[Script Info] Title: [Events] Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text Dialogue: 0,0:00:01.01,0:00:07.09,Default,,0000,0000,0000,,In Differentiation, when\Nwe differentiate A. Dialogue: 0,0:00:07.70,0:00:10.17,Default,,0000,0000,0000,,Function F of X. Dialogue: 0,0:00:11.17,0:00:18.46,Default,,0000,0000,0000,,So there's our function F of X\Nand this is a graph of it. Dialogue: 0,0:00:18.46,0:00:24.72,Default,,0000,0000,0000,,What do we actually doing? Well,\Nwhen we find the derivative were Dialogue: 0,0:00:24.72,0:00:29.93,Default,,0000,0000,0000,,actually finding the gradient of\Na tangent. So when we Dialogue: 0,0:00:29.93,0:00:35.14,Default,,0000,0000,0000,,differentiate that's F dashed of\NX, this represents the gradient. Dialogue: 0,0:00:36.07,0:00:38.58,Default,,0000,0000,0000,,Of. The curve. Dialogue: 0,0:00:40.56,0:00:47.64,Default,,0000,0000,0000,,Or since we have a curve\Nand the tangent at that point, Dialogue: 0,0:00:47.64,0:00:50.59,Default,,0000,0000,0000,,it is the gradient of. Dialogue: 0,0:00:51.69,0:00:55.04,Default,,0000,0000,0000,,The tangent so given. Dialogue: 0,0:00:55.75,0:00:59.64,Default,,0000,0000,0000,,A point where X is equal to Dialogue: 0,0:00:59.64,0:01:03.36,Default,,0000,0000,0000,,A. And therefore where? Dialogue: 0,0:01:03.91,0:01:10.90,Default,,0000,0000,0000,,The value of the function is F\Nof a, then F dashed at that Dialogue: 0,0:01:10.90,0:01:12.89,Default,,0000,0000,0000,,point is the gradient. Dialogue: 0,0:01:14.33,0:01:18.07,Default,,0000,0000,0000,,Of. The Tangent. Dialogue: 0,0:01:19.14,0:01:22.40,Default,,0000,0000,0000,,Where? X equals Dialogue: 0,0:01:22.40,0:01:27.90,Default,,0000,0000,0000,,a. And that's what we're going\Nto be using that the derivative. Dialogue: 0,0:01:29.31,0:01:35.26,Default,,0000,0000,0000,,The value of it at the point\Nwhere X equals a gives us the Dialogue: 0,0:01:35.26,0:01:40.36,Default,,0000,0000,0000,,gradient of the tangent, because\Nif we know the gradient of the Dialogue: 0,0:01:40.36,0:01:46.31,Default,,0000,0000,0000,,tangent and we know the point on\Nthe curve, we can then find the Dialogue: 0,0:01:46.31,0:01:50.98,Default,,0000,0000,0000,,equation of this tangent because\Nit's just a straight line. So Dialogue: 0,0:01:50.98,0:01:55.66,Default,,0000,0000,0000,,let's have a look at some\Nexamples. The first example will Dialogue: 0,0:01:55.66,0:01:59.48,Default,,0000,0000,0000,,take F of X is equal to X cubed. Dialogue: 0,0:01:59.54,0:02:07.23,Default,,0000,0000,0000,,Minus three X squared plus X\Nminus one, and what we want Dialogue: 0,0:02:07.23,0:02:11.08,Default,,0000,0000,0000,,is the equation of the Tangent. Dialogue: 0,0:02:11.77,0:02:16.74,Default,,0000,0000,0000,,To this curve at the point where\NX equals 3. Dialogue: 0,0:02:17.62,0:02:22.22,Default,,0000,0000,0000,,Well, at the point where X\Nequals 3, so the first thing Dialogue: 0,0:02:22.22,0:02:27.58,Default,,0000,0000,0000,,we're going to find is where is\Nthis point on the curve? So we Dialogue: 0,0:02:27.58,0:02:30.64,Default,,0000,0000,0000,,take X equals 3 and we'll find F Dialogue: 0,0:02:30.64,0:02:37.14,Default,,0000,0000,0000,,of three. So that's\Nthree cubed minus 3 Dialogue: 0,0:02:37.14,0:02:42.95,Default,,0000,0000,0000,,* 3 squared plus\N3 - 1. Dialogue: 0,0:02:44.03,0:02:50.47,Default,,0000,0000,0000,,3 cubed is 27. Three squared is\N9 and 3 nines are also 27. Dialogue: 0,0:02:50.98,0:02:58.68,Default,,0000,0000,0000,,So this builds the 27 -\N27 + 3 - 1 altogether. Dialogue: 0,0:02:58.68,0:03:05.10,Default,,0000,0000,0000,,We come up with two, so\Nthe point that we're Dialogue: 0,0:03:05.10,0:03:10.88,Default,,0000,0000,0000,,actually interested in\Non the curve is the .32. Dialogue: 0,0:03:11.96,0:03:18.06,Default,,0000,0000,0000,,Now the next thing we need is\Nthe gradient. We've now got a Dialogue: 0,0:03:18.06,0:03:23.22,Default,,0000,0000,0000,,point on our line on our\Ntangent. We now need its Dialogue: 0,0:03:23.22,0:03:28.38,Default,,0000,0000,0000,,gradient, so we need to\Ndifferentiate F dashed of X. So Dialogue: 0,0:03:28.38,0:03:33.06,Default,,0000,0000,0000,,let's differentiate three X\Nsquared for the derivative of X Dialogue: 0,0:03:33.06,0:03:38.69,Default,,0000,0000,0000,,cubed. Now we differentiate this\Nterm and so that's minus six X. Dialogue: 0,0:03:38.69,0:03:42.91,Default,,0000,0000,0000,,Now this term, so the derivative\Nof X is. Dialogue: 0,0:03:42.95,0:03:50.13,Default,,0000,0000,0000,,One and now the final term, the\Nderivative of one is 0 because Dialogue: 0,0:03:50.13,0:03:56.75,Default,,0000,0000,0000,,it's a constant. We need the\Ngradient when X is equal to Dialogue: 