1 00:00:00,000 --> 00:00:00,280 2 00:00:00,280 --> 00:00:02,299 I just got sent this problem, and it's a 3 00:00:02,299 --> 00:00:03,349 pretty meaty problem. 4 00:00:03,350 --> 00:00:04,910 A lot harder than what you'd normally find 5 00:00:04,910 --> 00:00:05,580 in most textbooks. 6 00:00:05,580 --> 00:00:08,470 So I thought it would help us all to work it out. 7 00:00:08,470 --> 00:00:10,190 And it's one of those problems that when you first read it, 8 00:00:10,189 --> 00:00:12,629 your eyes kind of glaze over, but when you understand what 9 00:00:12,630 --> 00:00:15,339 they're talking about, it's reasonably interesting. 10 00:00:15,339 --> 00:00:18,579 So they say, the curve in the figure above is the parabola 11 00:00:18,579 --> 00:00:20,189 y is equal to x squared. 12 00:00:20,190 --> 00:00:25,150 So this curve right there is y is equal to x squared. 13 00:00:25,149 --> 00:00:29,259 Let us define a normal line as a line whose first quadrant 14 00:00:29,260 --> 00:00:31,650 intersection with the parabola is perpendicular 15 00:00:31,649 --> 00:00:32,939 to the parabola. 16 00:00:32,939 --> 00:00:35,919 So this is the first quadrant, right here. 17 00:00:35,920 --> 00:00:38,230 And they're saying that a normal line is something, when 18 00:00:38,229 --> 00:00:40,569 the first quadrant intersection with the parabola is 19 00:00:40,570 --> 00:00:41,719 normal to the parabola. 20 00:00:41,719 --> 00:00:46,869 So if I were to draw a tangent line right there, this line is 21 00:00:46,869 --> 00:00:48,449 normal to that tangent line. 22 00:00:48,450 --> 00:00:49,070 That's all that's saying. 23 00:00:49,070 --> 00:00:52,469 So this is a normal line, right there. 24 00:00:52,469 --> 00:00:53,250 Normal line. 25 00:00:53,250 --> 00:00:54,090 Fair enough. 26 00:00:54,090 --> 00:00:56,410 5 normal lines are shown in the figure. 27 00:00:56,409 --> 00:00:59,750 1, 2, 3, 4, 5. 28 00:00:59,750 --> 00:01:00,200 Good enough. 29 00:01:00,200 --> 00:01:03,010 And these all look perpendicular, or normal to 30 00:01:03,009 --> 00:01:05,539 the parabola in the first quadrant intersection, 31 00:01:05,540 --> 00:01:07,100 so that makes sense. 32 00:01:07,099 --> 00:01:10,419 For a while, the x-coordinate of the second quadrant 33 00:01:10,420 --> 00:01:14,140 intersection of the normal line of the parabola gets smaller, 34 00:01:14,140 --> 00:01:16,969 as the x-coordinate of the first quadrant intersection 35 00:01:16,969 --> 00:01:17,569 gets smaller. 36 00:01:17,569 --> 00:01:19,759 So let's see what happens as the x-quadrant of the first 37 00:01:19,760 --> 00:01:21,570 intersection gets smaller. 38 00:01:21,569 --> 00:01:24,619 So this is where I left off in that dense text. 39 00:01:24,620 --> 00:01:29,050 So if I start at this point right here, my x-coordinate 40 00:01:29,049 --> 00:01:30,840 right there would look something like this. 41 00:01:30,840 --> 00:01:32,170 Let me go down. 42 00:01:32,170 --> 00:01:33,379 my x-coordinate is right around there. 43 00:01:33,379 --> 00:01:37,079 And then as I move to a smaller x-coordinate to, say, this one 44 00:01:37,079 --> 00:01:40,170 right here, what happened to the normal line? 45 00:01:40,170 --> 00:01:42,379 Or even more important, what happened to the intersection 46 00:01:42,379 --> 00:01:44,589 of the normal line in the second quadrant? 47 00:01:44,590 --> 00:01:46,740 This is the second quadrant, right here. 48 00:01:46,739 --> 00:01:50,519 So when I had a larger x-value here, my normal 49 00:01:50,519 --> 00:01:53,109 line intersected here, in the second quadrant. 50 00:01:53,109 --> 00:01:57,469 Then when I brought my x-value in, when I lowered my x-value, 51 00:01:57,469 --> 00:02:01,420 my x-value here, because this is the next point, right here, 52 00:02:01,420 --> 00:02:04,960 my x-value at the intersection here, went-- actually, 53 00:02:04,959 --> 00:02:05,909 their wording is bad. 54 00:02:05,909 --> 00:02:07,569 They're saying that the second quadrant 55 00:02:07,569 --> 00:02:09,509 intersection gets smaller. 56 00:02:09,509 --> 00:02:11,500 But actually, it's not really getting smaller. 57 00:02:11,500 --> 00:02:14,870 It's getting less negative. 58 00:02:14,870 --> 00:02:17,430 I guess smaller could be just absolute value or magnitude, 59 00:02:17,430 --> 00:02:19,379 but it's just getting less negative. 