1
00:00:00,860 --> 00:00:02,500
So we have the
function f of x is

2
00:00:02,500 --> 00:00:05,290
equal to e to the
x over x squared.

3
00:00:05,290 --> 00:00:07,630
And what I want to
do in this video

4
00:00:07,630 --> 00:00:10,960
is find the equation,
not of the tangent line,

5
00:00:10,960 --> 00:00:17,220
but the equation of the normal
line, when x is equal to 1.

6
00:00:17,220 --> 00:00:22,970
So we care about the
equation of the normal line.

7
00:00:22,970 --> 00:00:26,250
So I encourage you to pause this
video and try this on your own.

8
00:00:26,250 --> 00:00:29,690
And if you need a little bit
of a hint, the hint I will give

9
00:00:29,690 --> 00:00:33,800
you is, is that the
slope of a normal line

10
00:00:33,800 --> 00:00:38,920
is going to be the
negative reciprocal

11
00:00:38,920 --> 00:00:41,020
of the slope of
the tangent line.

12
00:00:41,020 --> 00:00:46,640
If you imagine a
curve like this,

13
00:00:46,640 --> 00:00:49,686
and we want to find a
tangent line at a point,

14
00:00:49,686 --> 00:00:51,310
it's going to look
something like this.

15
00:00:51,310 --> 00:00:54,600
So the tangent line is
going to look like this.

16
00:00:54,600 --> 00:00:59,220
A normal line is perpendicular
to the tangent line.

17
00:00:59,220 --> 00:01:01,470
This is the tangent line.

18
00:01:01,470 --> 00:01:05,710
The normal line is going to
be perpendicular to that.

19
00:01:05,710 --> 00:01:08,310
It's going to go just like that.

20
00:01:08,310 --> 00:01:11,450
And if this has a
slope of m, then this

21
00:01:11,450 --> 00:01:14,100
has a slope of the
negative reciprocal of m.

22
00:01:14,100 --> 00:01:17,350
So negative 1/m.

23
00:01:17,350 --> 00:01:19,200
So with that as a
little bit of a hint,

24
00:01:19,200 --> 00:01:21,880
I encourage you to find the
equation of the normal line

25
00:01:21,880 --> 00:01:25,410
to this curve, when x equals 1.

26
00:01:25,410 --> 00:01:29,082
So let's find the slope
of the tangent line.

27
00:01:29,082 --> 00:01:30,790
And then we take the
negative reciprocal,

28
00:01:30,790 --> 00:01:33,070
we can find the slope
of the normal line.

29
00:01:33,070 --> 00:01:35,190
So to find the slope
of the tangent line,

30
00:01:35,190 --> 00:01:38,660
we just take the derivative here
and evaluate it at x equals 1.

31
00:01:38,660 --> 00:01:41,180
So f prime of x,
and actually, let me

32
00:01:41,180 --> 00:01:42,400
rewrite this a little bit.

33
00:01:42,400 --> 00:01:47,222
So f of x is equal to e to the
x times x to the negative 2.

34
00:01:47,222 --> 00:01:49,180
I like to rewrite it this
way, because I always

35
00:01:49,180 --> 00:01:50,721
forget the whole
quotient rule thing.

36
00:01:50,721 --> 00:01:52,770
I like the power
rule a lot more.

37
00:01:52,770 --> 00:01:54,610
And this allows me to
use the power rule.

38
00:01:54,610 --> 00:01:57,792
I'm sorry, not the power
rule, the product rule.

39
00:01:57,792 --> 00:01:59,250
So this allows me
to do the product

40
00:01:59,250 --> 00:02:01,150
rule instead of
the quotient rule.

41
00:02:01,150 --> 00:02:03,650
So the derivative
of this, f prime

42
00:02:03,650 --> 00:02:05,930
of x, is going to be the
derivative of e to the x.

43
00:02:05,930 --> 00:02:11,500
Which is just e to the x times
x to the negative 2, plus

44
00:02:11,500 --> 00:02:14,780
e to the x times the derivative
of x to the negative 2.

45
00:02:14,780 --> 00:02:18,650
Which is negative 2x to
the negative 3 power.

46
00:02:18,650 --> 00:02:21,200
I just used the power
rule right over here.

47
00:02:21,200 --> 00:02:24,710
So if I want to evaluate
when x is equal to 1,

48
00:02:24,710 --> 00:02:26,450
this is going to
be equal to-- let

49
00:02:26,450 --> 00:02:28,570
me do that in that
yellow color like.