0,0:03:56.75,0:04:03.93,Default,,0000,0000,0000,,three, so we can take that value\NX equals 3 and substitute it Dialogue: 0,0:04:03.93,0:04:10.55,Default,,0000,0000,0000,,into our expression for the\Ngradient. And so we have 3 * Dialogue: 0,0:04:10.55,0:04:13.31,Default,,0000,0000,0000,,3 squared minus six times. Dialogue: 0,0:04:13.32,0:04:15.81,Default,,0000,0000,0000,,3 + 1. Dialogue: 0,0:04:16.35,0:04:20.33,Default,,0000,0000,0000,,3 squared is 9 and 3 nines Dialogue: 0,0:04:20.33,0:04:27.58,Default,,0000,0000,0000,,are 27. Minus 6 * 3\Nis minus 18 plus one, and so Dialogue: 0,0:04:27.58,0:04:34.50,Default,,0000,0000,0000,,this works out to be 27 takeaway\N18. That's nine an add on one Dialogue: 0,0:04:34.50,0:04:40.90,Default,,0000,0000,0000,,that's 10. And so we've got a\Npoint, and we've got a gradient, Dialogue: 0,0:04:40.90,0:04:45.31,Default,,0000,0000,0000,,so let's write those things\Ndown. We've got our point. Dialogue: 0,0:04:46.12,0:04:49.03,Default,,0000,0000,0000,,Which is 3 two. Dialogue: 0,0:04:49.53,0:04:52.55,Default,,0000,0000,0000,,And we've got our gradient. Dialogue: 0,0:04:53.42,0:05:00.25,Default,,0000,0000,0000,,Which is 10. What we want is the\Nequation of the tangent at this Dialogue: 0,0:05:00.25,0:05:05.62,Default,,0000,0000,0000,,point on the curve with this\Ngradient. So that's the equation Dialogue: 0,0:05:05.62,0:05:07.57,Default,,0000,0000,0000,,of a straight line. Dialogue: 0,0:05:08.11,0:05:14.54,Default,,0000,0000,0000,,The equation of a straight line\Nthat goes through a point X one Dialogue: 0,0:05:14.54,0:05:22.46,Default,,0000,0000,0000,,Y one is Y minus Y one over\NX Minus X one is equal to M Dialogue: 0,0:05:22.46,0:05:29.66,Default,,0000,0000,0000,,the gradient. So this is\Nour X one, Y1. Let's just write Dialogue: 0,0:05:29.66,0:05:36.45,Default,,0000,0000,0000,,that in over the top X1Y one\Nand this is our gradient M. Dialogue: 0,0:05:37.34,0:05:43.24,Default,,0000,0000,0000,,So now will substitute these\Nnumbers in so we have Y minus Dialogue: 0,0:05:43.24,0:05:46.69,Default,,0000,0000,0000,,two over X. Minus three is equal Dialogue: 0,0:05:46.69,0:05:53.48,Default,,0000,0000,0000,,to 10. Now will multiply it\Nby this X minus three here, so Dialogue: 0,0:05:53.48,0:06:00.25,Default,,0000,0000,0000,,we have Y minus two is equal to\N10 times X minus three. Now all Dialogue: 0,0:06:00.25,0:06:07.01,Default,,0000,0000,0000,,we need to do is multiply out\Nthe bracket Y minus two is 10 X Dialogue: 0,0:06:07.01,0:06:10.62,Default,,0000,0000,0000,,minus 30. Remembering to\Nmultiply everything inside this Dialogue: 0,0:06:10.62,0:06:16.48,Default,,0000,0000,0000,,bracket by what's outside. And\Nnow we just need to get this two Dialogue: 0,0:06:16.48,0:06:18.74,Default,,0000,0000,0000,,over here with this 30. Dialogue: 0,0:06:18.79,0:06:25.92,Default,,0000,0000,0000,,So we add two to both\Nsides. Y equals 10 X, now Dialogue: 0,0:06:25.92,0:06:33.05,Default,,0000,0000,0000,,minus 30. Adding onto is minus\N28 and there's the equation of Dialogue: 0,0:06:33.05,0:06:39.07,Default,,0000,0000,0000,,our tangent. To the curve at\Nthis point on the curve with Dialogue: 0,0:06:39.07,0:06:44.93,Default,,0000,0000,0000,,that gradient. OK, let's take\Nanother example. This time. Dialogue: 0,0:06:44.93,0:06:52.40,Default,,0000,0000,0000,,Let's say that what we want\Nto be able to do is Dialogue: 0,0:06:52.40,0:06:59.86,Default,,0000,0000,0000,,looking at this curve Y equals\NX cubed minus six X squared. Dialogue: 0,0:07:00.44,0:07:07.43,Default,,0000,0000,0000,,Plus X +3 and what we want\Nto find our what are the Dialogue: 0,0:07:07.43,0:07:12.81,Default,,0000,0000,0000,,tangents? What are the equations\Nof the tangents that are Dialogue: 0,0:07:12.81,0:07:16.58,Default,,0000,0000,0000,,parallel to the line Y equals X Dialogue: 0,0:07:16.58,0:07:23.86,Default,,0000,0000,0000,,+5? So here's our curve and what\Nwe want are the equations of the Dialogue: 0,0:07:23.86,0:07:26.21,Default,,0000,0000,0000,,tangents parallel to this line. Dialogue: 0,0:07:26.91,0:07:30.88,Default,,0000,0000,0000,,Well, we've got to look at this\Nline and we've got to extract Dialogue: 0,0:07:30.88,0:07:32.10,Default,,0000,0000,0000,,some information from it. Dialogue: 0,0:07:32.85,0:07:37.79,Default,,0000,0000,0000,,Information that we can get from\Nit is what is its gradient, Dialogue: 0,0:07:37.79,0:07:42.74,Default,,0000,0000,0000,,because if the tangents have to\Nbe parallel to this line then Dialogue: 0,0:07:42.74,0:07:48.09,Default,,0000,0000,0000,,they have to have the same\Ngradient. And what we can see is Dialogue: 0,0:07:48.09,0:07:53.04,Default,,0000,0000,0000,,that we've got One X here and\Nthe