0:00:01.010,0:00:07.088 In Differentiation, when[br]we differentiate A. 0:00:07.700,0:00:10.168 Function F of X. 0:00:11.170,0:00:18.464 So there's our function F of X[br]and this is a graph of it. 0:00:18.464,0:00:24.716 What do we actually doing? Well,[br]when we find the derivative were 0:00:24.716,0:00:29.926 actually finding the gradient of[br]a tangent. So when we 0:00:29.926,0:00:35.136 differentiate that's F dashed of[br]X, this represents the gradient. 0:00:36.070,0:00:38.580 Of. The curve. 0:00:40.560,0:00:47.640 Or since we have a curve[br]and the tangent at that point, 0:00:47.640,0:00:50.590 it is the gradient of. 0:00:51.690,0:00:55.038 The tangent so given. 0:00:55.750,0:00:59.635 A point where X is equal to 0:00:59.635,0:01:03.359 A. And therefore where? 0:01:03.910,0:01:10.896 The value of the function is F[br]of a, then F dashed at that 0:01:10.896,0:01:12.892 point is the gradient. 0:01:14.330,0:01:18.070 Of. The Tangent. 0:01:19.140,0:01:22.402 Where? X equals 0:01:22.402,0:01:27.904 a. And that's what we're going[br]to be using that the derivative. 0:01:29.310,0:01:35.260 The value of it at the point[br]where X equals a gives us the 0:01:35.260,0:01:40.360 gradient of the tangent, because[br]if we know the gradient of the 0:01:40.360,0:01:46.310 tangent and we know the point on[br]the curve, we can then find the 0:01:46.310,0:01:50.985 equation of this tangent because[br]it's just a straight line. So 0:01:50.985,0:01:55.660 let's have a look at some[br]examples. The first example will 0:01:55.660,0:01:59.485 take F of X is equal to X cubed. 0:01:59.540,0:02:07.232 Minus three X squared plus X[br]minus one, and what we want 0:02:07.232,0:02:11.078 is the equation of the Tangent. 0:02:11.770,0:02:16.740 To this curve at the point where[br]X equals 3. 0:02:17.620,0:02:22.216 Well, at the point where X[br]equals 3, so the first thing 0:02:22.216,0:02:27.578 we're going to find is where is[br]this point on the curve? So we 0:02:27.578,0:02:30.642 take X equals 3 and we'll find F 0:02:30.642,0:02:37.140 of three. So that's[br]three cubed minus 3 0:02:37.140,0:02:42.950 * 3 squared plus[br]3 - 1. 0:02:44.030,0:02:50.470 3 cubed is 27. Three squared is[br]9 and 3 nines are also 27. 0:02:50.980,0:02:58.684 So this builds the 27 -[br]27 + 3 - 1 altogether. 0:02:58.684,0:03:05.104 We come up with two, so[br]the point that we're 0:03:05.104,0:03:10.882 actually interested in[br]on the curve is the .32. 0:03:11.960,0:03:18.057 Now the next thing we need is[br]the gradient. We've now got a 0:03:18.057,0:03:23.216 point on our line on our[br]tangent. We now need its 0:03:23.216,0:03:28.375 gradient, so we need to[br]differentiate F dashed of X. So 0:03:28.375,0:03:33.065 let's differentiate three X[br]squared for the derivative of X 0:03:33.065,0:03:38.693 cubed. Now we differentiate this[br]term and so that's minus six X. 0:03:38.693,0:03:42.914 Now this term, so the derivative[br]of X is. 0:03:42.950,0:03:50.126 One and now the final term, the[br]derivative of one is 0 because 0:03:50.126,0:03:56.750 it's a constant. We need the[br]gradient when X is equal to 0:03:56.750,0:04:03.926 three, so we can take that value[br]X equals 3 and substitute it 0:04:03.926,0:04:10.550 into our expression for the[br]gradient. And so we have 3 * 0:04:10.550,0:04:13.310 3 squared minus six times. 0:04:13.320,0:04:15.810 3 + 1. 0:04:16.350,0:04:20.326 3 squared is 9 and 3 nines 0:04:20.326,0:04:27.575 are 27. Minus 6 * 3[br]is minus 18 plus one, and so 0:04:27.575,0:04:34.505 this works out to be 27 takeaway[br]18. That's nine an add on one 0:04:34.505,0:04:40.901 that's 10. And so we've got a[br]point, and we've got a gradient, 0:04:40.901,0:04:45.311 so let's write those things[br]down. We've got our point. 