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Parametric Differentiation: Example 2

by Batool Akmal
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    00:00 Let’s look at our next example. We are now looking at parametric equations but this time the question is asking us to find the equation of a tangent. It’s almost like the application of gradients that we’re going to use. Firstly, let's just try and understand this question. We have some curve and it’s given by the two parametric equations. It then says that there’s a point P that lies on the curve C.

    00:27 It’s telling us the parameters for that. So, it’s telling us what’s holding it together is little p. It then asks us to find the equation of the tangent C at that point P and it gives us the equation but we have to show this.

    00:39 Let’s try this out and explain, see if we understand what this question actually means.

    00:45 We’re given two equations, x = at, so two parametric equations, and y = b/t.

    00:55 It tells us that these equations define some type of curve. So, just imagine some form of a curve.

    01:03 We have a point P that lies on C. So, we’ve got point P here. The parameters of P are defined as p.

    01:13 This is the parameter of it. We have a tangent to this point here. We’re looking for the equation of this tangent. So, remember the equation of any tangent follows the form y = mx + c.

    01:28 So, that does mean that we need to find the gradient of this tangent. We need to know what this is.

    01:34 We’re just going to substitute it all into the equation y = mx + c. Now, you may question before we start, what are our x points and what are our y points. We do have them but in terms of parameters. So, our x point is this and our y point is this. You can change the parameters to find these at different points in the equation. The parameters for this point are p.

    01:57 So, all we have to do is change this t to p and we’ll find what the point y and x is. But if we’re given actual numbers, we can just substitute numbers in to find what this is as a numerical value.

    02:10 So, let’s just start. The first thing that we need to do is find the gradient. In order to calculate the gradient, we’re going to have to differentiate. We’re looking for dy/dx. We know the rules to find dy/dx when it comes to parametric equations. So, let’s make a start. Firstly dx, we’ll differentiate to dx/dt. In this case, it's t to the power of 1, so you bring the power down and you decrease the power by 1 to give you just a. If that’s confusing, just imagine trying to differentiate something like x = 5t. When you differentiate that dx/dt, that just gives you 5.

    02:56 So, you’re a is just a number. A is just any constant. If I’m looking at at, that just differentiates to a.

    03:04 Before we do the second equation, we need to make it a little bit easier. We can rewrite this as b to the t - 1 because remember the t is at the bottom. We can bring it up as a power of -1.

    03:18 We can now do dy/dt, just apply the rules of differentiation. Bring the power to the front and decrease the power by 1. So, that gives me -1bt to the minus 2, which is the same as -b over t². We’ve calculated our dx/dt and dy/dt. We now need to find out what dy/dx is.

    03:45 So, the definition for that was dy/dt multiplied by dt/dx. If this is over 1, we’re going to have to flip dx/dt to give us dt/dx. So, dy/dt is straightforward. It’s just -b over t². Then we multiply this with rather than a, we do it as 1 over a. So, that’s going to be 1 over a. Our gradient becomes -b over at².

    04:18 Now that we know the gradient, we can now substitute it all back into our equation.

    04:26 If we use the equation of a straight line, either y = mx + c or y - y1 = m(x - x1), let’s just identify where everything is. This is our gradient, dy/dx which we’ve just worked out.

    04:42 Our y value is this value here. So, this is my y which is just b over t. My x value is this value here.

    04:51 So, I have all the values that I need to put in. They’re all in terms of a’s, and b’s, and t’s but we’ll fix that in a minute. We can rewrite this as y minus. So, our y value is b over t. Our gradient is minus b over at² and x minus, my x coordinate is at. So, on this side of the equation, we can take t as a common denominator. We can rewrite this as yt - b over t. On the other side, we can just leave it as it is. We’ve got b/at²(x - at). Now, all we’re trying to do is simplify this hopefully and reach the equation that was given in the question. We’re trying to achieve that answer.

    05:41 To tidy this up, we can take the at² to that side and multiply and the t to that side and multiply.