standard equation for a Dialogue: 0,0:07:53.04,0:07:58.39,Default,,0000,0000,0000,,straight line is Y equals MX\Nplus. See where this M is the Dialogue: 0,0:07:58.39,0:08:03.45,Default,,0000,0000,0000,,gradient. So what we gain from\Nlooking at this standard Dialogue: 0,0:08:03.45,0:08:07.95,Default,,0000,0000,0000,,equation and comparing it with\Nthe straight line is that the Dialogue: 0,0:08:07.95,0:08:12.04,Default,,0000,0000,0000,,gradient of the straight line M\Nis equal to 1. Dialogue: 0,0:08:13.51,0:08:17.50,Default,,0000,0000,0000,,So now we know what the\Ngradients of the tangents have Dialogue: 0,0:08:17.50,0:08:23.40,Default,,0000,0000,0000,,to be. The gradients have to be\N1, so how can we calculate that? Dialogue: 0,0:08:23.40,0:08:27.21,Default,,0000,0000,0000,,Well, we know that if we\Ndifferentiate this curve its Dialogue: 0,0:08:27.21,0:08:31.02,Default,,0000,0000,0000,,equation we will get an\Nexpression for the gradients of Dialogue: 0,0:08:31.02,0:08:36.36,Default,,0000,0000,0000,,the tangents. Then we can put it\Nequal to 1 and solve an equation Dialogue: 0,0:08:36.36,0:08:41.69,Default,,0000,0000,0000,,that will give us the points or\Nat least will give us the X Dialogue: 0,0:08:41.69,0:08:46.26,Default,,0000,0000,0000,,coordinates of those points. So\Nlet's do that. We know what. Why Dialogue: 0,0:08:46.26,0:08:49.31,Default,,0000,0000,0000,,is. Let's differentiate the why\Nby DX is. Dialogue: 0,0:08:49.36,0:08:56.33,Default,,0000,0000,0000,,Equal to. The derivative of X\Ncubed is 3 X squared. The Dialogue: 0,0:08:56.33,0:09:02.09,Default,,0000,0000,0000,,derivative of minus six X\Nsquared is minus 12 X the Dialogue: 0,0:09:02.09,0:09:08.91,Default,,0000,0000,0000,,derivative of X is plus one, and\Nthe derivative of three is 0 Dialogue: 0,0:09:08.91,0:09:15.19,Default,,0000,0000,0000,,because it's a constant and we\Nwant this expression to be equal Dialogue: 0,0:09:15.19,0:09:22.39,Default,,0000,0000,0000,,to 1. So we can take\None away from each side and we Dialogue: 0,0:09:22.39,0:09:25.77,Default,,0000,0000,0000,,have three X squared minus 12 X Dialogue: 0,0:09:25.77,0:09:32.36,Default,,0000,0000,0000,,equals 0. This is a quadratic\Nequation for X, which we can Dialogue: 0,0:09:32.36,0:09:34.92,Default,,0000,0000,0000,,solve, so let's do that. Dialogue: 0,0:09:35.50,0:09:41.92,Default,,0000,0000,0000,,Three X squared minus\N12 X equals 0. Dialogue: 0,0:09:42.91,0:09:46.77,Default,,0000,0000,0000,,It's a quadratic. The first\Nquestion we've got to ask Dialogue: 0,0:09:46.77,0:09:51.79,Default,,0000,0000,0000,,ourselves is, does it factor\Neyes? And if we look we can see Dialogue: 0,0:09:51.79,0:09:56.81,Default,,0000,0000,0000,,that we've got a common factor\Nin the numbers of three and a Dialogue: 0,0:09:56.81,0:10:02.98,Default,,0000,0000,0000,,common factor in the ex terms of\NX, so we can take out three X as Dialogue: 0,0:10:02.98,0:10:04.14,Default,,0000,0000,0000,,a common factor. Dialogue: 0,0:10:04.89,0:10:09.29,Default,,0000,0000,0000,,3X times by something has to\Ngive us three X squared, so Dialogue: 0,0:10:09.29,0:10:14.06,Default,,0000,0000,0000,,that's got to be X and three\NX times by something has to Dialogue: 0,0:10:14.06,0:10:18.47,Default,,0000,0000,0000,,give us minus 12 eggs, so\Nthat's got to be minus four, Dialogue: 0,0:10:18.47,0:10:20.30,Default,,0000,0000,0000,,and so that equals 0. Dialogue: 0,0:10:21.71,0:10:27.29,Default,,0000,0000,0000,,2 numbers multiplied together.\NGive us 0, so one of them has Dialogue: 0,0:10:27.29,0:10:34.73,Default,,0000,0000,0000,,to be 0 or both of them have\Nto be 0. So we can say three Dialogue: 0,0:10:34.73,0:10:42.17,Default,,0000,0000,0000,,X equals 0 or X minus 4 equals\N0, so therefore X is 0 or X Dialogue: 0,0:10:42.17,0:10:48.75,Default,,0000,0000,0000,,equals 4. Now having got these\Ntwo values of X, we want the Dialogue: 0,0:10:48.75,0:10:54.33,Default,,0000,0000,0000,,points. Where these tangents\Nwere, what we've got now the X Dialogue: 0,0:10:54.33,0:10:59.36,Default,,0000,0000,0000,,coordinates. We now need the Y\Ncoordinates and to do that we Dialogue: 0,0:10:59.36,0:11:04.80,Default,,0000,0000,0000,,need the equation of the curve\Nagain. So let me just bring back Dialogue: 0,0:11:04.80,0:11:10.25,Default,,0000,0000,0000,,the page and it's X cubed minus\Nsix X squared plus X +3. Dialogue: 0,0:11:11.19,0:11:18.57,Default,,0000,0000,0000,,So Y equals X cubed minus six\NX squared plus X plus three and Dialogue: 0,0:11:18.57,0:11:25.95,Default,,0000,0000,0000,,first of all we want to point\NX equals 0 and so Y equals Dialogue: 0,0:11:25.95,0:11:33.32,Default,,0000,0000,0000,,we put X equals 0.0 cubed. We\Nput X equals 0 - 6 * Dialogue: 0,0:11:33.32,0:11:40.70,Default,,0000,0000,0000,,0 squared plus 0 + 3 each\Nof these three terms is 0, so Dialogue: 0,0:11:40.70,0:11:43.16,Default,,0000,0000,0000,,we. End up with three. Dialogue: 0,0:11:43.82,0:11:46.68,Default,,0000,0000,0000,,X equals Dialogue: 0,0:11:46.68,0:11:53.08,Default,,0000,0000,0000,,4. Why is\Nequal to 4 cubed? Dialogue: 0,0:11:53.75,0:12:01.69,Default,,0000,0000,0000,,Minus 6 * 4\Nsquared plus 4 + Dialogue: 0,0:12:01.69,0:12:09.08,Default,,0000,0000,0000,,3. Equals.