0:04:46.120,0:04:49.028 Which is 3 two. 0:04:49.530,0:04:52.550 And we've got our gradient. 0:04:53.420,0:05:00.252 Which is 10. What we want is the[br]equation of the tangent at this 0:05:00.252,0:05:05.620 point on the curve with this[br]gradient. So that's the equation 0:05:05.620,0:05:07.572 of a straight line. 0:05:08.110,0:05:14.545 The equation of a straight line[br]that goes through a point X one 0:05:14.545,0:05:22.465 Y one is Y minus Y one over[br]X Minus X one is equal to M 0:05:22.465,0:05:29.660 the gradient. So this is[br]our X one, Y1. Let's just write 0:05:29.660,0:05:36.446 that in over the top X1Y one[br]and this is our gradient M. 0:05:37.340,0:05:43.244 So now will substitute these[br]numbers in so we have Y minus 0:05:43.244,0:05:46.688 two over X. Minus three is equal 0:05:46.688,0:05:53.481 to 10. Now will multiply it[br]by this X minus three here, so 0:05:53.481,0:06:00.246 we have Y minus two is equal to[br]10 times X minus three. Now all 0:06:00.246,0:06:07.011 we need to do is multiply out[br]the bracket Y minus two is 10 X 0:06:07.011,0:06:10.619 minus 30. Remembering to[br]multiply everything inside this 0:06:10.619,0:06:16.482 bracket by what's outside. And[br]now we just need to get this two 0:06:16.482,0:06:18.737 over here with this 30. 0:06:18.790,0:06:25.918 So we add two to both[br]sides. Y equals 10 X, now 0:06:25.918,0:06:33.046 minus 30. Adding onto is minus[br]28 and there's the equation of 0:06:33.046,0:06:39.070 our tangent. To the curve at[br]this point on the curve with 0:06:39.070,0:06:44.934 that gradient. OK, let's take[br]another example. This time. 0:06:44.934,0:06:52.398 Let's say that what we want[br]to be able to do is 0:06:52.398,0:06:59.862 looking at this curve Y equals[br]X cubed minus six X squared. 0:07:00.440,0:07:07.434 Plus X +3 and what we want[br]to find our what are the 0:07:07.434,0:07:12.814 tangents? What are the equations[br]of the tangents that are 0:07:12.814,0:07:16.580 parallel to the line Y equals X 0:07:16.580,0:07:23.860 +5? So here's our curve and what[br]we want are the equations of the 0:07:23.860,0:07:26.210 tangents parallel to this line. 0:07:26.910,0:07:30.875 Well, we've got to look at this[br]line and we've got to extract 0:07:30.875,0:07:32.095 some information from it. 0:07:32.850,0:07:37.794 Information that we can get from[br]it is what is its gradient, 0:07:37.794,0:07:42.738 because if the tangents have to[br]be parallel to this line then 0:07:42.738,0:07:48.094 they have to have the same[br]gradient. And what we can see is 0:07:48.094,0:07:53.038 that we've got One X here and[br]the standard equation for a 0:07:53.038,0:07:58.394 straight line is Y equals MX[br]plus. See where this M is the 0:07:58.394,0:08:03.451 gradient. So what we gain from[br]looking at this standard 0:08:03.451,0:08:07.950 equation and comparing it with[br]the straight line is that the 0:08:07.950,0:08:12.040 gradient of the straight line M[br]is equal to 1. 0:08:13.510,0:08:17.503 So now we know what the[br]gradients of the tangents have 0:08:17.503,0:08:23.402 to be. The gradients have to be[br]1, so how can we calculate that? 0:08:23.402,0:08:27.212 Well, we know that if we[br]differentiate this curve its 0:08:27.212,0:08:31.022 equation we will get an[br]expression for the gradients of 0:08:31.022,0:08:36.356 the tangents. Then we can put it[br]equal to 1 and solve an equation 0:08:36.356,0:08:41.690 that will give us the points or[br]at least will give us the X 0:08:41.690,0:08:46.262 coordinates of those points. So[br]let's do that. We know what. Why 0:08:46.262,0:08:49.310 is. Let's differentiate the why[br]by DX