We're going to go into the notion of what is a tangent to the line and what is a, what's the normal, which a lot of you guys might not even know of. The tangent to the line is basically, we have a line like, like so, and we're given a specific spot and we want to know what the tangent is. It's basically the... instantaneous gradient at that spot.
So it'll look like this. It'll just be a straight line that just scrapes, it touches it only once, the actual curve. Okay? And the normal is pretty much the gradient except it's perpendicular. So 90 degrees to the tangent.
So it looks like this. So you would say that this is the gradient of the normal, and you would say that this is the gradient to the tangent, okay? So we know the formula of the gradient is rise over run. So when we're given the angle, we can apply basic trig to find our gradient, right?
So since tan theta is equal to opposite over adjacent. is equal to rise over one which is the gradient, which I'll give you an example of now. So find the value of tan beta from the following.
Okay I'm going to give you a sec to try this one on your own. It is something that we went over in year 10 and possibly again already in year 11. So hopefully this is just revision for you but if not I'll go through how to do it. do this again. If you need a little bit more time looking at my previous slide, you can actually find the download of the slides, I think, to your right. Your right's that way, isn't it?
Oops. Thank you. Okay, hopefully you've had enough time to give this one a go.
But let's take a look at this. So the gradient of the function is just the, oops, it's just the coefficient of 2. So the rise over run is equal to 2 on 1. Okay, so tan theta will be equal to 2. Theta will be tan inverse. of 2. Make sure that you're in degrees and not radians like I am and we'll go shift tan 2 and my angle of elevation will be equal to 63 degrees if well 63 degrees if you're rounding to degrees I'm going to round to minutes so 63 degrees in 26 minutes okay to the nearest minute okay Or, well, technically, I didn't even read the question properly. They just want the value of Chanthita. My actual answer is just this.
Okay, cool. Now let's go into a little bit more of the definitions of what we've gone through in maybe year 11. Sorry, maybe year 10 for some of you guys. I know some year 10 schools just don't teach everything that they should. I don't know.
But basically... The gradient of the tangent is equal to the gradient at the x coordinate at which the tangent intercepts with the curve. OK, that's like the definition of the tangent's gradient.
So the gradient of the normal is actually equal to the negative reciprocal of the gradient to the tangent. OK, so from year 10, you'll know that if two lines are perpendicular, then m1 times m2 is equal to negative 1, a.k.a. mt times mn is equal to negative 1. Therefore, m, oops, that's a big m. I want a small m.
And n is equal to negative 1 on mt. Okay, that's a rule that I want you all to write down. So I'm going to give you five seconds to write that down. Five, four, three, two, one.
Okay, hopefully you've written that down. Okay, cool. Yeah, so in order to find the equation of the tangent or normal of a curve, at a specific point of x, we have to use something called the point gradient formula. Which I hope you guys already know is y minus y1 is equal to m times x minus x1, where m is the gradient and x1, xy is the coordinates of a specific point.
I think I have, I've got the steps to work through it first. Let's do the actual methods. So the first thing that you need to do is to actually differentiate the curve.
Okay. From there, you're going to find the tangent at the, like a specific X coordinate. So you're going to sub in. So you find Y dash. You're going to sub X is equal to, I'm going to call it X1.
into y dash. Okay, your next step is to sub x is equal to x1 into y to find y1. And then you're going to use the point gradient formula to obtain the actual, like, equation to the tangent slash normal.
So y minus y1 is equal to m x minus x1. Okay so here is a question. I would like for you guys to try this on your own. I'm going to give you quite a few minutes to try it.
So find the equation of the normal to the function y is equal to x cubed plus 3x squared minus 12x plus 1 at x is equal to 2. If the gradient of the normal to the curve y is equal of the curve f of x is equal to 3x squared minus 4x is negative a quarter. We find the point of contact of the normal. Okay so that's kind of the inverse method of doing a. I'm going to give you a few minutes for this. Okay, hopefully everyone's had a start to question one.
I'm going to go through that now, and then I'll give you a few minutes to try b by yourself as well. All right, so I need to find the equation of the normal of the function y is equal to f of x at x cubed plus 3x squared minus 12x plus 1. Okay. I'm actually going to take a new slide for this because there's a lot to be done. Okay, sorry, I just needed to write it down so I don't forget all the numbers. So find m n.
of f of x is equal to 2x cubed plus 3x squared minus 12x plus 1 at x is equal to 2. So the first step is to find the gradient function so f dash of x which will be 6x squared plus 6x minus 12. Now I'm going to find the gradient of the tangent at x is equal to 2 because that's just the first step to do it. So I'm going to write f dash 2 so I'm going to sub in x is equal to 2, that's just another way of writing it, which is 6 times 2 squared plus 6 times 2 minus 12 which is 24 plus 12 minus 12. which is 24. I'm just going to double check with my calculator just in case I made a mistake. No, I'm good.