60 00:02:19,379 --> 00:02:23,060 It's moving there, but it's actually becoming 61 00:02:23,060 --> 00:02:25,039 a larger number, right? 62 00:02:25,039 --> 00:02:27,509 It's becoming less negative, but a larger number. 63 00:02:27,509 --> 00:02:29,109 But if we think in absolute value, I guess it's 64 00:02:29,110 --> 00:02:30,920 getting smaller, right? 65 00:02:30,919 --> 00:02:34,549 As we went from that point to that point, as we moved the x 66 00:02:34,550 --> 00:02:36,660 in for the intersection of the first quadrant, the second 67 00:02:36,659 --> 00:02:39,430 quadrant intersection also moved in a bit, from 68 00:02:39,430 --> 00:02:40,640 that line to that line. 69 00:02:40,639 --> 00:02:41,889 Fair enough. 70 00:02:41,889 --> 00:02:45,299 But eventually, a normal line second quadrant intersection 71 00:02:45,300 --> 00:02:47,270 gets as small as it can get. 72 00:02:47,270 --> 00:02:50,520 So if we keep lowering our x-value in the first quadrant, 73 00:02:50,520 --> 00:02:53,210 so we keep on pulling in the first quadrant, as we 74 00:02:53,210 --> 00:02:54,480 get to this point. 75 00:02:54,479 --> 00:02:57,060 And then this point intersects the second quadrant, 76 00:02:57,060 --> 00:02:58,390 right there. 77 00:02:58,389 --> 00:03:01,579 And then, if you go even smaller x-values in the first 78 00:03:01,580 --> 00:03:06,150 quadrant then your normal line starts intersecting in the 79 00:03:06,150 --> 00:03:09,240 second quadrant, further and further negative numbers. 80 00:03:09,240 --> 00:03:13,680 So you can kind of view this as the highest value, or the 81 00:03:13,680 --> 00:03:17,020 smallest absolute value, at which the normal line can 82 00:03:17,020 --> 00:03:19,719 intersect in the second quadrant 83 00:03:19,719 --> 00:03:20,870 Let me make that clear. 84 00:03:20,870 --> 00:03:23,700 Up here, you were intersecting when you had a large x in the 85 00:03:23,699 --> 00:03:26,799 first quadrant, you had a large negative x in the second 86 00:03:26,800 --> 00:03:28,120 quadrant intersection. 87 00:03:28,120 --> 00:03:30,890 And then as you lowered your x-value, here, you had a 88 00:03:30,889 --> 00:03:32,939 smaller negative value. 89 00:03:32,939 --> 00:03:37,409 Up until you got to this point, right here, you got this, which 90 00:03:37,409 --> 00:03:40,979 you can view as the smallest negative value could get, and 91 00:03:40,979 --> 00:03:43,929 then when you pulled in your x even more, these normal lines 92 00:03:43,930 --> 00:03:47,280 started to push out again, out in the second quadrant. 93 00:03:47,280 --> 00:03:49,580 That's, I think, what they're talking about. 94 00:03:49,580 --> 00:03:53,219 The extreme normal line is shown as a thick 95 00:03:53,219 --> 00:03:54,229 line in the figure. 96 00:03:54,229 --> 00:03:54,479 Right. 97 00:03:54,479 --> 00:03:57,709 This is the extreme normal line, right there. 98 00:03:57,710 --> 00:04:01,500 So this is the extreme one, that deep, bold one. 99 00:04:01,500 --> 00:04:04,039 Extreme normal line. 100 00:04:04,039 --> 00:04:08,599 After this point, when you pull in your x-values even more, the 101 00:04:08,599 --> 00:04:10,680 intersection in your second quadrant starts to 102 00:04:10,680 --> 00:04:11,900 push out some. 103 00:04:11,900 --> 00:04:14,770 And you can think of the extreme case, if you draw the 104 00:04:14,770 --> 00:04:18,360 normal line down here, your intersection with the second 105 00:04:18,360 --> 00:04:20,259 quadrant is going to be way out here someplace, although it 106 00:04:20,259 --> 00:04:22,430 seems like it's kind of asymptoting a little bit. 107 00:04:22,430 --> 00:04:22,990 But I don't know. 108 00:04:22,990 --> 00:04:25,060 Let's read the rest of the problem. 109 00:04:25,060 --> 00:04:27,384 Once the normal line passes the extreme normal line, the 110 00:04:27,384 --> 00:04:30,129 x-coordinates of their second quadrant intersections what the 111 00:04:30,129 --> 00:04:32,300 parabola start to increase. 112 00:04:32,300 --> 00:04:34,520 And they're really, when they say they start to increase, 113 00:04:34,519 --> 00:04:36,099 they're actually just becoming more negative. 