00:01
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.
00:36
We have x equals to cos of t and y equals to sin t. T is defined between 2π and zero.
00:47
So, we’re looking at t and gradients. Now, these are obviously parametric equations.
00:55
You have x which is defined in terms of t and then you have y which is defined in terms of t.
01:01
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.
01:32
So, the first thing that we could do is we could just differentiate them separately. We could do dx.
01:39
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.
01:59
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.
02:11
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.
02:19
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.
02:32
Let's ignore these numerical functions for now and look at this. We have dx/dt and we have dy/dt.
02:42
We still don't have dy/dx. But think if there is any way that we can combine these two to get dy/dx.
02:50
So, you want dy at the top and you want dx at the bottom. Here’s something you could do.
02:57
You could flip this. So, rather than writing dx/dt, you can write this as dt/dx.
03:03
Obviously, we're going to change the other side as well. So, that becomes 1/-sin t.
03:10
So, we flipped the entire fraction. So, rather than dx/dt, we’ve written it as dt/dx.
03:16
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.
03:35
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.
03:53
Some textbooks also like to define this as dy/dx equals to dy/dt. That’s divided by dx/dt.
04:04
So in this case, you don't really have to flip the fraction. You just write them the same way.
04:09
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.
04:59
Now, that we've defined what dy/dx is, we can now come back to our numbers or our question.
05:07
We are saying that dy/dx is dy/dt which in our case is this, cos t. So, we don’t change that.
05:17
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.
05:30
So, I’m multiplying it with 1/-sin t. This gives you cos of t/sin of t. Remember that this is negative.
05:41
We’re done with the differentiation. The only things we can do now is tidy this up.
05:46
So, if you remember one of your identities for tan, so tan of, let’s just use t equals to sin t/cos t.
05:54
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.
06:25
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.
06:56
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.
07:30
This is the equation of a circle. So, this is the equation of a circle with unit radius.
07:35
So, we have a circle with a radius of 1. It’s defined by parameters. So, we’ve got the parameter t.
07:49
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.
08:02
You can define what kind of function it is. You can imagine it. You can sketch it. You can find its gradient.
08:07
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.