1
00:00:00,000 --> 00:00:00,620


2
00:00:00,620 --> 00:00:04,510
Now that we have distance
explicitly as a function of

3
00:00:04,510 --> 00:00:07,040
the angle that we're shooting
the object at, we can use a

4
00:00:07,040 --> 00:00:10,550
little bit of calculus to figure
out the optimal angle,

5
00:00:10,550 --> 00:00:13,290
the angle that's going optimize
our distance.

6
00:00:13,290 --> 00:00:17,340
And since we only care about
angles from 0 degrees to

7
00:00:17,340 --> 00:00:19,990
really 90 degrees, let's
constrain ourselves.

8
00:00:19,990 --> 00:00:21,870
So we're going to optimize
things for

9
00:00:21,870 --> 00:00:24,540
angles between 0 degrees.

10
00:00:24,540 --> 00:00:28,310
So theta is going to be greater
than or equal to 0 and

11
00:00:28,310 --> 00:00:31,490
less than or equal to 90.

12
00:00:31,490 --> 00:00:33,150
So let's see how we can do it.

13
00:00:33,150 --> 00:00:35,970
And just to get an idea of what
we're even conceptually

14
00:00:35,970 --> 00:00:38,000
doing with the calculus,
remember when you take a

15
00:00:38,000 --> 00:00:42,460
derivative, you are finding
the slope of a line, an

16
00:00:42,460 --> 00:00:44,760
instantaneous slope of a line.

17
00:00:44,760 --> 00:00:46,730
And if you were to graph this--
and I encourage you to

18
00:00:46,730 --> 00:00:49,960
graph it on your own, maybe with
a graphing calculator--

19
00:00:49,960 --> 00:00:53,850
it will look something like
this over the interval.

20
00:00:53,850 --> 00:00:58,560
It will look like this where
that is the distance as a

21
00:00:58,560 --> 00:01:02,410
function of theta axis
and then this would

22
00:01:02,410 --> 00:01:04,769
be our theta axis.

23
00:01:04,769 --> 00:01:07,810
And we care about angles between
0 and 90 degrees.

24
00:01:07,810 --> 00:01:11,670
So if you were to graph this
thing, so this is 0 degrees,

25
00:01:11,670 --> 00:01:14,620
this is maybe 90 degrees
right here.

26
00:01:14,620 --> 00:01:18,780
The graph of this function
will look like this.

27
00:01:18,780 --> 00:01:21,760
It'll look something
like this.

28
00:01:21,760 --> 00:01:24,510
It will look something
like that.

29
00:01:24,510 --> 00:01:27,900
And what we want to do is find
the angle, there's some angle

30
00:01:27,900 --> 00:01:32,720
here that gives us the
optimal distance.

31
00:01:32,720 --> 00:01:40,020
So this is, right here, this
is the optimal distance.

32
00:01:40,020 --> 00:01:41,560
And what we want to do
is find that out.

33
00:01:41,560 --> 00:01:43,570
And when you look at the graph,
and you could do it on

34
00:01:43,570 --> 00:01:46,640
a graphing calculator if you
like, what happens to the

35
00:01:46,640 --> 00:01:50,520
instantaneous slope at that
optimal distance?

36
00:01:50,520 --> 00:01:51,580
Well it's flat.

37
00:01:51,580 --> 00:01:53,130
The slope there is 0.

38
00:01:53,130 --> 00:01:56,540


39
00:01:56,540 --> 00:01:59,580
So what we need to do is take
the derivative of this

40
00:01:59,580 --> 00:02:02,850
function and then just figure
out at what angle is the

41
00:02:02,850 --> 00:02:06,080
derivative or the instantaneous
slope of this

42
00:02:06,080 --> 00:02:08,538
function equal to 0?

43
00:02:08,538 --> 00:02:11,590
And then we're done We will know
this mystery angle, this

44
00:02:11,590 --> 00:02:14,220
optimal angle, to shoot
the object at.

45
00:02:14,220 --> 00:02:16,020
So let's take the derivative.

46
00:02:16,020 --> 00:02:19,860
So the derivative, we'll just
use our derivative rules here.

47
00:02:19,860 --> 00:02:23,840
The derivative of-- I will call
it d prime I guess, or we

48
00:02:23,840 --> 00:02:25,920
could say the derivative of the
distance with respect to

49
00:02:25,920 --> 00:02:29,470
theta is equal to-- we're
assuming that s and g are

50
00:02:29,470 --> 00:02:32,360
constants, so we don't have to
worry about them right now.

51
00:02:32,360 --> 00:02:35,220
We could just put them out front
since we're assuming

52
00:02:35,220 --> 00:02:36,170
they're constants.