is. 0:08:49.360,0:08:56.330 Equal to. The derivative of X[br]cubed is 3 X squared. The 0:08:56.330,0:09:02.094 derivative of minus six X[br]squared is minus 12 X the 0:09:02.094,0:09:08.906 derivative of X is plus one, and[br]the derivative of three is 0 0:09:08.906,0:09:15.194 because it's a constant and we[br]want this expression to be equal 0:09:15.194,0:09:22.392 to 1. So we can take[br]one away from each side and we 0:09:22.392,0:09:25.766 have three X squared minus 12 X 0:09:25.766,0:09:32.360 equals 0. This is a quadratic[br]equation for X, which we can 0:09:32.360,0:09:34.920 solve, so let's do that. 0:09:35.500,0:09:41.924 Three X squared minus[br]12 X equals 0. 0:09:42.910,0:09:46.770 It's a quadratic. The first[br]question we've got to ask 0:09:46.770,0:09:51.788 ourselves is, does it factor[br]eyes? And if we look we can see 0:09:51.788,0:09:56.806 that we've got a common factor[br]in the numbers of three and a 0:09:56.806,0:10:02.982 common factor in the ex terms of[br]X, so we can take out three X as 0:10:02.982,0:10:04.140 a common factor. 0:10:04.890,0:10:09.294 3X times by something has to[br]give us three X squared, so 0:10:09.294,0:10:14.065 that's got to be X and three[br]X times by something has to 0:10:14.065,0:10:18.469 give us minus 12 eggs, so[br]that's got to be minus four, 0:10:18.469,0:10:20.304 and so that equals 0. 0:10:21.710,0:10:27.290 2 numbers multiplied together.[br]Give us 0, so one of them has 0:10:27.290,0:10:34.730 to be 0 or both of them have[br]to be 0. So we can say three 0:10:34.730,0:10:42.170 X equals 0 or X minus 4 equals[br]0, so therefore X is 0 or X 0:10:42.170,0:10:48.749 equals 4. Now having got these[br]two values of X, we want the 0:10:48.749,0:10:54.330 points. Where these tangents[br]were, what we've got now the X 0:10:54.330,0:10:59.358 coordinates. We now need the Y[br]coordinates and to do that we 0:10:59.358,0:11:04.805 need the equation of the curve[br]again. So let me just bring back 0:11:04.805,0:11:10.252 the page and it's X cubed minus[br]six X squared plus X +3. 0:11:11.190,0:11:18.568 So Y equals X cubed minus six[br]X squared plus X plus three and 0:11:18.568,0:11:25.946 first of all we want to point[br]X equals 0 and so Y equals 0:11:25.946,0:11:33.324 we put X equals 0.0 cubed. We[br]put X equals 0 - 6 * 0:11:33.324,0:11:40.702 0 squared plus 0 + 3 each[br]of these three terms is 0, so 0:11:40.702,0:11:43.158 we. End up with three. 0:11:43.820,0:11:46.680 X equals 0:11:46.680,0:11:53.080 4. Why is[br]equal to 4 cubed? 0:11:53.750,0:12:01.694 Minus 6 * 4[br]squared plus 4 + 0:12:01.694,0:12:09.082 3. Equals.[br]Well, 4 cubed is 4 times 0:12:09.082,0:12:16.246 by 4 times by 4 four[br]416 and four times by 16 0:12:16.246,0:12:23.410 is 64 takeaway. Now we've got[br]6 times by 16 and so 0:12:23.410,0:12:26.992 that's 96 + 4 + 3. 0:12:27.580,0:12:34.292 Equals 64 takeaway. 96[br]is minus 32, and 0:12:34.292,0:12:41.004 then we're adding on[br]another Seven, so that's 0:12:41.004,0:12:47.716 minus 25. So our[br]two points are 03. 0:12:48.490,0:12:54.782 And 4 - 25 and those are the[br]two points where the gradients 0:12:54.782,0:13:01.074 of the tangent are equal to 1[br]and so where the tangents are 0:13:01.074,0:13:06.398 parallel to the line that we[br]started out with. That's Y 0:13:06.398,0:13:07.850 equals X +5. 0:13:09.000,0:13:14.680 Let's take another example[br]and this time I want to 0:13:14.680,0:13:19.792 introduce the word normal.[br]What's a normal? Well, we 0:13:19.792,0:13:26.608 tend to think of the word[br]normal in English as mean the 0:13:26.608,0:13:30.584 same everything's alright,[br]each usual. But in 0:13:30.584,0:13:35.696 mathematics, the word normal[br]has a very specific meaning. 