Parametric Differentiation: Example

by Batool Akmal

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    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.

    About the Lecture

    The lecture Parametric Differentiation: Example by Batool Akmal is from the course Parametric Differentiation.

    Included Quiz Questions

    1. 2t
    2. x
    3. t
    4. 2
    1. 6t²
    2. 158
    3. 6t
    4. 2t²
    5. 6t³
    1. dy/dx = [dy/dt].[dt/dx]
    2. dy/dx = [dy/dt].[dx/dt]²
    3. dy/dx = [dy/dx].[dx/dt]
    4. dy/dx = [dy/dt]².[dt/dx]
    5. dy/dx = [dy/dt].[dx/dt]
    1. 1/t
    2. 2
    3. 2/(2t+1)
    4. 1/(2t)
    5. 4t
    1. dy/dx = -tan(t²)
    2. dy/dx = -2tsin(t²)
    3. dy/dx = -cot(t²)
    4. dy/dx = -4t²sin(t²)cos(t²)
    5. dy/dx = -tan(2t)

    Author of lecture Parametric Differentiation: Example

     Batool Akmal

    Batool Akmal

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