114 00:04:36,100 --> 00:04:37,550 That wording is bad. 115 00:04:37,550 --> 00:04:41,560 I should change this to more, more negative. 116 00:04:41,560 --> 00:04:43,420 Or they're becoming larger negative numbers. 117 00:04:43,420 --> 00:04:46,449 Because once you get below this, then all of a sudden 118 00:04:46,449 --> 00:04:49,459 the x-intersections start to push out more in 119 00:04:49,459 --> 00:04:50,389 the second quadrant. 120 00:04:50,389 --> 00:04:51,930 Fair enough. 121 00:04:51,930 --> 00:04:54,639 The figures show 2 pairs of normal lines. 122 00:04:54,639 --> 00:04:55,469 Fair enough. 123 00:04:55,470 --> 00:04:58,740 The 2 normal lines of a pair have the same second quadrant 124 00:04:58,740 --> 00:05:01,730 intersection with the parabola, but 1 is above the extreme 125 00:05:01,730 --> 00:05:03,590 normal line, in the first quadrant, the other 126 00:05:03,589 --> 00:05:05,509 is below it. 127 00:05:05,509 --> 00:05:07,180 Right, fair enough. 128 00:05:07,180 --> 00:05:10,280 For example, this guy right here, this is when we 129 00:05:10,279 --> 00:05:11,359 had a large x-value. 130 00:05:11,360 --> 00:05:13,759 He intersects with the second quadrant there. 131 00:05:13,759 --> 00:05:15,939 Then if you lower and lower the x-value, if you lower it 132 00:05:15,939 --> 00:05:18,930 enough, you pass the extreme normal line, and then you get 133 00:05:18,930 --> 00:05:22,889 to this point, and then this point, he intersects, or 134 00:05:22,889 --> 00:05:24,009 actually, you go to this point. 135 00:05:24,009 --> 00:05:27,159 So if you pull in your x-value enough, you once again 136 00:05:27,160 --> 00:05:29,650 intersect at that same point in the second quadrant. 137 00:05:29,649 --> 00:05:32,509 So hopefully I'm making some sense to you, as I try to make 138 00:05:32,509 --> 00:05:34,409 some sense of this problem. 139 00:05:34,410 --> 00:05:34,980 OK. 140 00:05:34,980 --> 00:05:36,210 Now what do they want to know? 141 00:05:36,209 --> 00:05:38,469 And I think I only have time for the first part of this. 142 00:05:38,470 --> 00:05:41,030 Maybe I'll do the second part in the another video. 143 00:05:41,029 --> 00:05:45,849 Find the equation of the extreme normal line. 144 00:05:45,850 --> 00:05:48,790 Well, that seems very daunting at first, but I think our 145 00:05:48,790 --> 00:05:51,790 toolkit of derivatives, and what we know about equations 146 00:05:51,790 --> 00:05:54,710 of a line, should be able to get us there. 147 00:05:54,709 --> 00:05:58,469 So what's the slope of the tangent line at any 148 00:05:58,470 --> 00:05:59,490 point on this curve? 149 00:05:59,490 --> 00:06:01,540 Well, we just take the derivative of y equals 150 00:06:01,540 --> 00:06:04,910 x squared, and y prime is just equal to 2x. 151 00:06:04,910 --> 00:06:14,040 This is the slope of the tangent at any point x. 152 00:06:14,040 --> 00:06:18,290 So if I want to know the slope of the tangent at x0, at some 153 00:06:18,290 --> 00:06:22,250 particular x, I would just say, well, let me just say, 154 00:06:22,250 --> 00:06:25,009 slope, it would be 2 x0. 155 00:06:25,009 --> 00:06:29,490 Or let me just say, f of x0 is equal to 2 x0. 156 00:06:29,490 --> 00:06:32,759 This is the slope at any particular x0 of 157 00:06:32,759 --> 00:06:34,310 the tangent line. 158 00:06:34,310 --> 00:06:38,550 Now, the normal line slope is perpendicular to this. 159 00:06:38,550 --> 00:06:41,720 So the perpendicular line, and I won't review it here, but the 160 00:06:41,720 --> 00:06:44,730 perpendicular line has a negative inverse slope. 161 00:06:44,730 --> 00:06:54,460 So the slope of normal line at x0 will be the negative inverse 162 00:06:54,459 --> 00:06:57,589 of this, because this is the slope of the tangent line x0. 163 00:06:57,589 --> 00:07:03,089 So it'll be equal to minus 1 over 2 x0. 164 00:07:03,089 --> 00:07:04,509 Fair enough. 165 00:07:04,509 --> 00:07:10,099 Now, what is the equation of the normal line at x0 let's say 166 00:07:10,100 --> 00:07:12,570 that this is my x0 in question. 167 00:07:12,569 --> 00:07:16,500 What is the equation of the normal line there? 168 00:07:16,500 --> 00:07:18,959 Well, we can just use the point-slope form 169 00:07:18,959 --> 00:07:20,329 of our equation. 