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 

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