\NWell, 4 cubed is 4 times Dialogue: 0,0:12:09.08,0:12:16.25,Default,,0000,0000,0000,,by 4 times by 4 four\N416 and four times by 16 Dialogue: 0,0:12:16.25,0:12:23.41,Default,,0000,0000,0000,,is 64 takeaway. Now we've got\N6 times by 16 and so Dialogue: 0,0:12:23.41,0:12:26.99,Default,,0000,0000,0000,,that's 96 + 4 + 3. Dialogue: 0,0:12:27.58,0:12:34.29,Default,,0000,0000,0000,,Equals 64 takeaway. 96\Nis minus 32, and Dialogue: 0,0:12:34.29,0:12:41.00,Default,,0000,0000,0000,,then we're adding on\Nanother Seven, so that's Dialogue: 0,0:12:41.00,0:12:47.72,Default,,0000,0000,0000,,minus 25. So our\Ntwo points are 03. Dialogue: 0,0:12:48.49,0:12:54.78,Default,,0000,0000,0000,,And 4 - 25 and those are the\Ntwo points where the gradients Dialogue: 0,0:12:54.78,0:13:01.07,Default,,0000,0000,0000,,of the tangent are equal to 1\Nand so where the tangents are Dialogue: 0,0:13:01.07,0:13:06.40,Default,,0000,0000,0000,,parallel to the line that we\Nstarted out with. That's Y Dialogue: 0,0:13:06.40,0:13:07.85,Default,,0000,0000,0000,,equals X +5. Dialogue: 0,0:13:09.00,0:13:14.68,Default,,0000,0000,0000,,Let's take another example\Nand this time I want to Dialogue: 0,0:13:14.68,0:13:19.79,Default,,0000,0000,0000,,introduce the word normal.\NWhat's a normal? Well, we Dialogue: 0,0:13:19.79,0:13:26.61,Default,,0000,0000,0000,,tend to think of the word\Nnormal in English as mean the Dialogue: 0,0:13:26.61,0:13:30.58,Default,,0000,0000,0000,,same everything's alright,\Neach usual. But in Dialogue: 0,0:13:30.58,0:13:35.70,Default,,0000,0000,0000,,mathematics, the word normal\Nhas a very specific meaning. Dialogue: 0,0:13:35.70,0:13:37.40,Default,,0000,0000,0000,,It means perpendicular. Dialogue: 0,0:13:39.36,0:13:43.25,Default,,0000,0000,0000,,Or at right angles. Dialogue: 0,0:13:43.75,0:13:48.57,Default,,0000,0000,0000,,At right\Nangles. Dialogue: 0,0:13:49.65,0:13:52.74,Default,,0000,0000,0000,,So if I have a curve. Dialogue: 0,0:13:54.40,0:14:00.24,Default,,0000,0000,0000,,Let's say that's my curve and I\Nhave a tangent at that point Dialogue: 0,0:14:00.24,0:14:07.41,Default,,0000,0000,0000,,there. Then what's the normal to\Nthe curve while the normal is at Dialogue: 0,0:14:07.41,0:14:13.75,Default,,0000,0000,0000,,right angles to the curve, so\Nit's also at right angles to the Dialogue: 0,0:14:13.75,0:14:16.19,Default,,0000,0000,0000,,tangent, so there's the Tangent. Dialogue: 0,0:14:17.68,0:14:24.72,Default,,0000,0000,0000,,To the curve and this is\Nthe normal to the curve normal Dialogue: 0,0:14:24.72,0:14:30.01,Default,,0000,0000,0000,,because it's at right angles\Nperpendicular to the Tangent. Dialogue: 0,0:14:31.69,0:14:32.30,Default,,0000,0000,0000,,OK. Dialogue: 0,0:14:33.39,0:14:38.47,Default,,0000,0000,0000,,If we can find the equation\Nof a tangent, we can surely Dialogue: 0,0:14:38.47,0:14:42.70,Default,,0000,0000,0000,,find the equation of a\Nnormal, but there is one Dialogue: 0,0:14:42.70,0:14:47.35,Default,,0000,0000,0000,,little piece of information\Nthat we need. That is, if we Dialogue: 0,0:14:47.35,0:14:50.73,Default,,0000,0000,0000,,have two lines at right\Nangles, what's the Dialogue: 0,0:14:50.73,0:14:52.42,Default,,0000,0000,0000,,relationship between their\Ngradients? Dialogue: 0,0:14:53.70,0:14:59.08,Default,,0000,0000,0000,,So if I take 2 lines there,\Nright angles to each other, Dialogue: 0,0:14:59.08,0:15:02.66,Default,,0000,0000,0000,,and let's say that this line\Nhas gradient. Dialogue: 0,0:15:04.38,0:15:10.98,Default,,0000,0000,0000,,M1 and let's say\Nthis line has gradient. Dialogue: 0,0:15:11.88,0:15:17.86,Default,,0000,0000,0000,,M2 And\Nthe relationship between these Dialogue: 0,0:15:17.86,0:15:24.92,Default,,0000,0000,0000,,two gradients, because they are\Nat right angles, is that M1 Dialogue: 0,0:15:24.92,0:15:32.63,Default,,0000,0000,0000,,times by M2 is equal to\Nminus one, and that's for lines. Dialogue: 0,0:15:33.17,0:15:36.70,Default,,0000,0000,0000,,At right angles. Dialogue: 0,0:15:36.70,0:15:43.55,Default,,0000,0000,0000,,And so, since tangent normal at\Nright angles, we can use this Dialogue: 0,0:15:43.55,0:15:49.26,Default,,0000,0000,0000,,relationship. We