170 00:07:20,329 --> 00:07:23,569 So this point right here will be on the normal line. 171 00:07:23,569 --> 00:07:27,230 And that's the point x0 squared. 172 00:07:27,230 --> 00:07:31,030 Because this the graph of y equals x0, x squared. 173 00:07:31,029 --> 00:07:35,719 So this normal line will also have this point. 174 00:07:35,720 --> 00:07:38,970 So we could say that the equation of the normal line, 175 00:07:38,970 --> 00:07:48,300 let me write it down, would be equal to, this is just a 176 00:07:48,300 --> 00:07:50,480 point-slope definition of a line. 177 00:07:50,480 --> 00:07:55,480 You say, y minus the y-point, which is just x0 squared, 178 00:07:55,480 --> 00:07:59,810 that's that right there, is equal to the slope of the 179 00:07:59,810 --> 00:08:08,865 normal line minus 1 over 2 x0 times x minus the 180 00:08:08,865 --> 00:08:10,050 x-point that we're at. 181 00:08:10,050 --> 00:08:12,680 Minus x, minus x0. 182 00:08:12,680 --> 00:08:15,780 This is the equation of the normal line. 183 00:08:15,779 --> 00:08:16,259 So let's see. 184 00:08:16,259 --> 00:08:19,620 And what we care about is when x0 is greater than 0, right? 185 00:08:19,620 --> 00:08:22,740 We care about the normal line when we're in the first 186 00:08:22,740 --> 00:08:25,410 quadrant, we're in all of these values right there. 187 00:08:25,410 --> 00:08:28,710 So that's my equation of the normal line. 188 00:08:28,709 --> 00:08:32,149 And let's solve it explicitly in terms of x. 189 00:08:32,149 --> 00:08:33,799 So y is a function of x. 190 00:08:33,799 --> 00:08:38,779 Well, if I add x0 squared to both sides, I get y is 191 00:08:38,779 --> 00:08:42,759 equal to, actually, let me multiply this guy out. 192 00:08:42,759 --> 00:08:52,340 I get minus 1/2 x0 times x, and then I have plus, plus, because 193 00:08:52,340 --> 00:08:55,879 I have a minus times a minus, plus 1/2. 194 00:08:55,879 --> 00:08:59,110 The x0 and the over the x0, they cancel out. 195 00:08:59,110 --> 00:09:01,340 And then I have to add this x0 to both sides. 196 00:09:01,340 --> 00:09:05,560 So all I did so far, this is just this part right there. 197 00:09:05,559 --> 00:09:07,039 That's this right there. 198 00:09:07,039 --> 00:09:10,259 And then I have to add this to both sides of the equation, so 199 00:09:10,259 --> 00:09:14,129 then I have plus x0 squared. 200 00:09:14,129 --> 00:09:18,299 So this is the equation of the normal line, in mx plus b form. 201 00:09:18,299 --> 00:09:22,859 This is its slope, this is the m, and then this is its 202 00:09:22,860 --> 00:09:24,740 y-intercept right here. 203 00:09:24,740 --> 00:09:27,279 That's kind of the b. 204 00:09:27,279 --> 00:09:29,199 Now, what do we care about? 205 00:09:29,200 --> 00:09:32,600 We care about where this thing intersects. 206 00:09:32,600 --> 00:09:35,990 We care about where it intersects the parabola. 207 00:09:35,990 --> 00:09:38,129 And the parabola, that's pretty straightforward, that's just 208 00:09:38,129 --> 00:09:41,590 y is equal to x squared. 209 00:09:41,590 --> 00:09:43,960 So to figure out where they intersect, we just have 210 00:09:43,960 --> 00:09:46,660 to set the 2 y's to be equal to each other. 211 00:09:46,659 --> 00:09:50,110 So they intersect, the x-values where they intersect, x 212 00:09:50,110 --> 00:09:52,659 squared, this y would have to be equal to that y. 213 00:09:52,659 --> 00:09:55,939 Or we could just substitute this in for that y. 214 00:09:55,940 --> 00:10:02,870 So you get x squared is equal to minus 1 over 2 x0 times x, 215 00:10:02,870 --> 00:10:10,179 plus 1/2 plus x0 squared. 216 00:10:10,179 --> 00:10:11,019 Fair enough. 217 00:10:11,019 --> 00:10:15,350 And let's put this in a quadratic equation, or try to 218 00:10:15,350 --> 00:10:17,670 solve this, so we can apply the quadratic equation. 219 00:10:17,669 --> 00:10:20,409 So let's put all of this stuff on the lefthand side. 220 00:10:20,409 --> 00:10:29,419 So you get x squared plus 1 over 2 x0 times x minus 221 00:10:29,419 --> 00:10:35,679 all of this, 1/2 plus x0 squared is equal to 0. 222 00:10:35,679 --> 00:10:38,099 All I did is, I took all of this stuff and I put it on the 223 00:10:38,100 --> 00:10:39,769 lefthand side of the equation. 