[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.86,0:00:02.50,Default,,0000,0000,0000,,So we have the\Nfunction f of x is
Dialogue: 0,0:00:02.50,0:00:05.29,Default,,0000,0000,0000,,equal to e to the\Nx over x squared.
Dialogue: 0,0:00:05.29,0:00:07.63,Default,,0000,0000,0000,,And what I want to\Ndo in this video
Dialogue: 0,0:00:07.63,0:00:10.96,Default,,0000,0000,0000,,is find the equation,\Nnot of the tangent line,
Dialogue: 0,0:00:10.96,0:00:17.22,Default,,0000,0000,0000,,but the equation of the normal\Nline, when x is equal to 1.
Dialogue: 0,0:00:17.22,0:00:22.97,Default,,0000,0000,0000,,So we care about the\Nequation of the normal line.
Dialogue: 0,0:00:22.97,0:00:26.25,Default,,0000,0000,0000,,So I encourage you to pause this\Nvideo and try this on your own.
Dialogue: 0,0:00:26.25,0:00:29.69,Default,,0000,0000,0000,,And if you need a little bit\Nof a hint, the hint I will give
Dialogue: 0,0:00:29.69,0:00:33.80,Default,,0000,0000,0000,,you is, is that the\Nslope of a normal line
Dialogue: 0,0:00:33.80,0:00:38.92,Default,,0000,0000,0000,,is going to be the\Nnegative reciprocal
Dialogue: 0,0:00:38.92,0:00:41.02,Default,,0000,0000,0000,,of the slope of\Nthe tangent line.
Dialogue: 0,0:00:41.02,0:00:46.64,Default,,0000,0000,0000,,If you imagine a\Ncurve like this,
Dialogue: 0,0:00:46.64,0:00:49.69,Default,,0000,0000,0000,,and we want to find a\Ntangent line at a point,
Dialogue: 0,0:00:49.69,0:00:51.31,Default,,0000,0000,0000,,it's going to look\Nsomething like this.
Dialogue: 0,0:00:51.31,0:00:54.60,Default,,0000,0000,0000,,So the tangent line is\Ngoing to look like this.
Dialogue: 0,0:00:54.60,0:00:59.22,Default,,0000,0000,0000,,A normal line is perpendicular\Nto the tangent line.
Dialogue: 0,0:00:59.22,0:01:01.47,Default,,0000,0000,0000,,This is the tangent line.
Dialogue: 0,0:01:01.47,0:01:05.71,Default,,0000,0000,0000,,The normal line is going to\Nbe perpendicular to that.
Dialogue: 0,0:01:05.71,0:01:08.31,Default,,0000,0000,0000,,It's going to go just like that.
Dialogue: 0,0:01:08.31,0:01:11.45,Default,,0000,0000,0000,,And if this has a\Nslope of m, then this
Dialogue: 0,0:01:11.45,0:01:14.10,Default,,0000,0000,0000,,has a slope of the\Nnegative reciprocal of m.
Dialogue: 0,0:01:14.10,0:01:17.35,Default,,0000,0000,0000,,So negative 1/m.
Dialogue: 0,0:01:17.35,0:01:19.20,Default,,0000,0000,0000,,So with that as a\Nlittle bit of a hint,
Dialogue: 0,0:01:19.20,0:01:21.88,Default,,0000,0000,0000,,I encourage you to find the\Nequation of the normal line
Dialogue: 0,0:01:21.88,0:01:25.41,Default,,0000,0000,0000,,to this curve, when x equals 1.
Dialogue: 0,0:01:25.41,0:01:29.08,Default,,0000,0000,0000,,So let's find the slope\Nof the tangent line.
Dialogue: 0,0:01:29.08,0:01:30.79,Default,,0000,0000,0000,,And then we take the\Nnegative reciprocal,
Dialogue: 0,0:01:30.79,0:01:33.07,Default,,0000,0000,0000,,we can find the slope\Nof the normal line.
Dialogue: 0,0:01:33.07,0:01:35.19,Default,,0000,0000,0000,,So to find the slope\Nof the tangent line,
Dialogue: 0,0:01:35.19,0:01:38.66,Default,,0000,0000,0000,,we just take the derivative here\Nand evaluate it at x equals 1.
Dialogue: 0,0:01:38.66,0:01:41.18,Default,,0000,0000,0000,,So f prime of x,\Nand actually, let me
Dialogue: 0,0:01:41.18,0:01:42.40,Default,,0000,0000,0000,,rewrite this a little bit.
