Parametric Differentiation: Exercise 2

00:01 Our next example involves some trigonometry. Let’s write it out and discuss what we’d have to do here.

00:07 Our first equation, y = 5 cos t. Our second equation states that x = 5 sin t.

00:18 Now firstly, you look at these equations and you recognize that these are parametric equations because they’re given in terms of t. So, we have y as a function of t and x as a function of t.

00:31 The definition, we will apply when we’re finding the gradient. But in order to do this, we need to differentiate dy/dt. We also need to differentiate dx/dt. Now straightforward, there’s no chain rule or function of a function. Five is just a constant so that can stay. Cos of t or cos of x differentiates to -sin. So, if I put a minus out here then that gives me -sin t. Then in this case, the 5 stays because it’s just a constant. Sin of t differentiates to cos of t. Remember that my definition states that dy/dx is dy/dt multiplied by dt/dx. In order to do dt/dx, I need to change this to dt/dx.

01:20 With practice, you’ll just be able to do this so you don’t have to write this out every time.

01:25 But dt/dx, I need to flip because this is over 1. So, I’m just going to swap places.

01:31 So, that becomes 1 over 5 cos of t. I usually do this before just so that it eliminates a new risk of making mistakes. Dy/dt is -5 sin t. Dt/dx is 1 over 5 cos t. You can see here that the 5’s can just cancel out leaving you with -sin t over cos t. Here is where you have your chance to show that you've learnt all your identities. Sin t over cos t is just -tan t.

02:05 Put brackets around the t so you can see that as an angle. We're saying that when you differentiate y = 5 cos t and x = 5 sin t, this is two parametric equations. You get a gradient of -tan t.