224 00:10:39,769 --> 00:10:42,340 Now, this is just a standard quadratic equation, so we can 225 00:10:42,340 --> 00:10:47,680 figure out now where the x-values that satisfy this 226 00:10:47,679 --> 00:10:52,419 quadratic equation will tell us where our normal line 227 00:10:52,419 --> 00:10:54,889 and our parabola intersect. 228 00:10:54,889 --> 00:10:57,850 So let's just apply the quadratic equation here. 229 00:10:57,850 --> 00:11:00,690 So the potential x-values, where they intersect, x is 230 00:11:00,690 --> 00:11:02,950 equal to minus b, I'm just applying the 231 00:11:02,950 --> 00:11:04,400 quadratic equation. 232 00:11:04,399 --> 00:11:10,949 So minus b is minus 1 over 2 x0, plus or minus the 233 00:11:10,950 --> 00:11:14,060 square root of b squared. 234 00:11:14,059 --> 00:11:15,199 So that's that squared. 235 00:11:15,200 --> 00:11:22,520 So it's one over four x0 squared minus 4ac. 236 00:11:22,519 --> 00:11:26,569 So minus 4 times 1 times this minus thing. 237 00:11:26,570 --> 00:11:29,320 So I'm going to have a minus times a minus is a plus, so 238 00:11:29,320 --> 00:11:31,370 it's just 4 times this, because there was one there. 239 00:11:31,370 --> 00:11:35,720 So plus 4 times this, right here. 240 00:11:35,720 --> 00:11:42,940 4 times this is just 2 plus 4 x0 squared. 241 00:11:42,940 --> 00:11:46,840 All I did is, this is 4ac right here. 242 00:11:46,840 --> 00:11:48,190 Well, minus 4ac. 243 00:11:48,190 --> 00:11:50,295 The minus and the minus canceled out, so 244 00:11:50,294 --> 00:11:50,979 you got a plus. 245 00:11:50,980 --> 00:11:51,820 There's a 1. 246 00:11:51,820 --> 00:12:00,010 So 4 times c is just 2 plus 4x squared. 247 00:12:00,009 --> 00:12:03,409 I just multiply this by two, and of course all of this 248 00:12:03,409 --> 00:12:07,289 should be over 2 times a. a is just 2 there. 249 00:12:07,289 --> 00:12:09,000 So let's see if I can simplify this. 250 00:12:09,000 --> 00:12:09,889 Remember what we're doing. 251 00:12:09,889 --> 00:12:14,740 We're just figuring out where the normal line and the 252 00:12:14,740 --> 00:12:16,810 parabola intersect. 253 00:12:16,809 --> 00:12:18,539 Now, what do we get here. 254 00:12:18,539 --> 00:12:21,349 This looks like a little hairy beast here. 255 00:12:21,350 --> 00:12:24,980 Let me see if I can simplify this a little bit. 256 00:12:24,980 --> 00:12:31,310 So let us factor out-- let me rewrite this. 257 00:12:31,309 --> 00:12:34,649 258 00:12:34,649 --> 00:12:40,289 I can just divide everything by 1/2, so this is minus 1 over 4 259 00:12:40,289 --> 00:12:47,110 x0, I just divided this by 2, plus or minus 1/2, that's just 260 00:12:47,110 --> 00:12:51,720 this 1/2 right there, times the square root, let me see what 261 00:12:51,720 --> 00:12:53,560 I can simplify out of here. 262 00:12:53,559 --> 00:13:03,299 So if I factor out a 4 over x0 squared, then what does 263 00:13:03,299 --> 00:13:05,629 my expression become? 264 00:13:05,629 --> 00:13:12,000 This term right here will become an x to the fourth, x0 265 00:13:12,000 --> 00:13:17,139 to the fourth, plus, now, what does this term become? 266 00:13:17,139 --> 00:13:25,240 This term becomes a 1/2 x0 squared. 267 00:13:25,240 --> 00:13:28,110 And just to verify this, multiply 4 times 1/2, you 268 00:13:28,110 --> 00:13:31,110 get 2, and then the x0 squares cancel out. 269 00:13:31,110 --> 00:13:34,409 So write this term times that, will equal 2, and then you have 270 00:13:34,409 --> 00:13:38,279 plus-- now we factored a four out of this and the x0 271 00:13:38,279 --> 00:13:40,761 squared, so plus 1/16. 272 00:13:40,761 --> 00:13:45,059 Let me scroll over a little bit. 273 00:13:45,059 --> 00:13:46,839 And you can verify that this works out. 274 00:13:46,840 --> 00:13:50,280 If you were to multiply this out, you should get this 275 00:13:50,279 --> 00:13:51,339 business right here. 276 00:13:51,340 --> 00:13:54,200 277 00:13:54,200 --> 00:13:56,680 I see the home stretch here, because this should actually 278 00:13:56,679 --> 00:13:58,889 factor out quite neatly. 279 00:13:58,889 --> 00:14:01,090 So what does this equal? 280 00:14:01,090 --> 00:14:04,120 So the intersection of our normal line and our 281 00:14:04,120 --> 00:14:06,649 parabola is equal to this. 282 00:14:06,649 --> 00:14:12,509 Minus 1 over 4 x0 plus or minus 1/2 times the square 283 00:14:12,509 --> 00:14:13,289 root of this business. 