50
00:02:28,570 --> 00:02:30,000
I like switching colors.

51
00:02:30,000 --> 00:02:32,950
This is going to be
equal to, let's see,

52
00:02:32,950 --> 00:02:35,580
this is going to be
e to the first power.

53
00:02:35,580 --> 00:02:39,100
Which is just e times
1 to the negative 2,

54
00:02:39,100 --> 00:02:46,290
which is just 1 plus e to the
first power, which is just e,

55
00:02:46,290 --> 00:02:48,410
times negative 2.

56
00:02:48,410 --> 00:02:50,210
1 to the negative 3 is just 1.

57
00:02:50,210 --> 00:02:51,644
So e times negative 2.

58
00:02:51,644 --> 00:02:52,810
So let me write it this way.

59
00:02:52,810 --> 00:02:57,390
So minus 2e.

60
00:02:57,390 --> 00:03:02,220
And e minus 2e is just going
to be equal to negative e.

61
00:03:02,220 --> 00:03:12,190
So this right over here, this is
the slope of the tangent line.

62
00:03:12,190 --> 00:03:14,390
And so if we want the
slope of the normal,

63
00:03:14,390 --> 00:03:16,100
we just take the
negative reciprocal.

64
00:03:16,100 --> 00:03:19,702
So the negative reciprocal
of this is going to be,

65
00:03:19,702 --> 00:03:21,410
well the reciprocal
is 1 over negative e,

66
00:03:21,410 --> 00:03:22,785
but we want the
negative of that.

67
00:03:22,785 --> 00:03:24,240
So it's going to be 1/e.

68
00:03:24,240 --> 00:03:27,260
This is going to be the
slope of the normal line.

69
00:03:34,590 --> 00:03:37,066
And then if we, and
our goal isn't just

70
00:03:37,066 --> 00:03:38,524
to the slope of
the normal line, we

71
00:03:38,524 --> 00:03:40,340
want the equation
of the normal line.

72
00:03:40,340 --> 00:03:42,560
And we know the
equation of a line

73
00:03:42,560 --> 00:03:45,860
can be represented
as y is equal to mx

74
00:03:45,860 --> 00:03:47,970
plus b, where m is the slope.

75
00:03:47,970 --> 00:03:52,050
So we can say it's going to be
y is equal to 1/e-- remember,

76
00:03:52,050 --> 00:03:56,660
we're doing the normal
line here-- times x plus b.

77
00:03:56,660 --> 00:03:59,250
And to solve for b, we
just have to recognize

78
00:03:59,250 --> 00:04:01,210
that we know a point
that this goes through.

79
00:04:01,210 --> 00:04:04,180
This goes through
the point x equals 1.

80
00:04:04,180 --> 00:04:09,970
And when x equals 1, what is y?

81
00:04:09,970 --> 00:04:13,430
Well, y is e to the 1st
over 1, which is just e.

82
00:04:13,430 --> 00:04:16,820
So this goes to the
point 1 comma e.

83
00:04:16,820 --> 00:04:22,800
So we know that when x is
equal to 1, y is equal to e.

84
00:04:22,800 --> 00:04:25,020
And now we can just solve for b.

85
00:04:25,020 --> 00:04:34,960
So we get e is
equal to 1/e plus b.

86
00:04:34,960 --> 00:04:38,630
Or we could just subtract
1 over e from both sides,

87
00:04:38,630 --> 00:04:45,160
and we would get b is
equal to e minus 1/e.

88
00:04:48,190 --> 00:04:53,355
And we could obviously right
this as e squared minus 1/e

89
00:04:53,355 --> 00:04:54,730
if we want to
write it like that.

90
00:04:54,730 --> 00:04:56,355
But could just leave
it just like this.

91
00:04:56,355 --> 00:04:58,170
So the equation of
the normal line--

92
00:04:58,170 --> 00:05:00,630
so we deserve our drum
roll right over here--

93
00:05:00,630 --> 00:05:10,020
is going to be y is equal
to 1/e times x, plus b.

94
00:05:10,020 --> 00:05:13,400
And b, plus b, is all of this.

95
00:05:13,400 --> 00:05:18,830
So plus e minus 1/e.

96
00:05:18,830 --> 00:05:25,530
So that right there is our
equation of the normal line.