Okay, so that is the gradient of the tangent. All right, give me a second, how do we recall what the gradient of the normal is from the gradient of the tangent? Did you say that it is the negative reciprocal?
So, Mn is equal to negative one on empty, which is negative 1 on 24. Okay, from here, I need to sub in f of 2, so my y coordinate of, like my y1 basically, from our step 3. The f of 2, actually I'm going to do it on the side over here so I don't confuse anyone or myself, which will be equal to 2 times 2 cubed. plus 3 times 2 squared minus 12 times 2 plus 1 which my brain doesn't want to comprehend so 2 times 8 plus 3 times 4 minus 24 plus 1. I hope I got this right, I got 5. Okay so now we're just going to sub it into the formula y minus y1 is equal to m x minus x1 so y minus y1 is equal to mx minus x1. I'm going to take a sec to just write down all my variables at this point. So my x1 is equal to 2, my y1 is equal to 5, and my m is equal to negative 1 on 24. All right, so y minus 5 is equal to negative 1 on 24x minus 5. No, two.
I'm gonna show you how to do it using the like y is equal to mx plus b kind of method, and I'm going to show you how to do it in the general formula as well. Let's do mx plus b first. Actually no, let's do general first. Basically for the general equation you cannot have a fraction coefficient in front of x, and your coefficient in front of x cannot be negative.
So what I'm going to do is I'm going to times everything by negative 24 so that my x is kind of like free on its own. I'm gonna show my lines are working. Times minus 24 times minus 24 which will oops which will get me to minus 24 y. plus, whatever 24 times 5 is, 120. is equal to x minus 2. Now what I need to do is I need to just bring everything over to one side, so it'll be ax plus by plus c is equal to 0. That's the, that's like the, what's the word, the structure that it needs to be in.
So my x, as I was saying before, needs to always be positive, so everything needs to be brought over to the right hand side. Your right hand side's over here, sorry. So it'll be x plus 24, because I'm bringing the 24 to the other side, y, minus 2 minus 120, so minus 122 is equal to 0. And that is my final answer using the general equation. Now I'm going to do it using y is equal to mx plus b, which we've got y minus 5 is equal to negative 1 on 24, x minus 2. So I'm just going to expand those brackets. So minus 1 on 24, oops, 24x plus 1 on 12. I'm going to bring 5 over to the other side.
So it's y is equal to negative x on 24 plus whatever 5 plus 1 on 12 is, 61 on 12. Okay. And that's my answer using the y is equal to mx plus b like formula. Okay, now I'm going to give you a sec to try doing b on your own. So I'm only going to give you like 30 seconds or so. So if you need more time, please pause the video.
But if the gradient of the normal to the curve y is equal to f of x is equal to 3x squared minus 4x is negative a quarter, then find the point of contact of the normal. Okay, I am going to get started. Oops, I need to make more pages.
Give me a quick sec. Perfect. Okay. So the question is my f of x is equal to 3x squared minus 4x. And my gradient of the normal is equal to negative a quarter.
What is the x coordinate? Was it? No, it said what is x1 y1 equals what? Okay, that's the question. And I just need to double check that the actual equation I got right.
Perfect. Alright, so the first thing that I need to do is I need to differentiate my function. So f dash of x will be equal to 6x minus 4. Okay, now I know the gradient of the normal, which means that I can find the gradient of the tangent, right?
Using the same relationship that I had in my last question, in that mn is equal to negative 1 on mt. So likewise... mt is equal to 1 on mn, which would be negative 1 on negative a quarter, which is 4. Okay, so now I need to find the x value for which my tangent is equal to 4. So I'm going to write f dash of x is equal to 6x minus 4. I'm going to change my pen color so I can show my work to you guys a bit better. which is equal to four.
Six x will be equal to eight. So x will be equal to eight on six, aka four on three. Okay, so that's my x1. x1 is equal to 4 on 3. Now, unfortunately that's not my full answer. I need to also find my y1, aka I need the coordinates of the point of intersection.
So what can I do? I can substitute my x is equal to 4 on 3 into my f of x function. So what I'll write is f. of 4 on 3 is equal to 3 times 4 on 3 squared not cubed minus 4 times 4 on 3. Okay, which will get me 3 times 4 on 3 squared. 16 on 3 minus 16 on 3, which is 0. Okay, so my y1 is equal to 0. So therefore, the point of intersection is...
4 on 3, 0. And that is my final answer. Okay, so...