284 00:14:13,289 --> 00:14:16,349 And the square root, this thing right here is 285 00:14:16,350 --> 00:14:19,159 4 over x0 squared. 286 00:14:19,159 --> 00:14:19,879 Now what's this? 287 00:14:19,879 --> 00:14:23,679 This is actually, lucky for us, a perfect square. 288 00:14:23,679 --> 00:14:25,849 And I won't go into details, because then the video will get 289 00:14:25,850 --> 00:14:28,090 too long, but I think you can recognize that this is 290 00:14:28,090 --> 00:14:33,580 x0 squared, plus 1/4. 291 00:14:33,580 --> 00:14:36,410 If you don't believe me, square this thing right here. 292 00:14:36,409 --> 00:14:38,480 You'll get this expression right there. 293 00:14:38,480 --> 00:14:40,370 And luckily enough, this is a perfect square, so we 294 00:14:40,370 --> 00:14:42,649 can actually take the square root of it. 295 00:14:42,649 --> 00:14:46,110 And so we get, the point at which they intersect, 296 00:14:46,110 --> 00:14:48,730 our normal line and our parabola, and this is 297 00:14:48,730 --> 00:14:50,970 quite a hairy problem. 298 00:14:50,970 --> 00:14:54,960 The points where they intersect is minus 1 over 4 x0, plus 299 00:14:54,960 --> 00:14:56,990 or minus 1/2 times the square root of this. 300 00:14:56,990 --> 00:14:59,269 The square root of this is the square root of this, which is 301 00:14:59,269 --> 00:15:05,809 just 2 over x0 times the square root of this, which is 302 00:15:05,809 --> 00:15:09,589 x0 squared plus 1/4. 303 00:15:09,590 --> 00:15:17,860 And if I were to rewrite all of this, I'd get minus 1 over 4 x0 304 00:15:17,860 --> 00:15:21,090 plus, let's see, this 1/2 and this 2 cancel out, right? 305 00:15:21,090 --> 00:15:22,810 So these cancel out. 306 00:15:22,809 --> 00:15:26,649 So plus or minus, now I just have a one over 307 00:15:26,649 --> 00:15:27,959 x0 times x0 squared. 308 00:15:27,960 --> 00:15:35,360 So I have 1 over x0-- oh sorry, let me, we have to be very 309 00:15:35,360 --> 00:15:39,475 careful there-- x0 squared divided by x0 is just x0, let 310 00:15:39,475 --> 00:15:42,399 me do that in a yellow color so you know what I'm dealing with. 311 00:15:42,399 --> 00:15:46,939 This term multiplied by this term is just x0, and then 312 00:15:46,940 --> 00:15:51,560 you have a plus 1/4 x0. 313 00:15:51,559 --> 00:15:53,629 And this is all a parentheses here. 314 00:15:53,629 --> 00:15:57,279 So these are the two points at which the normal curve and 315 00:15:57,279 --> 00:15:58,610 our parabola intersect. 316 00:15:58,610 --> 00:15:59,810 Let me just be very clear. 317 00:15:59,809 --> 00:16:04,229 Those 2 points are, for if this is my x0 that we're 318 00:16:04,230 --> 00:16:05,379 dealing with, right there. 319 00:16:05,379 --> 00:16:07,404 It's this point and this point. 320 00:16:07,404 --> 00:16:10,069 And we have a plus or minus here, so this is going to be 321 00:16:10,070 --> 00:16:12,920 the plus version, and this is going to be the minus version. 322 00:16:12,919 --> 00:16:15,679 In fact, the plus version should simplify into x0. 323 00:16:15,679 --> 00:16:17,669 Let's see if that's the case. 324 00:16:17,669 --> 00:16:22,360 Let's see if the plus version actually simplifies to x0. 325 00:16:22,360 --> 00:16:23,830 So these are our two points. 326 00:16:23,830 --> 00:16:26,600 If I take the plus version, that should be our first 327 00:16:26,600 --> 00:16:27,769 quadrant intersection. 328 00:16:27,769 --> 00:16:35,980 So x is equal to minus 1/4 x0 plus x0 plus 1/4 x0. 329 00:16:35,980 --> 00:16:38,320 And, good enough, it does actually cancel out. 330 00:16:38,320 --> 00:16:39,530 That cancels out. 331 00:16:39,529 --> 00:16:42,139 So x0 is one of the points of intersection, which 332 00:16:42,139 --> 00:16:43,259 makes complete sense. 333 00:16:43,259 --> 00:16:45,939 Because that's how we even defined the problem. 334 00:16:45,940 --> 00:16:48,540 But, so this is the first quadrant intersection. 335 00:16:48,539 --> 00:16:50,769 So that's the first quadrant intersection. 336 00:16:50,769 --> 00:16:53,429 The second quadrant intersection will be 337 00:16:53,429 --> 00:16:55,689 where we take the minus sign right there. 