Dialogue: 0,0:01:42.40,0:01:47.22,Default,,0000,0000,0000,,So f of x is equal to e to the\Nx times x to the negative 2.
Dialogue: 0,0:01:47.22,0:01:49.18,Default,,0000,0000,0000,,I like to rewrite it this\Nway, because I always
Dialogue: 0,0:01:49.18,0:01:50.72,Default,,0000,0000,0000,,forget the whole\Nquotient rule thing.
Dialogue: 0,0:01:50.72,0:01:52.77,Default,,0000,0000,0000,,I like the power\Nrule a lot more.
Dialogue: 0,0:01:52.77,0:01:54.61,Default,,0000,0000,0000,,And this allows me to\Nuse the power rule.
Dialogue: 0,0:01:54.61,0:01:57.79,Default,,0000,0000,0000,,I'm sorry, not the power\Nrule, the product rule.
Dialogue: 0,0:01:57.79,0:01:59.25,Default,,0000,0000,0000,,So this allows me\Nto do the product
Dialogue: 0,0:01:59.25,0:02:01.15,Default,,0000,0000,0000,,rule instead of\Nthe quotient rule.
Dialogue: 0,0:02:01.15,0:02:03.65,Default,,0000,0000,0000,,So the derivative\Nof this, f prime
Dialogue: 0,0:02:03.65,0:02:05.93,Default,,0000,0000,0000,,of x, is going to be the\Nderivative of e to the x.
Dialogue: 0,0:02:05.93,0:02:11.50,Default,,0000,0000,0000,,Which is just e to the x times\Nx to the negative 2, plus
Dialogue: 0,0:02:11.50,0:02:14.78,Default,,0000,0000,0000,,e to the x times the derivative\Nof x to the negative 2.
Dialogue: 0,0:02:14.78,0:02:18.65,Default,,0000,0000,0000,,Which is negative 2x to\Nthe negative 3 power.
Dialogue: 0,0:02:18.65,0:02:21.20,Default,,0000,0000,0000,,I just used the power\Nrule right over here.
Dialogue: 0,0:02:21.20,0:02:24.71,Default,,0000,0000,0000,,So if I want to evaluate\Nwhen x is equal to 1,
Dialogue: 0,0:02:24.71,0:02:26.45,Default,,0000,0000,0000,,this is going to\Nbe equal to-- let
Dialogue: 0,0:02:26.45,0:02:28.57,Default,,0000,0000,0000,,me do that in that\Nyellow color like.
Dialogue: 0,0:02:28.57,0:02:30.00,Default,,0000,0000,0000,,I like switching colors.
Dialogue: 0,0:02:30.00,0:02:32.95,Default,,0000,0000,0000,,This is going to be\Nequal to, let's see,
Dialogue: 0,0:02:32.95,0:02:35.58,Default,,0000,0000,0000,,this is going to be\Ne to the first power.
Dialogue: 0,0:02:35.58,0:02:39.10,Default,,0000,0000,0000,,Which is just e times\N1 to the negative 2,
Dialogue: 0,0:02:39.10,0:02:46.29,Default,,0000,0000,0000,,which is just 1 plus e to the\Nfirst power, which is just e,
Dialogue: 0,0:02:46.29,0:02:48.41,Default,,0000,0000,0000,,times negative 2.
Dialogue: 0,0:02:48.41,0:02:50.21,Default,,0000,0000,0000,,1 to the negative 3 is just 1.
Dialogue: 0,0:02:50.21,0:02:51.64,Default,,0000,0000,0000,,So e times negative 2.
Dialogue: 0,0:02:51.64,0:02:52.81,Default,,0000,0000,0000,,So let me write it this way.
Dialogue: 0,0:02:52.81,0:02:57.39,Default,,0000,0000,0000,,So minus 2e.
Dialogue: 0,0:02:57.39,0:03:02.22,Default,,0000,0000,0000,,And e minus 2e is just going\Nto be equal to negative e.
Dialogue: 0,0:03:02.22,0:03:12.19,Default,,0000,0000,0000,,So this right over here, this is\Nthe slope of the tangent line.
Dialogue: 0,0:03:12.19,0:03:14.39,Default,,0000,0000,0000,,And so if we want the\Nslope of the normal,
Dialogue: 0,0:03:14.39,0:03:16.10,Default,,0000,0000,0000,,we just take the\Nnegative reciprocal.