can calculate\Nthe gradient of the tangent and Dialogue: 0,0:15:49.26,0:15:53.26,Default,,0000,0000,0000,,use this to find the gradient of Dialogue: 0,0:15:53.26,0:15:57.64,Default,,0000,0000,0000,,the normal. Now let's have a\Nlook at that practice. Dialogue: 0,0:15:59.04,0:16:05.88,Default,,0000,0000,0000,,Let's take the curve Y equals\NX plus one over X. Dialogue: 0,0:16:06.85,0:16:12.21,Default,,0000,0000,0000,,At the point where X equals 2,\Nlet's ask ourselves what's the Dialogue: 0,0:16:12.21,0:16:17.58,Default,,0000,0000,0000,,equation of the tangent at the\Npoint where X equals 2. What's Dialogue: 0,0:16:17.58,0:16:22.50,Default,,0000,0000,0000,,the equation of the normal? So\Nfirst of all, let's establish Dialogue: 0,0:16:22.50,0:16:29.20,Default,,0000,0000,0000,,what the point is, X equals 2, Y\Nequals 2 plus one over 2, which Dialogue: 0,0:16:29.20,0:16:30.54,Default,,0000,0000,0000,,is 2 1/2. Dialogue: 0,0:16:31.08,0:16:36.31,Default,,0000,0000,0000,,But thinking ahead, I think I\Nwould prefer to have that as an Dialogue: 0,0:16:36.31,0:16:40.73,Default,,0000,0000,0000,,improper fraction, as five over\N2. That's because I'm going to Dialogue: 0,0:16:40.73,0:16:46.36,Default,,0000,0000,0000,,have to do some algebra with it\Nlater, and I'd rather keep it is Dialogue: 0,0:16:46.36,0:16:49.97,Default,,0000,0000,0000,,5 over 2. Then keep it as 2 1/2. Dialogue: 0,0:16:50.55,0:16:57.21,Default,,0000,0000,0000,,OK, next we want the gradient at\Nthis point X equals 2. Dialogue: 0,0:16:57.75,0:17:00.09,Default,,0000,0000,0000,,So let me just write down. Dialogue: 0,0:17:00.63,0:17:05.76,Default,,0000,0000,0000,,The equation of the curve Y\Nequals X plus and in order to Dialogue: 0,0:17:05.76,0:17:09.32,Default,,0000,0000,0000,,differentiate, be ready to\Ndifferentiate the one over X. Dialogue: 0,0:17:09.32,0:17:14.06,Default,,0000,0000,0000,,I'm going to write it as X to\Nthe power minus one. Dialogue: 0,0:17:14.81,0:17:21.31,Default,,0000,0000,0000,,And now we can differentiate it\NDY by DX is equal to the Dialogue: 0,0:17:21.31,0:17:26.31,Default,,0000,0000,0000,,derivative of X is one plus\Ndifferentiate this we multiply Dialogue: 0,0:17:26.31,0:17:33.81,Default,,0000,0000,0000,,by the minus one and we take one\Naway from the minus one to give Dialogue: 0,0:17:33.81,0:17:37.31,Default,,0000,0000,0000,,us a power of minus two. So Dialogue: 0,0:17:37.31,0:17:43.68,Default,,0000,0000,0000,,that's one. Plus and A minus\Ngives us a minus there one over Dialogue: 0,0:17:43.68,0:17:47.11,Default,,0000,0000,0000,,X to the minus 2 means one over Dialogue: 0,0:17:47.11,0:17:50.87,Default,,0000,0000,0000,,X squared. X is equal Dialogue: 0,0:17:50.87,0:17:57.81,Default,,0000,0000,0000,,to 2. And so my\Ngradient D why by DX is Dialogue: 0,0:17:57.81,0:18:05.14,Default,,0000,0000,0000,,equal to 1 - 1 over 4,\Ntwo squared is 4 and one Dialogue: 0,0:18:05.14,0:18:10.21,Default,,0000,0000,0000,,take away a quarter leaves\Nme with three quarters. Dialogue: 0,0:18:11.35,0:18:18.97,Default,,0000,0000,0000,,So we have got a .25\Nover 2 and we've got a Dialogue: 0,0:18:18.97,0:18:26.59,Default,,0000,0000,0000,,gradient of 3/4 and so we\Ncan find the equation of the Dialogue: 0,0:18:26.59,0:18:33.60,Default,,0000,0000,0000,,Tangent. So let's just list\Nwhat we know. We've got the Dialogue: 0,0:18:33.60,0:18:35.91,Default,,0000,0000,0000,,tangents at the point. Dialogue: 0,0:18:36.48,0:18:43.48,Default,,0000,0000,0000,,And this is 2, five over 2 and\Nwe know that the gradient at Dialogue: 0,0:18:43.48,0:18:45.48,Default,,0000,0000,0000,,that point is 3/4. Dialogue: 0,0:18:46.27,0:18:53.03,Default,,0000,0000,0000,,So the standard equation for a\Nstraight line Y minus Y one over Dialogue: 0,0:18:53.03,0:19:00.31,Default,,0000,0000,0000,,X Minus X one is equal to\NM the gradient. This is X one Dialogue: 0,0:19:00.31,0:19:06.55,Default,,0000,0000,0000,,and This is why one. So now\Nwe can substitute these things Dialogue: 0,0:19:06.55,0:19:13.83,Default,,0000,0000,0000,,into here, Y minus five over 2\Nover X minus two is equal to Dialogue: 0,0:19:13.83,0:19:20.44,Default,,0000,0000,0000,,3/4. Let me multiply both\Nsides by X minus 2. Dialogue: 0,0:19:20.44,0:19:27.46,Default,,0000,0000,0000,,And then let's\Nmultiply both sides Dialogue: 0,0:19:27.46,0:19:30.97,Default,,0000,0000,0000,,by this form. Dialogue: 0,0:19:31.47,0:19:38.30,Default,,0000,0000,0000,,Now if I\Nmultiply out, the Dialogue: 0,0:19:38.30,0:19:45.14,Default,,0000,0000,0000,,brackets that I've\Ngot for, why? Dialogue: 0,0:19:45.68,0:19:51.34,Default,,0000,0000,0000,,Now 4 times by minus five over 2\Nfour times by the five