0:13:35.696,0:13:37.400 It means perpendicular. 0:13:39.360,0:13:43.248 Or at right angles. 0:13:43.750,0:13:48.568 At right[br]angles. 0:13:49.650,0:13:52.740 So if I have a curve. 0:13:54.400,0:14:00.237 Let's say that's my curve and I[br]have a tangent at that point 0:14:00.237,0:14:07.406 there. Then what's the normal to[br]the curve while the normal is at 0:14:07.406,0:14:13.750 right angles to the curve, so[br]it's also at right angles to the 0:14:13.750,0:14:16.190 tangent, so there's the Tangent. 0:14:17.680,0:14:24.724 To the curve and this is[br]the normal to the curve normal 0:14:24.724,0:14:30.007 because it's at right angles[br]perpendicular to the Tangent. 0:14:31.690,0:14:32.300 OK. 0:14:33.390,0:14:38.466 If we can find the equation[br]of a tangent, we can surely 0:14:38.466,0:14:42.696 find the equation of a[br]normal, but there is one 0:14:42.696,0:14:47.349 little piece of information[br]that we need. That is, if we 0:14:47.349,0:14:50.733 have two lines at right[br]angles, what's the 0:14:50.733,0:14:52.425 relationship between their[br]gradients? 0:14:53.700,0:14:59.076 So if I take 2 lines there,[br]right angles to each other, 0:14:59.076,0:15:02.660 and let's say that this line[br]has gradient. 0:15:04.380,0:15:10.980 M1 and let's say[br]this line has gradient. 0:15:11.880,0:15:17.860 M2 And[br]the relationship between these 0:15:17.860,0:15:24.922 two gradients, because they are[br]at right angles, is that M1 0:15:24.922,0:15:32.626 times by M2 is equal to[br]minus one, and that's for lines. 0:15:33.170,0:15:36.698 At right angles. 0:15:36.700,0:15:43.552 And so, since tangent normal at[br]right angles, we can use this 0:15:43.552,0:15:49.262 relationship. We can calculate[br]the gradient of the tangent and 0:15:49.262,0:15:53.259 use this to find the gradient of 0:15:53.259,0:15:57.644 the normal. Now let's have a[br]look at that practice. 0:15:59.040,0:16:05.882 Let's take the curve Y equals[br]X plus one over X. 0:16:06.850,0:16:12.214 At the point where X equals 2,[br]let's ask ourselves what's the 0:16:12.214,0:16:17.578 equation of the tangent at the[br]point where X equals 2. What's 0:16:17.578,0:16:22.495 the equation of the normal? So[br]first of all, let's establish 0:16:22.495,0:16:29.200 what the point is, X equals 2, Y[br]equals 2 plus one over 2, which 0:16:29.200,0:16:30.541 is 2 1/2. 0:16:31.080,0:16:36.306 But thinking ahead, I think I[br]would prefer to have that as an 0:16:36.306,0:16:40.728 improper fraction, as five over[br]2. That's because I'm going to 0:16:40.728,0:16:46.356 have to do some algebra with it[br]later, and I'd rather keep it is 0:16:46.356,0:16:49.974 5 over 2. Then keep it as 2 1/2. 0:16:50.550,0:16:57.210 OK, next we want the gradient at[br]this point X equals 2. 0:16:57.750,0:17:00.090 So let me just write down. 0:17:00.630,0:17:05.765 The equation of the curve Y[br]equals X plus and in order to 0:17:05.765,0:17:09.320 differentiate, be ready to[br]differentiate the one over X. 0:17:09.320,0:17:14.060 I'm going to write it as X to[br]the power minus one. 0:17:14.810,0:17:21.310 And now we can differentiate it[br]DY by DX is equal to the 0:17:21.310,0:17:26.310 derivative of X is one plus[br]differentiate this we multiply 0:17:26.310,0:17:33.810 by the minus one and we take one[br]away from the minus one to give 0:17:33.810,0:17:37.310 us a power of minus two. So 0:17:37.310,0:17:43.679 that's one. Plus and A minus[br]gives us a minus there one over 0:17:43.679,0:17:47.111 X to the minus 2 means one over 0:17:47.111,0:17:50.870 X squared. X is equal 0:17:50.870,0:17:57.806 to 2. And so my[br]gradient D why by DX is 0:17:57.806,0:18:05.138 equal to 1 - 1 over 4,[br]two squared is 4 and one 0:18:05.138,0:18:10.214 take away a quarter leaves[br]me with three quarters. 