Parametric Differentiation: Exercise 4 (1)

by Batool Akmal

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    00:01 Our next example is broken down into two parts. We have to find the gradient of the tangent to C.

    00:08 Then if you look at the next part, it also asks you to find the equation of the tangent to C at point P.

    00:15 Now, we’ve done an example like this before. So, it shouldn’t be too difficult. But we’re obviously dealing with different functions this time. It’s still got a’s in it which are constant, so treat them as any numbers. Don’t get confused with the a’s. It’s just the t’s and the parameters that we’re interested in. So, let’s try this out. We have a curve C. That is defined by these parametric equations. We have x = at² and y = 2at. We have a point P that lies on C and has parameter p. So, any type of curve here, we have our point P there and it has parameter p. The first question is asking us simply to find the gradient of the tangent.

    00:59 So, we’re just finding the gradient. That we can use for the second part as well where we’re actually finding the equation of the tangent. So, the gradient of the tangent is shared with the gradient of the curve at point P. If it touches P at that point, that means that they have the same gradient.

    01:18 So, we simply have to find the gradient here. We do dx/dt. A is any constant. So, you can imagine a number there. So, when you’re differentiating any number t², you just bring the power down and then you decrease the power by 1. That’s dx/dt. Dy/dt, when you differentiate say 10t or 15t, you’re just going to end up with 15. So because t just has a power of 1, you bring the power down, decrease it by 1, you’re just left with 2a. Our definition states that dy/dx is dy/dt multiplied by dt/dx.

    02:04 So, when we put dy/dt in here, that’s this term. We’ve got 2a. Don’t forget that we don’t want dx/dt.

    02:13 We want dt/dx. So, you can flip that. You can write that as 1 over 2at. Dt/dx is 1 over 2at which we can then multiply as 1 over 2at. You’ll notice that the 2a and the 2a cancels out, leaving us with a gradient of just 1 over t. We can say the gradient of the tangent, so dy/dx is the gradient of the tangent which is m, at point P is 1 over t. Well, as a parameter is 1 over t. At point P, dy/dx is equal to 1 over P. Remember what I said? All you have to do is change your parameter to whatever letter or any number that you are looking at. So the gradient 1 over t at point P is in fact 1 over P.

    About the Lecture

    The lecture Parametric Differentiation: Exercise 4 (1) by Batool Akmal is from the course Parametric Differentiation.

    Included Quiz Questions

    1. [1 - e^t] / [3te^(3t) + e^(3t)]
    2. [1 - e^t] / [3te^(3t) - e^(3t)]
    3. [1 + e^t] / [3te^(3t) + e^(3t)]
    4. [1 - e^t] [3te^(3t) + e^(3t)]
    5. -e^t / [3te^(3t) + e^(3t)]
    1. 15t/4
    2. 15/4
    3. 15at²
    4. 60a²t³
    5. 5t/4
    1. 15p/4
    2. 15/4
    3. 15ap²
    4. 5p/4
    5. 60a²p³

    Author of lecture Parametric Differentiation: Exercise 4 (1)

     Batool Akmal

    Batool Akmal

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