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Differentiation: Exercise 3 – Calculus Methods

by Batool Akmal
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    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.


    Included Quiz Questions

    1. y=4x-2
    2. y=4x
    3. y=4x+2
    4. x=4y-2
    5. y=2x-4
    1. y=(-1/4)x+(9/4)
    2. y=(-1/4)x-(9/4)
    3. y=(-9/4)x+(1/4)
    4. y=(-9/4)x-(1/4)
    5. y=(9/4)x+(1/4)
    1. Gradient of Tangent passes through given point
    2. Gradient of normal passes through given point
    3. Gradient of any curve passes through given point
    4. Gradient of Tangent passes through any point
    5. Gradient of normal passes through any point

    Author of lecture Differentiation: Exercise 3 – Calculus Methods

     Batool Akmal

    Batool Akmal


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