is 20 Dialogue: 0,0:19:51.34,0:19:56.24,Default,,0000,0000,0000,,divided by the two is 10 and\Nthen the minus sign minus 10 Dialogue: 0,0:19:56.24,0:20:02.27,Default,,0000,0000,0000,,equals 3 X minus six and it will\Nbe nice to be able to get all Dialogue: 0,0:20:02.27,0:20:06.79,Default,,0000,0000,0000,,these numbers together. So I'm\Ngoing to add 10 to each side Dialogue: 0,0:20:06.79,0:20:09.81,Default,,0000,0000,0000,,will give me 4 Y equals 3X plus Dialogue: 0,0:20:09.81,0:20:17.45,Default,,0000,0000,0000,,4. So the equation of\Nthe curve is 4 Y Dialogue: 0,0:20:17.45,0:20:20.65,Default,,0000,0000,0000,,equals 3X plus 4. Dialogue: 0,0:20:20.65,0:20:28.35,Default,,0000,0000,0000,,Now we need to find the\Nequation of normal to the curve. Dialogue: 0,0:20:28.39,0:20:33.46,Default,,0000,0000,0000,,Let's say the gradient of\Nthe normal. Dialogue: 0,0:20:35.72,0:20:41.49,Default,,0000,0000,0000,,Is M2 and that the gradient\Nof the tangent? Dialogue: 0,0:20:42.33,0:20:49.59,Default,,0000,0000,0000,,Is M1.\NAnd let's recall that normal Dialogue: 0,0:20:49.59,0:20:52.45,Default,,0000,0000,0000,,and tangent are perpendicular. Dialogue: 0,0:20:53.07,0:20:57.97,Default,,0000,0000,0000,,And the thing that we said about\N2 lines that were perpendicular Dialogue: 0,0:20:57.97,0:21:02.86,Default,,0000,0000,0000,,at right angles to each other\Nwas that if we multiplied their Dialogue: 0,0:21:02.86,0:21:06.53,Default,,0000,0000,0000,,gradients together, the answer\Nwe got was minus one. Dialogue: 0,0:21:07.11,0:21:13.64,Default,,0000,0000,0000,,Now we do know what the value of\NM1 is. We do know the gradient Dialogue: 0,0:21:13.64,0:21:20.16,Default,,0000,0000,0000,,of the tangent is 3/4. So if we\Nput that into here 3/4 times by Dialogue: 0,0:21:20.16,0:21:22.77,Default,,0000,0000,0000,,M2 is equal to minus one. Dialogue: 0,0:21:23.57,0:21:30.16,Default,,0000,0000,0000,,So if we multiply it by the four\Nand divide by the three we get Dialogue: 0,0:21:30.16,0:21:36.30,Default,,0000,0000,0000,,M2 is equal to minus four over\Nthree and so now we know the Dialogue: 0,0:21:36.30,0:21:42.01,Default,,0000,0000,0000,,gradient of the normal. We also\Nknow the point on the curve that Dialogue: 0,0:21:42.01,0:21:45.96,Default,,0000,0000,0000,,still hasn't changed. That still\Nthe .25 over 2. Dialogue: 0,0:21:47.10,0:21:53.83,Default,,0000,0000,0000,,So we now want the equation of\Nthe normal. Let's just write Dialogue: 0,0:21:53.83,0:22:01.69,Default,,0000,0000,0000,,down what we know. We know the\Npoint that's 25 over 2 an we Dialogue: 0,0:22:01.69,0:22:06.17,Default,,0000,0000,0000,,know the gradient that's minus\Nfour over 3. Dialogue: 0,0:22:06.78,0:22:12.40,Default,,0000,0000,0000,,So our standard equation for\Nstraight line Y minus Y one over Dialogue: 0,0:22:12.40,0:22:16.14,Default,,0000,0000,0000,,X Minus X one is equal to the Dialogue: 0,0:22:16.14,0:22:23.30,Default,,0000,0000,0000,,gradient. And this is the\NPoint X one Y one. So Dialogue: 0,0:22:23.30,0:22:30.26,Default,,0000,0000,0000,,now we substitute that in Y\Nminus five over 2 over X. Dialogue: 0,0:22:30.26,0:22:35.48,Default,,0000,0000,0000,,Minus two is equal to minus\Nfour over 3. Dialogue: 0,0:22:36.14,0:22:43.76,Default,,0000,0000,0000,,Let's multiply up by X\Nminus two and by three. Dialogue: 0,0:22:44.35,0:22:51.84,Default,,0000,0000,0000,,So we have three times Y minus\Nfive over 2 is equal to minus Dialogue: 0,0:22:51.84,0:22:54.52,Default,,0000,0000,0000,,four times X minus 2. Dialogue: 0,0:22:55.21,0:23:03.19,Default,,0000,0000,0000,,Multiply out the brackets 3 Y\Nminus 15 over 2 is equal Dialogue: 0,0:23:03.19,0:23:06.52,Default,,0000,0000,0000,,to minus four X +8. Dialogue: 0,0:23:07.44,0:23:12.70,Default,,0000,0000,0000,,Now let's get things together.\NIt's awkward having this minus Dialogue: 0,0:23:12.70,0:23:20.06,Default,,0000,0000,0000,,4X here, so we'll add 4X to\Neach side, and we'll add 15 over Dialogue: 0,0:23:20.06,0:23:27.43,Default,,0000,0000,0000,,2 to each side, so will have\N3 Y plus 4X is equal to Dialogue: 0,0:23:27.43,0:23:34.79,Default,,0000,0000,0000,,8 + 15 over 2. Now eight\Nis 16 over 2. So here I'll Dialogue: 0,0:23:34.79,0:23:37.42,Default,,0000,0000,0000,,have 30 one over 2. Dialogue: 0,0:23:37.73,0:23:44.80,Default,,0000,0000,0000,,So we'll have 3 Y plus 4X is\N31 over 2 and this tools and Dialogue: 0,0:23:44.80,0:23:49.98,Default,,0000,0000,0000,,awkward thing. So let's multiply\Nthroughout by two in order to Dialogue: 0,0:23:49.98,0:23:56.10,Default,,0000,0000,0000,,get rid of it. 6 Y plus 8X\Nequals 31 and that's the Dialogue: 0,0:23:56.10,0:23:57.98,Default,,0000,0000,0000,,equation of our normal. Dialogue: 