Once again, just to recap over everything that we've done and we do it more formally now.
If you have a function of x defined as f of t or any function that includes t and a function of y
defined by a different function g of t, you can bring them together firstly by differentiating
each part separately. So the x, we have to differentiate with respect to t so we’ll get dx/dt.
Then the y, we have to differentiate with respect to t as well to get dy of dt. Then to get your dy/dx,
you flip dx/dt to give you dt/dx, so that you can cancel the dt’s and you end up with dy/dx.
So, both methods that I mentioned earlier are now here. You can see that you could either learn dy/dx
as dy/dt divided by dx/dt or if you prefer to multiply, you'd have to remember that it's dy/dt left as it is
and then you flip the other fraction. So, dx/dt, you flip to make it into dt/dx so you can end up with dy/dx.
Let’s look at another example now just to consolidate the ideas. We are looking for dy/dx.
Again, you can see that you have two equations. You can see that this is a parametric equation
because you have not only x's and y's but a third variable, t. Don’t ever worry about what this letter is.
It could be p, or q, or z, or anything. But as long as you spot a third variable, you have a third parameter
that’s holding it together. Let’s try and differentiate this using the rules we've just learned.
We have y = t² - 5 and x = t³ + 5. Remember what we said. Firstly, spot what kind of equation it is.
I know that you know a lot of different methods and lots of different types of equations.
You can question all of them. Is it chain rule or product rule or quotient rule, implicit differentiation?
Hopefully, when you come to the part of thinking about parametric equations,
you'd recognize that this is a parametric equation. So, in order to differentiate it,
we're going to have to do this separately. We'll do dy/dt here. So, that just gives me 2t
and the constant 5 just disappears. We have dx/dt. When you differentiate that, that gives you 3t².
Bring the power to the front and decrease the power by 1 and then the 5 just disappears.
Now, our definition was this. Dy/dx is dy/dt multiplied by dt/dx or if you wanted to do it the other way,
you could do dy/dt divided by dx/dt. Both of them are the same thing. If I’m using this form of it,
I obviously need to flip this before. So, dx/dt needs to become dt/dx. You can't just do that
to one side of the equation. We have to do it to this side as well. So, this is 3t²/1.
So, you can write it as 1/3t². We just flipped the entire equation. Put it together here
in the formula. Dy/dt is 2t. Dt/dx is 1/3t². Nice and straightforward. We can now just
simplify this a little bit to say that dy/dx. You can cancel this one t with one of them is 2/3t.
That’s your gradient using parametric equations on parametric differentiation.
Moving on then. We now might wonder how to find the second differential of a parametric equation.
It’s not as straightforward as just differentiating a function again. So, when you're doing the second
differential, we have to use the next part of this definition. In order to do d²y/dx²,
we essentially need to differentiate dy/dx again as we do with any other differentiation. If you have to do
a second differential of any function, you just differentiate it again using the same rules.
However, because this time you have t’s in your equation, you can’t just dy/dx with respect to x.
So, we use this definition. We use d²y/dx². We differentiate dy/dx which we've already found
in the first part with respect to t because that's important. Then to make up for that dt
that we've just done there, we divide it by dx/dt or we multiply it with dx/dt flipped, so by dt/dx.
It looks fairly complicated but it really isn't when you start to use this with numbers
or you start to apply it to real questions. So, let’s have a look at an example.