    05:46 So, we’re getting rid of the denominators. Let’s make some space here. You can multiply this side with at². You have yt - b. On the other side, you are multiplying t. So, let’s just put this as -tb. Just multiply this t with that b there because that will eventually multiply everything.

    06:07 We have -bt and then we’re multiplying this with x - at. Let’s times this through.

    06:16 So that gives me at³y. When you times that with this term, multiply this here.

    06:24 We have -abt². On this side of the equation, we have -tbx. Then when you multiply that, you get +abt². We’re getting there. The first thing you can spot is that you have a t in every term.

    06:47 So, we have a t, a t, a t, and a t. You can divide this entire equation by t just to simplify things a little bit. That leaves you with at²y - ab. I’m also going to bring everything over. So, imagine that we’ve cancelled. So, we’ve got two of those, one of them, that t cancels, and one of that t. I will bring that over to this side. So that gives me +bx and then -ab. That should have a t there, an abt equals to zero. You can see that we can add these two terms. So, we have -abt and -abt. So, we end up with at²y + bx -2abt = 0.

    07:38 Now, you might think that this isn’t quite looking like the equation that's been given, only because the t’s are your p’s. Remember what we said that all you need to do to change it to that parameter at that point, you just need to change your t’s to that number.

    07:54 So, if you were looking at the point t equals to 2, you can just substitute 2 into this value here and you’d get your answer or you’d get the equation of the tangent of a point. In our case, they don’t specify what the number is. They just tell us that the parameter is p.

    08:11 So you can say that at p, you’ve replaced all your t’s with p’s. We have ap²y + bx - 2abp = 0.

    08:25 That is your equation of the tangent. Now, you may prefer to change your parameters sooner in the question. By that, I mean that you could have changed your t to p here, so when we calculated the gradient and you could then substitute the gradient as -p over ap². Then you may have chosen to change your x to ap. You may have chosen to change your y to b over p. Then substitute it into this equation here. It really doesn't matter. So, you could change all your t’s to p’s first before you put them in or you could just do all your calculations with t’s or any other letter and just change the parameters right at the end. In my case, I left all these t’s so I didn’t really bring in the parameters until the very last step where I’m done, I've finished my solution. I’ve got lots of t’s left.

    09:21 All I had to do is change my t to any parameter. So that gives me the freedom then to change it to anything. So, you could ask me the equation of the tangent at 3. All I'd have to do is change my t to 3 or you can ask me the equation of the tangent at parameter p and all I had to do is change my t’s to p’s.

    09:40 Now, you have some parametric equation questions to try out for yourselves. Each one of them is building up in difficulty but we’ve done some very similar examples. Especially when you come to calculating the first and second differential, make sure that you remember the equation of the second differential and you take it slow, one step at a time. I’ll let you have a little try and then we’ll go through the solutions together.


    About the Lecture

    The lecture Parametric Differentiation: Example 2 by Batool Akmal is from the course Parametric Differentiation.


    Included Quiz Questions

    1. 2bt/a
    2. bt/a
    3. b/2a
    4. t/a
    5. bt²/a
    1. -bt²/a
    2. bt²/a
    3. bt/a
    4. t²/ab
    5. abt²
    1. y-t³=(3t/2)(x-t²)
    2. y-t³=(t/2)(x-t²)
    3. y-t³=(-3t/2)(x-t²)
    4. y-t³=(-t/2)(x-t²)
    5. y-t³= t(x-t²)
    1. y-t³=(-2/3t)(x-t²)
    2. y-t³=(3t/2)(x-t²)
    3. y-t³=(-3t/2)(x-t²)
    4. y-t³=(2/3t)(x-t²)
    5. y-t³=(1/3t)(x+t²)
    1. 14/361
    2. 361/14
    3. -361/14
    4. -14/361
    5. -14

    Author of lecture Parametric Differentiation: Example 2

     Batool Akmal

    Batool Akmal


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