0:18:11.350,0:18:18.970 So we have got a .25[br]over 2 and we've got a 0:18:18.970,0:18:26.590 gradient of 3/4 and so we[br]can find the equation of the 0:18:26.590,0:18:33.600 Tangent. So let's just list[br]what we know. We've got the 0:18:33.600,0:18:35.908 tangents at the point. 0:18:36.480,0:18:43.480 And this is 2, five over 2 and[br]we know that the gradient at 0:18:43.480,0:18:45.480 that point is 3/4. 0:18:46.270,0:18:53.030 So the standard equation for a[br]straight line Y minus Y one over 0:18:53.030,0:19:00.310 X Minus X one is equal to[br]M the gradient. This is X one 0:19:00.310,0:19:06.550 and This is why one. So now[br]we can substitute these things 0:19:06.550,0:19:13.830 into here, Y minus five over 2[br]over X minus two is equal to 0:19:13.830,0:19:20.438 3/4. Let me multiply both[br]sides by X minus 2. 0:19:20.440,0:19:27.460 And then let's[br]multiply both sides 0:19:27.460,0:19:30.970 by this form. 0:19:31.470,0:19:38.304 Now if I[br]multiply out, the 0:19:38.304,0:19:45.138 brackets that I've[br]got for, why? 0:19:45.680,0:19:51.335 Now 4 times by minus five over 2[br]four times by the five is 20 0:19:51.335,0:19:56.236 divided by the two is 10 and[br]then the minus sign minus 10 0:19:56.236,0:20:02.268 equals 3 X minus six and it will[br]be nice to be able to get all 0:20:02.268,0:20:06.792 these numbers together. So I'm[br]going to add 10 to each side 0:20:06.792,0:20:09.808 will give me 4 Y equals 3X plus 0:20:09.808,0:20:17.450 4. So the equation of[br]the curve is 4 Y 0:20:17.450,0:20:20.650 equals 3X plus 4. 0:20:20.650,0:20:28.354 Now we need to find the[br]equation of normal to the curve. 0:20:28.390,0:20:33.465 Let's say the gradient of[br]the normal. 0:20:35.720,0:20:41.489 Is M2 and that the gradient[br]of the tangent? 0:20:42.330,0:20:49.590 Is M1.[br]And let's recall that normal 0:20:49.590,0:20:52.454 and tangent are perpendicular. 0:20:53.070,0:20:57.966 And the thing that we said about[br]2 lines that were perpendicular 0:20:57.966,0:21:02.862 at right angles to each other[br]was that if we multiplied their 0:21:02.862,0:21:06.534 gradients together, the answer[br]we got was minus one. 0:21:07.110,0:21:13.635 Now we do know what the value of[br]M1 is. We do know the gradient 0:21:13.635,0:21:20.160 of the tangent is 3/4. So if we[br]put that into here 3/4 times by 0:21:20.160,0:21:22.770 M2 is equal to minus one. 0:21:23.570,0:21:30.155 So if we multiply it by the four[br]and divide by the three we get 0:21:30.155,0:21:36.301 M2 is equal to minus four over[br]three and so now we know the 0:21:36.301,0:21:42.008 gradient of the normal. We also[br]know the point on the curve that 0:21:42.008,0:21:45.959 still hasn't changed. That still[br]the .25 over 2. 0:21:47.100,0:21:53.832 So we now want the equation of[br]the normal. Let's just write 0:21:53.832,0:22:01.686 down what we know. We know the[br]point that's 25 over 2 an we 0:22:01.686,0:22:06.174 know the gradient that's minus[br]four over 3. 0:22:06.780,0:22:12.396 So our standard equation for[br]straight line Y minus Y one over 0:22:12.396,0:22:16.140 X Minus X one is equal to the 0:22:16.140,0:22:23.300 gradient. And this is the[br]Point X one Y one. So 0:22:23.300,0:22:30.260 now we substitute that in Y[br]minus five over 2 over X. 0:22:30.260,0:22:35.480 Minus two is equal to minus[br]four over 3. 0:22:36.140,0:22:43.760 Let's multiply up by X[br]minus two and by three. 0:22:44.350,0:22:51.840 So we have three times Y minus[br]five over 2 is equal to minus 0:22:51.840,0:22:54.515 four times X minus 2. 