338 00:16:55,690 --> 00:16:59,890 So x, I'll just call it in the second quadrant intersection, 339 00:16:59,889 --> 00:17:06,240 it'd be equal to minus 1/4 x0 minus this stuff over here, 340 00:17:06,240 --> 00:17:08,019 minus the stuff there. 341 00:17:08,019 --> 00:17:19,309 So minus x0 minus 1 over 4 mine x0. 342 00:17:19,309 --> 00:17:20,849 Now what do we have? 343 00:17:20,849 --> 00:17:21,269 So let's see. 344 00:17:21,269 --> 00:17:24,150 We have a minus 1 over 4 x0, minus 1 over 4 x0. 345 00:17:24,150 --> 00:17:33,019 So this is equal to minus x0, minus x0, minus 1 over 2 x0. 346 00:17:33,019 --> 00:17:36,759 So if I take minus 1/4 minus 1/4, I get minus 1/2. 347 00:17:36,759 --> 00:17:39,779 And so my second quadrant intersection, all this work 348 00:17:39,779 --> 00:17:41,899 I did got me this result. 349 00:17:41,900 --> 00:17:43,990 My second quadrant intersection, I hope I 350 00:17:43,990 --> 00:17:46,089 don't run out of space. 351 00:17:46,089 --> 00:17:49,730 My second quadrant intersection, of the normal 352 00:17:49,730 --> 00:17:59,730 line and the parabola, is minus x0 minus 1 over 2 x0. 353 00:17:59,730 --> 00:18:02,970 Now this by itself is a pretty neat result we just got, but 354 00:18:02,970 --> 00:18:05,880 we're unfortunately not done with the problem. 355 00:18:05,880 --> 00:18:10,550 Because the problem wants us to find that point, the maximum 356 00:18:10,549 --> 00:18:11,619 point of intersection. 357 00:18:11,619 --> 00:18:13,419 They call this the extreme normal line. 358 00:18:13,420 --> 00:18:16,930 The extreme normal line is when our second quadrant 359 00:18:16,930 --> 00:18:19,560 intersection essentially achieves a maximum point. 360 00:18:19,559 --> 00:18:21,750 I know they call it the smallest point, but it's the 361 00:18:21,750 --> 00:18:24,869 smallest negative value, so it's really a maximum point. 362 00:18:24,869 --> 00:18:27,439 So how do we figure out that maximum point? 363 00:18:27,440 --> 00:18:31,940 Well, we have our second quadrant intersection as 364 00:18:31,940 --> 00:18:35,460 a function of our first quadrant x. 365 00:18:35,460 --> 00:18:38,950 I could rewrite this as, my second quadrant intersection as 366 00:18:38,950 --> 00:18:45,210 a function of x0 is equal to minus x minus 1 over 2 x0. 367 00:18:45,210 --> 00:18:48,269 So this is going to reach a minimum or a maximum point when 368 00:18:48,269 --> 00:18:50,789 its derivative is equal to 0. 369 00:18:50,789 --> 00:18:52,909 This is a very unconventional notation, and that's 370 00:18:52,910 --> 00:18:54,529 probably the hardest thing about this problem. 371 00:18:54,529 --> 00:18:58,389 But let's take this derivative with respect to x0. 372 00:18:58,390 --> 00:19:01,470 So my second quadrant intersection, the derivative of 373 00:19:01,470 --> 00:19:05,910 that with respect to x0, is equal to, this is pretty 374 00:19:05,910 --> 00:19:06,580 straightforward. 375 00:19:06,579 --> 00:19:12,859 It's equal to minus 1, and then I have a minus 1/2 times, this 376 00:19:12,859 --> 00:19:14,799 is the same thing as x to the minus 1. 377 00:19:14,799 --> 00:19:20,629 So it's minus 1 times x0 to the minus 2, right? 378 00:19:20,630 --> 00:19:23,320 I could have rewritten this as minus 1/2 times 379 00:19:23,319 --> 00:19:24,849 x0 to the minus 1. 380 00:19:24,849 --> 00:19:26,559 So you just put its exponent out front and 381 00:19:26,559 --> 00:19:28,089 decrement it by 1. 382 00:19:28,089 --> 00:19:31,869 And so this is the derivative with respect to my first 383 00:19:31,869 --> 00:19:33,769 quadrant intersection. 384 00:19:33,769 --> 00:19:35,759 So let me simplify this. 385 00:19:35,759 --> 00:19:41,819 So x, my second quadrant intersection, the derivative of 386 00:19:41,819 --> 00:19:45,669 it with respect to my first quadrant intersection, is equal 387 00:19:45,670 --> 00:19:52,110 to minus 1, the minus 1/2 and the minus 1 become a positive 388 00:19:52,109 --> 00:19:58,189 when you multiply them, and so plus 1/2 over x0 squared. 389 00:19:58,190 --> 00:20:01,279 Now, this'll reach a maximum or minimum when it equals 0. 390 00:20:01,279 --> 00:20:06,599 So let's set that equal to 0, and then solve this 391 00:20:06,599 --> 00:20:08,409 problem right there. 392 00:20:08,410 --> 00:20:10,000 Well, we add one to both sides. 