0,0:23:59.42,0:24:03.61,Default,,0000,0000,0000,,Now when we've got tangents and\Nnormals because the normally Dialogue: 0,0:24:03.61,0:24:08.64,Default,,0000,0000,0000,,sort of inside the curve if you\Nlike or passing through the Dialogue: 0,0:24:08.64,0:24:12.83,Default,,0000,0000,0000,,curve, sometimes a normal can\Nactually meet the curve again, Dialogue: 0,0:24:12.83,0:24:17.86,Default,,0000,0000,0000,,and we might be interested to\Nknow where it meets that curve. Dialogue: 0,0:24:18.44,0:24:24.55,Default,,0000,0000,0000,,So. Question we're going to ask\Nis if we have the curve XY Dialogue: 0,0:24:24.55,0:24:30.66,Default,,0000,0000,0000,,equals 4. And we look at\Nthe point, X equals 2. Dialogue: 0,0:24:32.02,0:24:36.82,Default,,0000,0000,0000,,Find the equation of the normal.\NWhere does the normal meet the Dialogue: 0,0:24:36.82,0:24:42.02,Default,,0000,0000,0000,,curve? Again? If it does, well,\Nlet's have a look at a picture Dialogue: 0,0:24:42.02,0:24:47.22,Default,,0000,0000,0000,,why equals 4 over X? What does\Nthis look like as a curve? Dialogue: 0,0:24:50.63,0:24:57.59,Default,,0000,0000,0000,,OK, if we have a very large\Nvalue of X, say 104, / 100 is Dialogue: 0,0:24:57.59,0:25:04.09,Default,,0000,0000,0000,,very small, so we got a little\Nbit of curve down here. If we Dialogue: 0,0:25:04.09,0:25:09.19,Default,,0000,0000,0000,,have a very small value of X,\Nsay point nor one. Dialogue: 0,0:25:09.86,0:25:16.04,Default,,0000,0000,0000,,Then 4 divided by Point N 1.1 is\N100, so 4 / 100 is 400 very Dialogue: 0,0:25:16.04,0:25:21.05,Default,,0000,0000,0000,,large, so we're very small\Nvalues of X. We've got a bit of Dialogue: 0,0:25:21.05,0:25:24.91,Default,,0000,0000,0000,,curve there and we can join it\Nup like that. Dialogue: 0,0:25:25.49,0:25:30.10,Default,,0000,0000,0000,,Now if we take negative values\Nof X, the same thing happens, Dialogue: 0,0:25:30.10,0:25:35.09,Default,,0000,0000,0000,,except we get negative values of\NY and so the curve looks like Dialogue: 0,0:25:35.09,0:25:40.47,Default,,0000,0000,0000,,that. Notice we do not allow X\Nto be 0 because we cannot divide Dialogue: 0,0:25:40.47,0:25:46.23,Default,,0000,0000,0000,,by zero, so there is a hole in\Nthis curve. There is a gap here Dialogue: 0,0:25:46.23,0:25:51.22,Default,,0000,0000,0000,,at X equals 0. Well, here's the\Npoint. Let's say X equals 2. Dialogue: 0,0:25:52.62,0:25:55.96,Default,,0000,0000,0000,,That's the point on the curve. Dialogue: 0,0:25:56.60,0:26:02.26,Default,,0000,0000,0000,,There is the tangent to the\Ncurve, and there's the normal to Dialogue: 0,0:26:02.26,0:26:07.93,Default,,0000,0000,0000,,the curve, and as we can see,\Nthis normal goes through there. Dialogue: 0,0:26:07.93,0:26:13.12,Default,,0000,0000,0000,,So the question is where is it\Nwhere is that point? Dialogue: 0,0:26:14.41,0:26:21.37,Default,,0000,0000,0000,,So first of all, let's establish\Nwhat this point is up here so we Dialogue: 0,0:26:21.37,0:26:29.32,Default,,0000,0000,0000,,know that X is equal to two and\NY is equal to 4 over 2 equals Dialogue: 0,0:26:29.32,0:26:32.80,Default,,0000,0000,0000,,2. So our point is the .22. Dialogue: 0,0:26:33.37,0:26:38.05,Default,,0000,0000,0000,,Next, we want the gradient of\Nthe tangent in order that we can Dialogue: 0,0:26:38.05,0:26:40.21,Default,,0000,0000,0000,,find the gradient of the normal. Dialogue: 0,0:26:41.02,0:26:48.92,Default,,0000,0000,0000,,So here we have Y equals 4 over\NX, which is 4 times X to the Dialogue: 0,0:26:48.92,0:26:54.85,Default,,0000,0000,0000,,minus one. So I'll differentiate\Nthat the why by DX equals minus Dialogue: 0,0:26:54.85,0:27:01.27,Default,,0000,0000,0000,,4 multiplied by the minus one\Nand then taking one of the index Dialogue: 0,0:27:01.27,0:27:08.19,Default,,0000,0000,0000,,that gets it to be minus two. So\Nwe have minus four over X Dialogue: 0,0:27:08.19,0:27:12.12,Default,,0000,0000,0000,,squared. X is equal Dialogue: 0,0:27:12.12,0:27:19.70,Default,,0000,0000,0000,,to 2. And so DY\Nby the X is minus 4 Dialogue: 0,0:27:19.70,0:27:22.94,Default,,0000,0000,0000,,over 4 is minus one. Dialogue: 0,0:27:23.87,0:27:30.67,Default,,0000,0000,0000,,So what have I got? I've got\Nthe .22 and I've got the Dialogue: 0,0:27:30.67,0:27:35.38,Default,,0000,0000,0000,,gradient of the tangent and\Nremember tangent and normal Dialogue: 0,0:27:35.38,0:27:36.94,Default,,0000,0000,0000,,are right angles. Dialogue: 0,0:27:38.62,0:27:45.42,Default,,0000,0000,0000,,So for two lines at right\Nangles, M1 times by M2 is minus Dialogue: 0,0:27:45.42,0:27:52.74,Default,,0000,0000,0000,,one. I know the value of this,\Nit's minus one times by N 2 Dialogue: 