Parametric Differentiation: Exercise 4 (2)

by Batool Akmal

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    00:01 Moving on to our second part where we're actually finding the equation of the tangent.

    00:06 Remember that we had some form of a curve, P. We have the tangent. We have calculated that the gradient at this point is 1 over P using the parameter. We also know our points.

    00:20 So, we know that x = at² and y = 2at. All we really need to do now is put this into our equation.

    00:32 The equation of a straight line is y = mx + c. That’s its general form or the form that we’ve been using through this course is y - y1 = m(x - x1). So, all we need to do is substitute our x values.

    00:48 So, this is my x. This is my y here. Then, I also know my m which I’ve just calculated.

    00:55 In order to find the equation of the tangent, we just put it all in. Now, because I’ve calculated my gradient to be 1 over P which was initially 1 over t, you can if you want to have a little play around here.

    01:08 Change your x’s into parameters. So, you can say from here that at P, x = ap² and y = 2ap.

    01:20 I hope this doesn't confuse you too much. Just look at what I’ve done. I’ve just changed the y, the t to a p, and in the x, I’ve just changed this t to a p. That can work for any point on the curve.

    01:35 So, we’re looking at parameters p. If we’re looking at what it’s doing at point 2 when t = 2, all you have to do is substitute 2 in there. It could work with any numbers or any letters.

    01:46 Now, in the previous example that I did, I capped it all in terms of t’s until the very last moment which is sometimes preferred. But because I’ve already calculated my gradient in terms of p’s, I’m just going to change it all to p’s. It makes no difference when you do it as long as they’re all p’s or all t’s. So, let’s put that in. So, we’ve got y minus and I’ll use my point at parameter p.

    02:13 So, y - 2ap = m, which is 1/p(x - ap²). We’ve got a p here that you can multiply on the other side.

    02:25 So, I’ve got py - 2ap = x - ap². We’ve got py, when I times this through, -2ap² = x - ap².

    02:42 Take it all to one side just so we can equal it to zero. We’ve got py - 2ap² + ap² - x = 0.

    02:54 Lastly, you can combine these two terms here. You’ve got -2ap² + 1ap². So that gives you py.

    03:03 Let’s put the x first. Then you’ve got -2ap² + ap² to give you -ap² = 0. That's the equation of your tangent.

    03:13 In this question, they broke it down into two stages. They asked you for the gradient then the equation of the tangent. Even if they went straight into asking you for the equation of the tangent, you’d have to find the gradient first. So, it’s the same process every time. You also remember that you can find the equations of the normals if you wanted. So, the normals are the lines that meet at 90 degrees. If I just remind you, this is my tangent. This is my normal here at 90 degrees.

    03:40 Remember, all you have to do is flip the gradient. If this gradient is 1 over p, then this gradient here will just be -p. So remember that they multiply together to give you -1. When we studied equations of tangents and normals, we went over the gradients and the combination between the two.

    About the Lecture

    The lecture Parametric Differentiation: Exercise 4 (2) by Batool Akmal is from the course Parametric Differentiation.

    Included Quiz Questions

    1. y - b = m(x - a)
    2. y - b = m(x + a)
    3. y + b = m(x + a)
    4. y - b = (1/m)(x - a)
    5. y - b = (-1/m)(x - a)
    1. 4/(3p)
    2. -4/(3p)
    3. 4ap
    4. -4/3
    5. 4/p
    1. y - 2ap² = 4/(3p) (x - ap³)
    2. y - 2ap² = 1/(3p) (x - ap³)
    3. y - ap³ = 4/(3p) (x - 2ap²)
    4. y - 2ap² = (-3p/4)(x - ap³)
    5. y - ap³ = -4/(3p)(x - 2ap²)

    Author of lecture Parametric Differentiation: Exercise 4 (2)

     Batool Akmal

    Batool Akmal

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