393 00:20:10,000 --> 00:20:16,180 We get 1 over 2 x0 squared is equal to 1, or you could just 394 00:20:16,180 --> 00:20:20,509 say that that means that 2 x0 squared must be equal to 395 00:20:20,509 --> 00:20:23,750 1, if we just invert both sides of this equation. 396 00:20:23,750 --> 00:20:32,740 Or we could say that x0 squared is equal to 1/2, or if we take 397 00:20:32,740 --> 00:20:35,779 the square roots of both sides of that equation, we get x0 is 398 00:20:35,779 --> 00:20:40,029 equal to 1 over the square root of 2. 399 00:20:40,029 --> 00:20:42,470 So we're really, really, really close now. 400 00:20:42,470 --> 00:20:46,000 We've just figured out the x0 value that gives us 401 00:20:46,000 --> 00:20:47,529 our extreme normal line. 402 00:20:47,529 --> 00:20:48,910 This value right here. 403 00:20:48,910 --> 00:20:51,080 Let me do it in a nice deeper color. 404 00:20:51,079 --> 00:20:53,639 This value right here, that gives us the extreme normal 405 00:20:53,640 --> 00:20:57,410 line, that over there is x0 is equal to 1 over 406 00:20:57,410 --> 00:20:58,930 the square root of 2. 407 00:20:58,930 --> 00:21:01,390 Now, they want us to figure out the equation of the 408 00:21:01,390 --> 00:21:02,910 extreme normal line. 409 00:21:02,910 --> 00:21:05,710 Well, the equation of the extreme normal line we already 410 00:21:05,710 --> 00:21:06,670 figured out right here. 411 00:21:06,670 --> 00:21:07,910 It's this. 412 00:21:07,910 --> 00:21:11,900 The equation of the normal line is that thing, right there. 413 00:21:11,900 --> 00:21:15,400 So if we want the equation of the normal line at this extreme 414 00:21:15,400 --> 00:21:18,120 point, right here, the one that creates the extreme normal 415 00:21:18,119 --> 00:21:21,139 line, I just substitute 1 over the square root of 2 in for x0. 416 00:21:21,140 --> 00:21:22,400 So what do I get? 417 00:21:22,400 --> 00:21:26,800 I get, and this is the home stretch, and this is quite 418 00:21:26,799 --> 00:21:28,440 a beast of a problem. 419 00:21:28,440 --> 00:21:31,670 y minus x0 squared. 420 00:21:31,670 --> 00:21:34,289 x0 squared is 1/2, right? 421 00:21:34,289 --> 00:21:36,430 1 over the square root of 2 squared is 1/2. 422 00:21:36,430 --> 00:21:42,509 Is equal to minus 1 over 2 x0. 423 00:21:42,509 --> 00:21:43,660 So let's be careful here. 424 00:21:43,660 --> 00:21:48,560 So minus 1/2 times 1 over x0. 425 00:21:48,559 --> 00:21:52,669 One over x0 is the square root of 2, right? 426 00:21:52,670 --> 00:21:57,529 All of that times x minus x0. 427 00:21:57,529 --> 00:22:02,079 So that's 1 over the square root of 2. x0 is one 428 00:22:02,079 --> 00:22:02,869 over square root of 2. 429 00:22:02,869 --> 00:22:04,849 So let's simplify this a little bit. 430 00:22:04,849 --> 00:22:07,569 So the equation of our normal line, assuming I haven't made 431 00:22:07,569 --> 00:22:13,399 any careless mistakes, is equal to, so y minus 1/2 432 00:22:13,400 --> 00:22:15,060 is equal to, let's see. 433 00:22:15,059 --> 00:22:21,409 If we multiply this minus square root of 2 over 2x, and 434 00:22:21,410 --> 00:22:25,310 then if I multiply these square root of 2 over 435 00:22:25,309 --> 00:22:26,690 this, it becomes one. 436 00:22:26,690 --> 00:22:29,120 And then I have a minus and a minus, so that 437 00:22:29,119 --> 00:22:31,289 I have a plus 1/2. 438 00:22:31,289 --> 00:22:32,230 I think that's right. 439 00:22:32,230 --> 00:22:37,589 Yeah, plus 1/2, this times this times that 440 00:22:37,589 --> 00:22:39,169 is equal to plus 1/2. 441 00:22:39,170 --> 00:22:41,190 And then, we're at the home stretch. 442 00:22:41,190 --> 00:22:44,650 So we just add 1/2 to both sides of this equation, and we 443 00:22:44,650 --> 00:22:51,150 get our extreme normal line equation, which is y is 444 00:22:51,150 --> 00:22:54,670 equal to minus square root of 2 over 2x. 445 00:22:54,670 --> 00:23:00,100 If you add 1/2 to both sides of this equation, you get plus 1. 446 00:23:00,099 --> 00:23:01,299 And there you go. 447 00:23:01,299 --> 00:23:04,829 That's the equation of that line there, assuming I haven't 448 00:23:04,829 --> 00:23:06,019 made any careless mistakes. 449 00:23:06,019 --> 00:23:09,319 But even if I have, I think you get the idea of hopefully how 450 00:23:09,319 --> 00:23:12,710 to do this problem, which is quite a beastly one.