0,0:27:52.74,0:27:59.02,Default,,0000,0000,0000,,equals minus one. The only\Nnumber M2 can be there is one. Dialogue: 0,0:27:59.80,0:28:06.37,Default,,0000,0000,0000,,So now I have got for the normal\NI want its equation. I've got Dialogue: 0,0:28:06.37,0:28:12.93,Default,,0000,0000,0000,,the point that it goes through\Nwhich is 2 two and I've got its Dialogue: 0,0:28:12.93,0:28:14.81,Default,,0000,0000,0000,,gradient which is one. Dialogue: 0,0:28:15.58,0:28:20.87,Default,,0000,0000,0000,,So I can write down the standard\Nequation of a straight line Y Dialogue: 0,0:28:20.87,0:28:25.76,Default,,0000,0000,0000,,minus Y one over X Minus X one\Nis equal to M. Dialogue: 0,0:28:26.46,0:28:29.49,Default,,0000,0000,0000,,This is my X One Y1. Dialogue: 0,0:28:30.46,0:28:37.64,Default,,0000,0000,0000,,And I can substitute those in so\NI have Y minus two over X, minus Dialogue: 0,0:28:37.64,0:28:40.04,Default,,0000,0000,0000,,two is equal to 1. Dialogue: 0,0:28:41.29,0:28:48.42,Default,,0000,0000,0000,,Multiply it by X minus two, so\Nwe have Y minus two is equal Dialogue: 0,0:28:48.42,0:28:54.02,Default,,0000,0000,0000,,to X minus two, and so why\Nis equal to X? Dialogue: 0,0:28:54.02,0:29:01.79,Default,,0000,0000,0000,,And so now we have the equation\Nof the normal Y equals X and Dialogue: 0,0:29:01.79,0:29:07.34,Default,,0000,0000,0000,,we have the equation of the\Ncurve XY equals 4. Dialogue: 0,0:29:07.98,0:29:13.91,Default,,0000,0000,0000,,Now. Where does this line meet\Nthis curve? That's what we're Dialogue: 0,0:29:13.91,0:29:17.92,Default,,0000,0000,0000,,asking ourselves. Where does the\Nnormal intersect the curve? Dialogue: 0,0:29:19.19,0:29:24.07,Default,,0000,0000,0000,,At the points of intersection,\Nboth of these equations are true Dialogue: 0,0:29:24.07,0:29:30.73,Default,,0000,0000,0000,,at the same time, so that means\NI can take the value of Y, which Dialogue: 0,0:29:30.73,0:29:36.95,Default,,0000,0000,0000,,is X and substituted into here.\NSo therefore I have X times by X Dialogue: 0,0:29:36.95,0:29:42.28,Default,,0000,0000,0000,,equals 4 at these points. In\Nother words, X squared is equal Dialogue: 0,0:29:42.28,0:29:48.39,Default,,0000,0000,0000,,to 4. If I take the square root,\Nthat will give me the value of Dialogue: 0,0:29:48.39,0:29:50.56,Default,,0000,0000,0000,,X. We might think that's just Dialogue: 0,0:29:50.56,0:29:56.96,Default,,0000,0000,0000,,two. But remember, when you take\Na square root you get a plus and Dialogue: 0,0:29:56.96,0:30:01.63,Default,,0000,0000,0000,,A minus. So we have X equals 2\Nor minus 2. Dialogue: 0,0:30:02.97,0:30:09.54,Default,,0000,0000,0000,,And so if X equals 2, then why\Nmust be equal to X? So that Dialogue: 0,0:30:09.54,0:30:16.55,Default,,0000,0000,0000,,tells us Y equals 2, or and if\Nwe take minus two, Y is equal to Dialogue: 0,0:30:16.55,0:30:23.12,Default,,0000,0000,0000,,X. That gives us minus two and\Nso our two points are two 2 and Dialogue: 0,0:30:23.12,0:30:24.87,Default,,0000,0000,0000,,minus 2 - 2. Dialogue: 0,0:30:24.96,0:30:28.93,Default,,0000,0000,0000,,Those are the two points where\Nthe normal meets the curve. Dialogue: 0,0:30:28.93,0:30:32.90,Default,,0000,0000,0000,,Notice this is the first point\Nthat we started off with. Dialogue: 0,0:30:33.47,0:30:36.63,Default,,0000,0000,0000,,And indeed, when we're doing\Nthis kind of question, the point Dialogue: 0,0:30:36.63,0:30:40.64,Default,,0000,0000,0000,,where we started off with always\Nis going to be a part of the Dialogue: 0,0:30:40.64,0:30:45.27,Default,,0000,0000,0000,,solution. So we've dealt with\Napplications to tangents and Dialogue: 0,0:30:45.27,0:30:50.38,Default,,0000,0000,0000,,normals. We've seen that in\Norder to find the gradient of Dialogue: 0,0:30:50.38,0:30:55.04,Default,,0000,0000,0000,,the tangent you differentiate,\Nsubstituting the value of X and Dialogue: 0,0:30:55.04,0:31:00.62,Default,,0000,0000,0000,,that gives you the gradient of\Nthe curve and hence the gradient Dialogue: 0,0:31:00.62,0:31:05.73,Default,,0000,0000,0000,,of the tangent and the other\Nrelationship that we found was Dialogue: 0,0:31:05.73,0:31:10.84,Default,,0000,0000,0000,,that a normal was perpendicular\Nto the tangent and that the Dialogue: 0,0:31:10.84,0:31:15.23,Default,,0000,0000,0000,,product result. Of multiplying\Ntwo gradients together, where Dialogue: 0,0:31:15.23,0:31:20.94,Default,,0000,0000,0000,,the two lines are perpendicular\Nwas minus one. That's an Dialogue: 0,0:31:20.94,0:31:26.08,Default,,0000,0000,0000,,important relationship when\Nwe're looking at 2 lines that Dialogue: 0,0:31:26.08,0:31:31.78,Default,,0000,0000,0000,,are perpendicular, as is the\Ncase for tangent and normal.