Let’s look at an example and then we’ll derive our formula for parametric differentiation
and then we’ll come back to the rules. But just to show you what a parametric equation
might look like, you have our first example. You have x as a function of t so in our case,
it's cos t. Then you have y as a function of t and in our case, it’s sin of t. It’s important
when we’re dealing with cos and sines to have a limit. So, it’s telling us that t, the range of t
is between 2π and zero. So, it’s the full circle for sine and cos. Let’s have a look and write them down.
We have x equals to cos of t and y equals to sin t. T is defined between 2π and zero.
So, we’re looking at t and gradients. Now, these are obviously parametric equations.
You have x which is defined in terms of t and then you have y which is defined in terms of t.
Now, remember if someone asks you to find the gradient or asks you to find the steepness or m,
we are always looking at dy/dx. Now, at the moment looking at these two equations here,
it seems quite impossible to derive this little expression of dy/dx. But as we go through it,
you'll see that it's fairly straightforward. We have an x and we have a y. We are looking for dy/dx.
So, the first thing that we could do is we could just differentiate them separately. We could do dx.
We could differentiate dx with respect to t because that will work because we have cos of t
that we're differentiating. So, we need to differentiate it with respect to t. Then also, we can do
the same here. We can do dy/dt and then find our answer there. So, these are two important steps.
We are taking our x. We’re differentiating x with respect to t. We’re taking our y and we’re just
differentiating y with respect to t. Let's actually look at our functions and differentiate it.
Cos of t or cos of x, remember, goes to -sin of x. In our case, it goes to -sin t or sin of t.
Then sin of t differentiates to cos of t. These are our rules. Look back at them. But remember
with practice, we should be able to learn what these general functions differentiate to.
Let's ignore these numerical functions for now and look at this. We have dx/dt and we have dy/dt.
We still don't have dy/dx. But think if there is any way that we can combine these two to get dy/dx.
So, you want dy at the top and you want dx at the bottom. Here’s something you could do.
You could flip this. So, rather than writing dx/dt, you can write this as dt/dx.
Obviously, we're going to change the other side as well. So, that becomes 1/-sin t.
So, we flipped the entire fraction. So, rather than dx/dt, we’ve written it as dt/dx.
Watch what happens if I multiply the two. If I do dy/dt and I multiply this with dt/dx,
by simple algebra, you can see that the dt’s here can cancel out. Then that leaves you with dy/dx.
We've almost defined how to differentiate parametric equations. The formula for differentiating
parametric equations is simply this. Dy/dt multiplied by dt/dx equals to dy/dx.
Some textbooks also like to define this as dy/dx equals to dy/dt. That’s divided by dx/dt.
So in this case, you don't really have to flip the fraction. You just write them the same way.
In that case, like I said before, you don’t have to flip the fraction. You can just divide it
because if you look at this closely, we are saying dy/dt divided by dx/dt, which when you change
to a multiplication sign, you get dy/dt multiplied by dt/dx, which is exactly the same thing
as we've said here. So, it’s really up to you whatever you find easier. If you prefer to flip dx/dt first,
do dt/dx and then times it, that’s fine. If you prefer to just divide it and not flip,
so it’s simply just take your dx/dt here or this term there and you just divide dy/dt
by this term here. So, either way is fine. You’ll get exactly the same answer.
Now, that we've defined what dy/dx is, we can now come back to our numbers or our question.
We are saying that dy/dx is dy/dt which in our case is this, cos t. So, we don’t change that.
It’s cos of t. Then we are multiplying it with dt/dx or dividing it with dx/dt. I’ve done dt/dx here.
So, I’m multiplying it with 1/-sin t. This gives you cos of t/sin of t. Remember that this is negative.
We’re done with the differentiation. The only things we can do now is tidy this up.
So, if you remember one of your identities for tan, so tan of, let’s just use t equals to sin t/cos t.
So, if you had 1 over tan of t, that is cos of t over sin of t. So, you can see here
that this is just -1/tan of t. If you really try and remember, how you remember that 1 over tan
is the same as cot of t. So, what we've done here is actually found what dy/dx is.
So, the gradient of dy/dx is -cot t just by using parametric differentiation. Now, interestingly enough,
just a little extra point here, we can also combine these two equations together just to see
what this actually is. So, just to show you, when you have x = cos t and y = sin t,
if you square both the equations, so let’s just square this. So then, we square this.
Then if we square this, we square this. So, we've just squared everything. Now, if you try to add them,
if we have x² + y², this is the same as cos² t + sin² t. Look at your identities. Hopefully, you’ll remember
that cos² x + sin² x is the same as 1. So now, if I just write this part of the equation on this part
of the equation, you will get x² + y² = 1. Hopefully, you’ll recognize what this is.
This is the equation of a circle. So, this is the equation of a circle with unit radius.
So, we have a circle with a radius of 1. It’s defined by parameters. So, we’ve got the parameter t.
Also, we have found that the gradient of this circle is -cot t. So, it’s really quite interesting,
with very little information and everything that you know, the kind of things that you could do.
You can define what kind of function it is. You can imagine it. You can sketch it. You can find its gradient.
You can find the gradients at different points. We can extend this. If someone asked us to find
the equation of a tangent or a normal, we can do that. So, with very little information given to us,
you can find a lot about these functions.