Differentiation: Exercise 3 – Calculus Methods

by Batool Akmal

Questions about the lecture
My Notes
  • Required.
Save Cancel
    Learning Material 2
    • PDF
      DLM Differentiation Exercise Calculus Akmal.pdf
    • PDF
      Download Lecture Overview
    Report mistake

    00:01 Right, let's have a look at the last question on this exercise.

    00:05 It states that a tangent meets at curve C y equals 3x squared minus 5 at point x equals to 2.

    00:12 And then it's asking us to find the the equation of the line perpendicular to the tangent at x equals to 2.

    00:19 Now if you remember these perpendicular lines to tangents are called normals.

    00:24 So if I just quickly sketch this so I have something that looks like 3x squared - 5.

    00:32 So it's a quadratic and it's going through minus 5 here.

    00:35 So this is what it would look like roughly.

    00:40 We are saying at the point x equals to 2 so I don't quite know where x equals to 2 is yet.

    00:46 But let's just say for now here and we'll correct it as we go through.

    00:50 So we have a tangent line there.

    00:54 But the question is now asking you to find the equation of the line perpendicular to the tangent.

    01:00 So we are looking for this equation here.

    01:03 That makes it 90 degrees.

    01:04 So this here is the normal.

    01:07 Or the equation that is perpendicular to the tangent.

    01:11 Now this is really important that you remember as you go through the solution.

    01:15 Because we will have to change the gradient of the normal, of the tangent, into the gradient of the normal.

    01:22 So let's come to our calculations.

    01:24 Our curve here is y equals is to 3x squared minus 5.

    01:30 Remember that the equation of any line is y = mx + c.

    01:38 Or y - y1 = mx - x1.

    01:44 You can use anyone that you want.

    01:47 If for example we use the second equation here.

    01:49 You can start to think of what you have and what you need.

    01:53 So we're looking at x.

    01:55 We have an x point here.

    01:56 We need to find the gradient m.

    01:59 And we also need to find a value of y.

    02:02 Straight forward but let's start finding all the things so that we can substitute it back into the equation.

    02:09 Let's start with m.

    02:11 Remember m for a curve is dy by dx.

    02:15 When you differentiate this that gives you 6x.

    02:18 Nice and easy.

    02:20 You want to know the gradient.

    02:22 Or you want to know m at x equals to 2.

    02:26 So you can say dy by dx for x equals to 2 is just 6 times 2 which is 12.

    02:34 So at x equals to I have a positive gradient.

    02:38 But remember that you are now looking for the equation of the normal.

    02:42 So the gradient of the tangent will be 12.

    02:47 However, the gradient of the normal would be minus 1 over 12.

    02:56 And that is because I'm using the rule that m1 multiplied by m2 should give me minus 1.

    03:02 So if I do 12 multiplied by minus 1 over 12 that gives me an answer of minus 1.

    03:09 So this gradient must be correct.

    03:12 There's a faster way of you looking at it.

    03:14 If you are given a gradient of tangent, all you do is you change the sign which we have done here.

    03:18 And you flip the fraction.

    03:20 So we've just done 1 over 12.

    03:22 And we've changed the sign.

    03:24 And it should give you the same result.

    03:27 So we now know the gradient of the normal.

    03:29 So we've calculated this.

    03:31 We also need to know y.

    03:33 So in order to find y we are now going to substitute.

    03:38 So let's just say we're going to find y.

    03:42 So we can substitute x equals to 2 into the original equation.

    03:46 y equals to 3x squared minus 5.

    03:50 So y will be 3 times two squared, minus 5, which gives you 3 times 4.

    03:56 So that's 12 minus 5 to give you a y value of 7.

    04:00 So our point here which obviously I can see now, I haven't drawn correctly.

    04:05 I'll just draw it correctly.

    04:06 So the point we are looking at is 2 and 7.

    04:09 And the reason that I know it can't be down here is because this might be 2 but I'm getting a negative value there.

    04:15 So if I just quickly correct this.

    04:18 We have here our graph, our curve and then 2 and 7 must be somewhere around here.

    04:29 So if that's 2 and this is 7.

    04:31 So the only error that I made earlier was that I put that point down here.

    04:35 And obviously now I can see now that's not the case.

    04:38 Because I've got two values 2 and 7.

    04:40 So here is my tangent and here is my normal.

    04:47 So we're almost there.

    04:48 We know that the point y, we've also calculated now.

    04:51 So we can put it all into the equation remembering that the gradient that we use now is the gradient of the normal not the gradient of the tangent.

    04:59 So we have y minus 7 equals to our gradient which is minus 1 over 12.

    05:05 And x minus 2.

    05:09 And basically this is the equation of the straight line.

    05:12 You can rearrange it if you would like to.

    05:16 So we can rearrange it to write in the form y = mx + c.

    05:20 So you can take that 12 over to the other side.

    05:22 So we've got 12, y minus 7 equals to minus 1, x minus 2.

    05:31 Again you have worked it out.

    05:33 So you don't have to do this any further.

    05:36 But just to make it look more like this form y = mx + c.

    05:40 So we can actually get all the information from y will continue just tidying this up.

    05:45 So that gives me 12y minus 84 equals to minus x plus 2.

    05:53 You can take everything over.

    05:55 So we've got 12y = -x + 2 + 84.

    06:02 So you can have 12y = -x + 86.

    06:09 So we've come down to an equation.

    06:11 12y = -x + 86.

    06:15 If you want to find what the actual gradient is, which we already know.

    06:19 But you could divide the other side of the equation with 12, to give you the gradient of -1 over 12.

    06:25 And a point of intersection on the y axis as 86 over 12.

    06:30 So you'll be able to extend the line or you'll be able to find out where the line crosses the y axis.

    About the Lecture

    The lecture Differentiation: Exercise 3 – Calculus Methods by Batool Akmal is from the course Calculus Methods: Differentiation.

    Author of lecture Differentiation: Exercise 3 – Calculus Methods

     Batool Akmal

    Batool Akmal

    Customer reviews

    5,0 of 5 stars
    5 Stars
    4 Stars
    3 Stars
    2 Stars
    1  Star