Lectures

Gradient of a Point on a Curve

by Batool Akmal
(1)

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

    00:01 So let's not wait any longer and let's go straight into an example.

    00:04 Here we have a question where you are given a curve c.

    00:08 And curve c has the equation 2x squared - 6x + 4.

    00:12 So it's going to be a quadratic, a curve that we've been dealing with.

    00:17 And it's asking us to find the gradient of the tangent to see at point x equals to 3.

    00:22 Now before we start, let's just see if we can model this.

    00:26 So we can kind of figure out what's actually happening.

    00:29 So we're not quite answering the question yet.

    00:32 We're just going to see if we can think about, what this question is actually asking.

    00:36 So I have here, y = 2x2 - 6x + 4.

    00:42 Just like we've mentioned already is a quadratic.

    00:46 We can solve this.

    00:48 So if we say y equals to 0.

    00:50 We can try to factorize it.

    00:51 And see if we can find the two factors.

    00:54 So those points would be the point of intersection on the x axis.

    00:57 If I try that here x and x.

    01:00 And if we try our two factors, 2 and 2.

    01:03 And if you double check this that gives me 2x squared minus 4, minus 2 to give me minus 6x.

    01:12 And then I've got minus 2 times minus 2 to give me plus 4.

    01:15 So I can factorize it.

    01:17 It's fairly nice equation.

    01:19 From here you can say you got two solutions.

    01:21 So you can say that either 2x - 2 = 0.

    01:24 Or x - 2 = 0.

    01:27 When you solve this, you get x equals to 1 you get x = 2 here.

    01:33 So remember I haven't quite started answering the question yet.

    01:36 I'm just trying to imagine what this actually looks like.

    01:39 And for mathematicians it's really important for us to be able to visualize to have graphical and algebraic interpretations of the questions we're looking at.

    01:48 So here is what I think this looks like.

    01:54 So I've got x = 1 and x = 2.

    01:57 So those are my two solutions from there.

    02:00 I know it's a quadratic so it has a shape like so.

    02:05 And now coming back to the question.

    02:08 It says find the gradient of the tangent to see at point x to 3.

    02:12 So let's just say 3 is somewhere around here.

    02:14 So we are looking for the gradient at that point.

    02:18 And as a result, we are also looking for the tangent here.

    02:25 So this here is our tangent line.

    02:30 So we're actually looking for the equation of this line which you know takes the form, y = mx + c.

    02:36 Or the alternate form that I mentioned earlier.

    02:39 And we could see if we could work with both.

    02:41 So now that we understand the problem, let's try and answer the question.

    02:45 The question says find the equation of the tangent.

    02:49 So we're looking for the equation of this blue line here on this side.

    02:53 Right, we've already mentioned that the equation of a tangent takes this form y = mx + c.

    02:59 Or we're using y - y1 = mx - x1.

    03:07 Both methods again are valid.

    03:09 The first thing that we need to do is to actually calculate the gradient.

    03:14 Whether you're using this equation or whether you're using this equation we need to calculate the gradient of the tangent.

    03:20 And here's what we start to use all the knowledge that we have just discussed.

    03:23 The gradient of the tangent, this tangent, I can calculate by finding the gradient of the curve at this point x = 3.

    03:32 And it will exactly the same gradient of the tangent as it is of the curve at x = 3.

    03:38 And we know how to do that.

    03:40 So to calculate the gradient of a curve, I want to take my curve.

    03:44 I'm going to dy by dx it.

    03:47 So that's how I calculate my gradient.

    03:49 That gives me 4x - 6.

    03:52 This is the general gradient.

    03:55 So everywhere around this curve I can use that gradient.

    03:58 However, if I need the gradient at the point that the question asks, so we want to know dy by dx at x = 3.

    04:07 You simply substitute 3 into here.

    04:12 So we will do 4 times 3 minus 6 which gives you 12 minus 6 to give you a positive gradient of 6.

    04:20 Which is good news because from our picture it looks like this graph has a positive gradient anyway.

    04:26 So it's a nice little way of checking.

    04:28 And that you can see it graphically.

    04:31 You expect a positive gradient.

    04:32 And so you've got a positive gradient.

    04:34 Now we can put into our equation.

    04:38 So you could put it into y = mx + c or we can substitute this into y - y1 = mx - x1.

    04:49 Both of them are fine.

    04:52 But you might find this a little bit easier to use.

    04:55 Because you don't have to do it into two parts.

    04:58 So you don't have to find your m first and then find your c at the end.

    05:03 Okay, let's just talk about what we have and what we need.

    05:08 So we found the gradient.

    05:09 We know where m is.

    05:10 We know an x point here.

    05:12 So we know that, that point here is 3.

    05:14 However, we don't know this yet.

    05:16 So we're missing a y value.

    05:19 Now how could you find a y value.

    05:22 You go back to your original equation here.

    05:24 Which says that y equals to something.

    05:27 So all we have to do here is replace our x with 3.

    05:31 So using your original function we got y = 2x squared minus 6x plus 4.

    05:37 We want to know what this is at x = 3.

    05:40 So we're going to substitute that n, 3 squared minus 6 times 3 plus 4.

    05:47 That gives you two times 9 to give you 18 minus 6 times 3 which also gives you 18 + 4 to give you an answer of 4.

    05:57 Therefore, your y value, y equals to 4, 1x equals to 3.

    06:06 Now we have everything.

    06:08 So we've also calculated at this y value.

    06:11 You can put it all into the equation.

    06:13 So we have the y minus 4 equals to the gradient which was 6x minus 3.

    06:20 The equation has been found.

    06:23 But we can tidy up and make it look more like y = mx + c.

    06:28 So we make it look more into that from there.

    06:30 We've got y minus 4 equals to 6x minus 18.

    06:35 Take the 4 over to the other side, you get y equals to 6x minus 18 + 4 to give you minus 14.

    06:43 And this equation now follows the form, y = mx + c.

    06:50 So the plus c is going to be minus 14.

    06:52 So it meets this y axis here at -14.

    06:57 So it gives you everything you need to know about that equation.


    About the Lecture

    The lecture Gradient of a Point on a Curve by Batool Akmal is from the course Calculus Methods: Differentiation.


    Author of lecture Gradient of a Point on a Curve

     Batool Akmal

    Batool Akmal


    Customer reviews

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