53
00:02:36,170 --> 00:02:38,580
And then we can do the product
rule to take the derivative of

54
00:02:38,580 --> 00:02:42,700
this part with respect
to theta.

55
00:02:42,700 --> 00:02:45,510
In the product rule, we take
the derivative of the first

56
00:02:45,510 --> 00:02:48,150
function times the
second function.

57
00:02:48,150 --> 00:02:51,070
So the derivative of
cosine of theta is

58
00:02:51,070 --> 00:02:54,820
negative sine of theta.

59
00:02:54,820 --> 00:02:56,010
And we're going to
multiply that

60
00:02:56,010 --> 00:02:58,320
times the second function.

61
00:02:58,320 --> 00:03:01,290
So that's times the
sine of theta.

62
00:03:01,290 --> 00:03:06,880
And to that we're going to add
the first function, which is

63
00:03:06,880 --> 00:03:10,580
cosine of theta times
the derivative

64
00:03:10,580 --> 00:03:12,170
of the second function.

65
00:03:12,170 --> 00:03:17,250
The derivative of sine theta
is cosine of theta.

66
00:03:17,250 --> 00:03:18,610
I know it's a little
bit confusing.

67
00:03:18,610 --> 00:03:20,820
All we did is we took the
derivative of the first one

68
00:03:20,820 --> 00:03:22,190
times the second one.

69
00:03:22,190 --> 00:03:24,770
And then we took the derivative
of the second one

70
00:03:24,770 --> 00:03:26,440
times the first one.

71
00:03:26,440 --> 00:03:28,450
Let me make it even more
explicitly clear.

72
00:03:28,450 --> 00:03:31,290
We took the derivative of this
guy here, so this is the

73
00:03:31,290 --> 00:03:33,520
derivative with respect
to theta.

74
00:03:33,520 --> 00:03:37,530
And we took the derivative of
this guy over here with

75
00:03:37,530 --> 00:03:39,720
respect to theta.

76
00:03:39,720 --> 00:03:40,960
We took the derivative
of cosine there and

77
00:03:40,960 --> 00:03:42,190
multiplied it by sine.

78
00:03:42,190 --> 00:03:43,510
Took the derivative
of sine here and

79
00:03:43,510 --> 00:03:44,860
multiplied it by cosine.

80
00:03:44,860 --> 00:03:46,290
Just the product rule.

81
00:03:46,290 --> 00:03:47,130
Now what does this give us?

82
00:03:47,130 --> 00:03:49,500
We can simplify this
a good bit.

83
00:03:49,500 --> 00:03:54,440
So we could write the derivative
d prime is equal

84
00:03:54,440 --> 00:03:57,170
to-- we could keep this constant
out there-- 2s

85
00:03:57,170 --> 00:04:01,910
squared over g-- times-- now
negative sine of theta times

86
00:04:01,910 --> 00:04:03,800
sine of theta, that's
just negative

87
00:04:03,800 --> 00:04:06,790
sine squared of theta.

88
00:04:06,790 --> 00:04:10,110
And then, cosine theta times
cosine theta, that's just plus

89
00:04:10,110 --> 00:04:12,440
cosine squared of theta.

90
00:04:12,440 --> 00:04:16,339
Now, what we just said is we
want to figure out the point,

91
00:04:16,339 --> 00:04:19,170
the angle at which this
derivative or the

92
00:04:19,170 --> 00:04:21,620
instantaneous slope is 0.

93
00:04:21,620 --> 00:04:25,430
So let's set this thing
equal to 0.

94
00:04:25,430 --> 00:04:27,450
So we just have to solve
for theta now.

95
00:04:27,450 --> 00:04:33,760
Now the first thing I do to
solve for theta is just divide

96
00:04:33,760 --> 00:04:37,280
both sides by 2s
squared over g.

97
00:04:37,280 --> 00:04:39,900
If you divide the left-hand side
by that, it cancels out

98
00:04:39,900 --> 00:04:41,350
with 2s squared over g.

99
00:04:41,350 --> 00:04:45,040
And if you divide 0 by that,
assuming this isn't 0, which

100
00:04:45,040 --> 00:04:47,550
it shouldn't be, then
you'll still get 0.

101
00:04:47,550 --> 00:04:51,660
So this equation simplifies to--
I'll write it in blue--

102
00:04:51,660 --> 00:04:57,540
negative sine squared of theta
plus cosine squared of theta

103
00:04:57,540 --> 00:04:59,630
is equal to 0.

104
00:04:59,630 --> 00:05:03,520
Now, if we add sine squared of
theta of both sides of this

105
00:05:03,520 --> 00:05:06,580
equation, let's add
sine squared of

106
00:05:06,580 --> 00:05:09,660
theta to both sides.

