Parametric differentiation

In a nutshell

Sometimes functions are defined parametrically, meaning that both $y$ and $x$ are given in terms of some parameter, say, $t$ . While it may be possible to re-express a given function in terms of $y$ and $x$ only, you can use the chain rule to find the gradient of the function without doing this.

Using the chain rule

If your function is given parametrically, where the parameter is $t$ , you will have two functions $x(t)$ and $y(t)$ . To find the gradient of this function, you seek $\frac{\text dy}{\text dx}$ . Currently, you can find $\frac{\text dx}{\text dt}$ and $\frac{\text dy}{\text dt}$ by differentiating the parametric equations with respect to $t$ . Using the chain rule, you have

$\boxed{\frac{\text dy}{\text dx}=\dfrac{\frac{\text dy}{\text dt}}{\frac{\text dx}{\text dt}}}$

You can justify this to yourself by considering the following:

$\frac{\text dy}{\text dx}\times\frac{\text dx}{\text dt}=\frac{\text dy}{\text dt}$

This rearranges to the formula above.

Example 1

Find the gradient of the curve given by the parametric equations

$\begin{aligned}x&=2t+1\\y&=4t^2+10t+4\end{aligned}$

Start by differentiating each of these functions with respect to $t$ :

$\begin{aligned}\frac{\text dx}{\text dt}&=2\\\frac{\text dy}{\text dt}&=8t+10\end{aligned}$ 

Thus, by the chain rule:

$\begin{aligned}\frac{\text dy}{\text dx}&=\frac{8t+10}{2}\\&=4t+5\end{aligned}$ 

Finally, you express this in terms of $x$ using the original parametric equation $x=2t+1$ . This implies that $t=\frac{x-1}{2}$ and hence:

$\begin{aligned}\frac{\text dy}{\text dx}&=4\left(\frac{x-1}{2}\right)+5\\&=\underline{2x+3}\end{aligned}$ 

Note: It is possible to express this function non-parametrically: $y=x^2+3x$ . See that differentiating this gives the same as concluded above.

Example 2

Find the gradient at the point where $x=2$ of the curve given by the parametric equations

$\begin{aligned}x&=3\cos(\theta)\\y&=\sin(\theta)\end{aligned}$ 

Differentiate each of these with respect to $\theta$ :

$\begin{aligned}\frac{\text dx}{\text d\theta}&=-3\sin(\theta)\\\frac{\text dy}{\text d\theta}&=\cos(\theta)\end{aligned}$ 

Therefore, the chain rule says:

$\begin{aligned}\frac{\text dy}{\text dx}&=\frac{\cos(\theta)}{-3\sin(\theta)}\end{aligned}$ 

Now use the parametric equations to express this in terms of $x$ .

You already have that $\cos(\theta)=\frac{x}3$ . But you can also use this to find an expression for $\sin(\theta)$ :

$\begin{aligned}\cos(\theta)&=\frac{x}3\\\cos^2(\theta)&=\frac{x^2}{9}\\1-\cos^2(\theta)&=1-\frac{x^2}9\\&=\frac{9-x^2}9\\\sin^2(\theta)&=\frac{9-x^2}9\\\sin(\theta)&=\frac{\sqrt{9-x^2}}3\end{aligned}$

So

$\begin{aligned}\frac{\text dy}{\text dx}&=\frac{\frac{x}3}{-3\frac{\sqrt{9-x^2}}3}\\&={-\frac{x}{3\sqrt{9-x^2}}}\end{aligned}$ 

When $x=2$ , the gradient is

$\begin{aligned}\frac{\text dy}{\text dx}&=-\frac{x}{3\sqrt{9-x^2}}\\&=-\frac{2}{3\sqrt{9-2^2}}\\&=\underline{-\frac{2\sqrt{15}}{15}}\\&\approx\underline{-0.298}\end{aligned}$ 