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Differentiation II

Parametric differentiation

Parametric differentiation

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Explainer Video

Tutor: Toby

Summary

Parametric differentiation

​​In a nutshell

Sometimes functions are defined parametrically, meaning that both yyy and xxx are given in terms of some parameter, say, ttt. While it may be possible to re-express a given function in terms of yyy and xxx 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 ttt, you will have two functions x(t)x(t)x(t) and y(t)y(t)y(t). To find the gradient of this function, you seek dydx\frac{\text dy}{\text dx}dxdy​. Currently, you can find dxdt\frac{\text dx}{\text dt}dtdx​ and dydt\frac{\text dy}{\text dt}dtdy​ by differentiating the parametric equations with respect to ttt. Using the chain rule, you have

​dydx=dydtdxdt\boxed{\frac{\text dy}{\text dx}=\dfrac{\frac{\text dy}{\text dt}}{\frac{\text dx}{\text dt}}}dxdy​=dtdx​dtdy​​​​​


You can justify this to yourself by considering the following:

​dydx×dxdt=dydt\frac{\text dy}{\text dx}\times\frac{\text dx}{\text dt}=\frac{\text dy}{\text dt}dxdy​×dtdx​=dtdy​​​


This rearranges to the formula above. 


Example 1

Find the gradient of the curve given by the parametric equations

​x=2t+1y=4t2+10t+4\begin{aligned}x&=2t+1\\y&=4t^2+10t+4\end{aligned}xy​=2t+1=4t2+10t+4​ ​


Start by differentiating each of these functions with respect to ttt:

dxdt=2dydt=8t+10\begin{aligned}\frac{\text dx}{\text dt}&=2\\\frac{\text dy}{\text dt}&=8t+10\end{aligned}dtdx​dtdy​​=2=8t+10​​​


Thus, by the chain rule:

dydx=8t+102=4t+5\begin{aligned}\frac{\text dy}{\text dx}&=\frac{8t+10}{2}\\&=4t+5\end{aligned}dxdy​​=28t+10​=4t+5​​​


Finally, you express this in terms of xxx using the original parametric equation x=2t+1x=2t+1x=2t+1. This implies that t=x−12t=\frac{x-1}{2}t=2x−1​ and hence:

dydx=4(x−12)+5=2x+3‾\begin{aligned}\frac{\text dy}{\text dx}&=4\left(\frac{x-1}{2}\right)+5\\&=\underline{2x+3}\end{aligned}dxdy​​=4(2x−1​)+5=2x+3​​​​


Note: It is possible to express this function non-parametrically: y=x2+3xy=x^2+3xy=x2+3x. See that differentiating this gives the same as concluded above.


Example 2

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

x=3cos⁡(θ)y=sin⁡(θ)\begin{aligned}x&=3\cos(\theta)\\y&=\sin(\theta)\end{aligned}xy​=3cos(θ)=sin(θ)​​​


Differentiate each of these with respect to θ\thetaθ:​

dxdθ=−3sin⁡(θ)dydθ=cos⁡(θ)\begin{aligned}\frac{\text dx}{\text d\theta}&=-3\sin(\theta)\\\frac{\text dy}{\text d\theta}&=\cos(\theta)\end{aligned}dθdx​dθdy​​=−3sin(θ)=cos(θ)​​​


Therefore, the chain rule says:

dydx=cos⁡(θ)−3sin⁡(θ)\begin{aligned}\frac{\text dy}{\text dx}&=\frac{\cos(\theta)}{-3\sin(\theta)}\end{aligned}dxdy​​=−3sin(θ)cos(θ)​​​​


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


You already have that cos⁡(θ)=x3\cos(\theta)=\frac{x}3cos(θ)=3x​. But you can also use this to find an expression for sin⁡(θ)\sin(\theta)sin(θ):

cos⁡(θ)=x3cos⁡2(θ)=x291−cos⁡2(θ)=1−x29=9−x29sin⁡2(θ)=9−x29sin⁡(θ)=9−x23\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}cos(θ)cos2(θ)1−cos2(θ)sin2(θ)sin(θ)​=3x​=9x2​=1−9x2​=99−x2​=99−x2​=39−x2​​​​​


So

dydx=x3−39−x23=−x39−x2\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}dxdy​​=−339−x2​​3x​​=−39−x2​x​​​​


When x=2x=2x=2, the gradient is 

dydx=−x39−x2=−239−22=−21515‾≈−0.298‾\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}dxdy​​=−39−x2​x​=−39−22​2​=−15215​​​≈−0.298​​​​

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FAQs - Frequently Asked Questions

How do you differentiate parametrically?

Given the functions y(t) and x(t), the chain rule says that dy/dx=(dy/dt)/(dx/dt).

What is a parameter?

A separate variable that your function is given in terms of.

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