0:22:55.210,0:23:03.190 Multiply out the brackets 3 Y[br]minus 15 over 2 is equal 0:23:03.190,0:23:06.515 to minus four X +8. 0:23:07.440,0:23:12.700 Now let's get things together.[br]It's awkward having this minus 0:23:12.700,0:23:20.064 4X here, so we'll add 4X to[br]each side, and we'll add 15 over 0:23:20.064,0:23:27.428 2 to each side, so will have[br]3 Y plus 4X is equal to 0:23:27.428,0:23:34.792 8 + 15 over 2. Now eight[br]is 16 over 2. So here I'll 0:23:34.792,0:23:37.422 have 30 one over 2. 0:23:37.730,0:23:44.795 So we'll have 3 Y plus 4X is[br]31 over 2 and this tools and 0:23:44.795,0:23:49.976 awkward thing. So let's multiply[br]throughout by two in order to 0:23:49.976,0:23:56.099 get rid of it. 6 Y plus 8X[br]equals 31 and that's the 0:23:56.099,0:23:57.983 equation of our normal. 0:23:59.420,0:24:03.610 Now when we've got tangents and[br]normals because the normally 0:24:03.610,0:24:08.638 sort of inside the curve if you[br]like or passing through the 0:24:08.638,0:24:12.828 curve, sometimes a normal can[br]actually meet the curve again, 0:24:12.828,0:24:17.856 and we might be interested to[br]know where it meets that curve. 0:24:18.440,0:24:24.548 So. Question we're going to ask[br]is if we have the curve XY 0:24:24.548,0:24:30.655 equals 4. And we look at[br]the point, X equals 2. 0:24:32.020,0:24:36.820 Find the equation of the normal.[br]Where does the normal meet the 0:24:36.820,0:24:42.020 curve? Again? If it does, well,[br]let's have a look at a picture 0:24:42.020,0:24:47.220 why equals 4 over X? What does[br]this look like as a curve? 0:24:50.630,0:24:57.590 OK, if we have a very large[br]value of X, say 104, / 100 is 0:24:57.590,0:25:04.086 very small, so we got a little[br]bit of curve down here. If we 0:25:04.086,0:25:09.190 have a very small value of X,[br]say point nor one. 0:25:09.860,0:25:16.036 Then 4 divided by Point N 1.1 is[br]100, so 4 / 100 is 400 very 0:25:16.036,0:25:21.054 large, so we're very small[br]values of X. We've got a bit of 0:25:21.054,0:25:24.914 curve there and we can join it[br]up like that. 0:25:25.490,0:25:30.098 Now if we take negative values[br]of X, the same thing happens, 0:25:30.098,0:25:35.090 except we get negative values of[br]Y and so the curve looks like 0:25:35.090,0:25:40.466 that. Notice we do not allow X[br]to be 0 because we cannot divide 0:25:40.466,0:25:46.226 by zero, so there is a hole in[br]this curve. There is a gap here 0:25:46.226,0:25:51.218 at X equals 0. Well, here's the[br]point. Let's say X equals 2. 0:25:52.620,0:25:55.956 That's the point on the curve. 0:25:56.600,0:26:02.264 There is the tangent to the[br]curve, and there's the normal to 0:26:02.264,0:26:07.928 the curve, and as we can see,[br]this normal goes through there. 0:26:07.928,0:26:13.120 So the question is where is it[br]where is that point? 0:26:14.410,0:26:21.368 So first of all, let's establish[br]what this point is up here so we 0:26:21.368,0:26:29.320 know that X is equal to two and[br]Y is equal to 4 over 2 equals 0:26:29.320,0:26:32.799 2. So our point is the .22. 0:26:33.370,0:26:38.050 Next, we want the gradient of[br]the tangent in order that we can 0:26:38.050,0:26:40.210 find the gradient of the normal. 0:26:41.020,0:26:48.924 So here we have Y equals 4 over[br]X, which is 4 times X to the 0:26:48.924,0:26:54.852 minus one. So I'll differentiate[br]that the why by DX equals minus 0:26:54.852,0:27:01.274 4 multiplied by the minus one[br]and then taking one of the index 0:27:01.274,0:27:08.190 that gets it to be minus two. So[br]we have minus four over X 0:27:08.190,0:27:12.118 squared. X is equal 0:27:12.118,0:27:19.702 to 2. And so DY[br]by the X is minus 4 0:27:19.702,0:27:22.942 over 4 is minus one. 0:27:23.870,0:27:30.669 So what have I got? I've got[br]the .22 and I've got the 0:27:30.669,0:27:35.376 gradient of the tangent and[br]remember tangent and normal 0:27:35.376,0:27:36.945 are right angles. 