Parametric Differentiation: Exercise 1

00:00 I hoped you've managed to solve these questions. Now, it’s our chance to check how you've done.

00:06 So, let’s go through them together. Our first example is giving us two equations. It’s asking us to differentiate. Obviously, it’s important that you recognize that this is a parametric equation.

00:19 So, we’ll have to differentiate this parametrically. Let’s have a look at our two equations.

00:24 We have y = 3t + 5 and then we have x = 3t - 5. The rules here are straightforward.

00:37 You differentiate each one of them separately, so dy/dt and dx/dt and then you bring them together to give you dy/dx. So, dy/dt gives you 3. We’re differentiating with respect to t, so bring the power down and decrease the power by 1. Dx/dt also gives me 3. Now remember that when I used my definition dy/dx, that is the same as dy/dt multiplied by dt/dx. So, I want to flip this. Rather than dx/dt, I want to write this as dt/dx which is just 1 over 3. Now, when I put it all here, I’ll get dy/dt which is 3 and dt/dx which is 1 over 3 giving me a nice, easy gradient of just 1.

01:29 This is fairly straightforward. It’s just telling us the steepness of this curve is just 1 at this point.