pi$

at $t=\pi ∕4$ .

Solution

The semicircle is drawn in Figure 9. We have also drawn the tangent line at $t=\pi ∕4$ (or, equivalently, at $x=cos\frac{\pi }{4}=\frac{1}{\sqrt{2}},y=sin\frac{\pi }{4}=\frac{1}{\sqrt{2}}.$ )

Now

$\frac{dy}{dx}=\frac{dy}{dt}÷\frac{dx}{dt}=\frac{cost}{-sint}=-cott.$

Thus at $t=\frac{\pi }{4}$ we have $\frac{dy}{dx}=-cot\left(\frac{\pi }{4}\right)=-1.$

The equation of the tangent line is

$y=mx+c$

where $m$ is the gradient of the line and $c$ is a constant.

Clearly $m=-1$ (since, at the point $P$ the line and the circle have the same gradient).

To find $c$ we note that the line passes through the point $P$ with coordinates $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$ . Hence

$\frac{1}{\sqrt{2}}=\left(-1\right)\frac{1}{\sqrt{2}}+c\therefore c=\frac{2}{\sqrt{2}}$

Finally,

$y=-x+\frac{2}{\sqrt{2}}$

is the equation of the tangent line at the point in question.

We should note, before proceeding, that a derivative with respect to the parameter $t$ is often denoted by a ‘dot’. Thus

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

Task!

Find the value of $\frac{dy}{dx}$ if   $x=3t,y=t^2-4t+1$ .

Check your result by finding $\frac{dy}{dx}$ in the normal way.

First find $\frac{dx}{dt},\frac{dy}{dt}$ :

Answer

Answer

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Task!

Find the value of $\frac{dy}{dx}$ at $t=2$ if   $x=3t-4sin\pi t,y=t^2+tcos\pi t,0\le t\le 4$

First find $\frac{dx}{dt},\frac{dy}{dt}$ :

Answer

Answer

Answer