Dialogue: 0,0:03:16.10,0:03:19.70,Default,,0000,0000,0000,,So the negative reciprocal\Nof this is going to be,
Dialogue: 0,0:03:19.70,0:03:21.41,Default,,0000,0000,0000,,well the reciprocal\Nis 1 over negative e,
Dialogue: 0,0:03:21.41,0:03:22.78,Default,,0000,0000,0000,,but we want the\Nnegative of that.
Dialogue: 0,0:03:22.78,0:03:24.24,Default,,0000,0000,0000,,So it's going to be 1/e.
Dialogue: 0,0:03:24.24,0:03:27.26,Default,,0000,0000,0000,,This is going to be the\Nslope of the normal line.
Dialogue: 0,0:03:34.59,0:03:37.07,Default,,0000,0000,0000,,And then if we, and\Nour goal isn't just
Dialogue: 0,0:03:37.07,0:03:38.52,Default,,0000,0000,0000,,to the slope of\Nthe normal line, we
Dialogue: 0,0:03:38.52,0:03:40.34,Default,,0000,0000,0000,,want the equation\Nof the normal line.
Dialogue: 0,0:03:40.34,0:03:42.56,Default,,0000,0000,0000,,And we know the\Nequation of a line
Dialogue: 0,0:03:42.56,0:03:45.86,Default,,0000,0000,0000,,can be represented\Nas y is equal to mx
Dialogue: 0,0:03:45.86,0:03:47.97,Default,,0000,0000,0000,,plus b, where m is the slope.
Dialogue: 0,0:03:47.97,0:03:52.05,Default,,0000,0000,0000,,So we can say it's going to be\Ny is equal to 1/e-- remember,
Dialogue: 0,0:03:52.05,0:03:56.66,Default,,0000,0000,0000,,we're doing the normal\Nline here-- times x plus b.
Dialogue: 0,0:03:56.66,0:03:59.25,Default,,0000,0000,0000,,And to solve for b, we\Njust have to recognize
Dialogue: 0,0:03:59.25,0:04:01.21,Default,,0000,0000,0000,,that we know a point\Nthat this goes through.
Dialogue: 0,0:04:01.21,0:04:04.18,Default,,0000,0000,0000,,This goes through\Nthe point x equals 1.
Dialogue: 0,0:04:04.18,0:04:09.97,Default,,0000,0000,0000,,And when x equals 1, what is y?
Dialogue: 0,0:04:09.97,0:04:13.43,Default,,0000,0000,0000,,Well, y is e to the 1st\Nover 1, which is just e.
Dialogue: 0,0:04:13.43,0:04:16.82,Default,,0000,0000,0000,,So this goes to the\Npoint 1 comma e.
Dialogue: 0,0:04:16.82,0:04:22.80,Default,,0000,0000,0000,,So we know that when x is\Nequal to 1, y is equal to e.
Dialogue: 0,0:04:22.80,0:04:25.02,Default,,0000,0000,0000,,And now we can just solve for b.
Dialogue: 0,0:04:25.02,0:04:34.96,Default,,0000,0000,0000,,So we get e is\Nequal to 1/e plus b.
Dialogue: 0,0:04:34.96,0:04:38.63,Default,,0000,0000,0000,,Or we could just subtract\N1 over e from both sides,
Dialogue: 0,0:04:38.63,0:04:45.16,Default,,0000,0000,0000,,and we would get b is\Nequal to e minus 1/e.
Dialogue: 0,0:04:48.19,0:04:53.36,Default,,0000,0000,0000,,And we could obviously right\Nthis as e squared minus 1/e
Dialogue: 0,0:04:53.36,0:04:54.73,Default,,0000,0000,0000,,if we want to\Nwrite it like that.
Dialogue: 0,0:04:54.73,0:04:56.36,Default,,0000,0000,0000,,But could just leave\Nit just like this.
Dialogue: 0,0:04:56.36,0:04:58.17,Default,,0000,0000,0000,,So the equation of\Nthe normal line--
Dialogue: 0,0:04:58.17,0:05:00.63,Default,,0000,0000,0000,,so we deserve our drum\Nroll right over here--
Dialogue: 0,0:05:00.63,0:05:10.02,Default,,0000,0000,0000,,is going to be y is equal\Nto 1/e times x, plus b.
Dialogue: 0,0:05:10.02,0:05:13.40,Default,,0000,0000,0000,,And b, plus b, is all of this.
Dialogue: 0,0:05:13.40,0:05:18.83,Default,,0000,0000,0000,,So plus e minus 1/e.
Dialogue: 0,0:05:18.83,0:05:25.53,Default,,0000,0000,0000,,So that right there is our\Nequation of the normal line.