0:27:38.620,0:27:45.419 So for two lines at right[br]angles, M1 times by M2 is minus 0:27:45.419,0:27:52.741 one. I know the value of this,[br]it's minus one times by N 2 0:27:52.741,0:27:59.017 equals minus one. The only[br]number M2 can be there is one. 0:27:59.800,0:28:06.366 So now I have got for the normal[br]I want its equation. I've got 0:28:06.366,0:28:12.932 the point that it goes through[br]which is 2 two and I've got its 0:28:12.932,0:28:14.808 gradient which is one. 0:28:15.580,0:28:20.871 So I can write down the standard[br]equation of a straight line Y 0:28:20.871,0:28:25.755 minus Y one over X Minus X one[br]is equal to M. 0:28:26.460,0:28:29.490 This is my X One Y1. 0:28:30.460,0:28:37.645 And I can substitute those in so[br]I have Y minus two over X, minus 0:28:37.645,0:28:40.040 two is equal to 1. 0:28:41.290,0:28:48.416 Multiply it by X minus two, so[br]we have Y minus two is equal 0:28:48.416,0:28:54.015 to X minus two, and so why[br]is equal to X? 0:28:54.020,0:29:01.790 And so now we have the equation[br]of the normal Y equals X and 0:29:01.790,0:29:07.340 we have the equation of the[br]curve XY equals 4. 0:29:07.980,0:29:13.910 Now. Where does this line meet[br]this curve? That's what we're 0:29:13.910,0:29:17.915 asking ourselves. Where does the[br]normal intersect the curve? 0:29:19.190,0:29:24.074 At the points of intersection,[br]both of these equations are true 0:29:24.074,0:29:30.734 at the same time, so that means[br]I can take the value of Y, which 0:29:30.734,0:29:36.950 is X and substituted into here.[br]So therefore I have X times by X 0:29:36.950,0:29:42.278 equals 4 at these points. In[br]other words, X squared is equal 0:29:42.278,0:29:48.393 to 4. If I take the square root,[br]that will give me the value of 0:29:48.393,0:29:50.559 X. We might think that's just 0:29:50.559,0:29:56.955 two. But remember, when you take[br]a square root you get a plus and 0:29:56.955,0:30:01.630 A minus. So we have X equals 2[br]or minus 2. 0:30:02.970,0:30:09.540 And so if X equals 2, then why[br]must be equal to X? So that 0:30:09.540,0:30:16.548 tells us Y equals 2, or and if[br]we take minus two, Y is equal to 0:30:16.548,0:30:23.118 X. That gives us minus two and[br]so our two points are two 2 and 0:30:23.118,0:30:24.870 minus 2 - 2. 0:30:24.960,0:30:28.931 Those are the two points where[br]the normal meets the curve. 0:30:28.931,0:30:32.902 Notice this is the first point[br]that we started off with. 0:30:33.470,0:30:36.627 And indeed, when we're doing[br]this kind of question, the point 0:30:36.627,0:30:40.645 where we started off with always[br]is going to be a part of the 0:30:40.645,0:30:45.270 solution. So we've dealt with[br]applications to tangents and 0:30:45.270,0:30:50.385 normals. We've seen that in[br]order to find the gradient of 0:30:50.385,0:30:55.035 the tangent you differentiate,[br]substituting the value of X and 0:30:55.035,0:31:00.615 that gives you the gradient of[br]the curve and hence the gradient 0:31:00.615,0:31:05.730 of the tangent and the other[br]relationship that we found was 0:31:05.730,0:31:10.845 that a normal was perpendicular[br]to the tangent and that the 0:31:10.845,0:31:15.226 product result. Of multiplying[br]two gradients together, where 0:31:15.226,0:31:20.936 the two lines are perpendicular[br]was minus one. That's an 0:31:20.936,0:31:26.075 important relationship when[br]we're looking at 2 lines that 0:31:26.075,0:31:31.785 are perpendicular, as is the[br]case for tangent and normal.