107
00:05:09,660 --> 00:05:14,260
We are left with--
these cancel out.

108
00:05:14,260 --> 00:05:17,800
Cosine squared of theta
is equal to

109
00:05:17,800 --> 00:05:19,270
sine squared of theta.

110
00:05:19,270 --> 00:05:22,470


111
00:05:22,470 --> 00:05:25,430
Now, both of these are going
to be positive over the

112
00:05:25,430 --> 00:05:27,470
interval, so we're going to just
take the positive square

113
00:05:27,470 --> 00:05:30,260
root of both of them, or the
principal root of both sides

114
00:05:30,260 --> 00:05:31,680
of this equation.

115
00:05:31,680 --> 00:05:32,990
So let's do that.

116
00:05:32,990 --> 00:05:35,300
So you take the principal
roots of both

117
00:05:35,300 --> 00:05:37,740
sides of this equation.

118
00:05:37,740 --> 00:05:38,510
You could do it that way.

119
00:05:38,510 --> 00:05:42,590
Actually, a more interesting way
than doing it that way, is

120
00:05:42,590 --> 00:05:45,830
to divide both sides of this
equation by cosine squared of

121
00:05:45,830 --> 00:05:51,200
theta assuming that it's not
equal to 0 over this interval.

122
00:05:51,200 --> 00:05:55,990
So cosine squared of theta.

123
00:05:55,990 --> 00:05:58,330
You could also do it using the
positive square root, the

124
00:05:58,330 --> 00:06:00,070
principal root, either
one will work.

125
00:06:00,070 --> 00:06:02,950
But this is interesting because
the left-hand side

126
00:06:02,950 --> 00:06:07,110
cancels out to 1, and the 1 will
be equal to-- what's sine

127
00:06:07,110 --> 00:06:09,270
squared over cosine
squared of theta?

128
00:06:09,270 --> 00:06:17,090
Well that's the same thing as
sine of theta over cosine of

129
00:06:17,090 --> 00:06:24,740
theta squared.

130
00:06:24,740 --> 00:06:27,300
You have a square divided
by another square.

131
00:06:27,300 --> 00:06:30,190
That's the same thing as the
numerator divided by the

132
00:06:30,190 --> 00:06:30,500
denominator.

133
00:06:30,500 --> 00:06:31,820
That whole thing squared.

134
00:06:31,820 --> 00:06:34,590
And what's sine of theta divided
by cosine of theta?

135
00:06:34,590 --> 00:06:36,270
Well that's just the
tangent of theta.

136
00:06:36,270 --> 00:06:39,090


137
00:06:39,090 --> 00:06:45,670
So we have 1 is equal to the
tangent squared of theta.

138
00:06:45,670 --> 00:06:49,470
Or, we could take the positive
square root of both sides of

139
00:06:49,470 --> 00:06:50,380
this equation.

140
00:06:50,380 --> 00:06:54,100
Tangent is positive over the
interval from 0 to 90 degrees,

141
00:06:54,100 --> 00:06:55,710
so that's cool to do.

142
00:06:55,710 --> 00:06:58,250
So if you take the positive
square root of both sides, you

143
00:06:58,250 --> 00:07:00,800
get the positive square
root of 1 is 1.

144
00:07:00,800 --> 00:07:04,410
1 is equal to tangent
of theta.

145
00:07:04,410 --> 00:07:06,900
And then you take the inverse
tan of both sides or the arc

146
00:07:06,900 --> 00:07:12,400
tan of both sides and you
get the arc tan of

147
00:07:12,400 --> 00:07:16,510
1 is equal to theta.

148
00:07:16,510 --> 00:07:19,660
And this is just a very fancy
way of saying theta's the

149
00:07:19,660 --> 00:07:23,270
angle that if you were take
its tangent, you get 1.

150
00:07:23,270 --> 00:07:26,780
And you could use a calculator
to solve that or you might

151
00:07:26,780 --> 00:07:28,900
just know that by memory.

152
00:07:28,900 --> 00:07:36,390
This theta, the arc tangent
of 1 is 45 degrees.

153
00:07:36,390 --> 00:07:39,030
Or if you are dealing
in radians, it

154
00:07:39,030 --> 00:07:43,750
is pi over 4 radians.

155
00:07:43,750 --> 00:07:45,910
Either one of those
is going to work.

156
00:07:45,910 --> 00:07:48,960
So our optimal angle when we
shoot this thing is going to

157
00:07:48,960 --> 00:07:52,160
be at 45 degrees.

158
00:07:52,160 --> 00:07:55,280
Now, what is that optimal
distance going to be when we

159
00:07:55,280 --> 00:07:57,370
shoot it off at 45 degrees?

160
00:07:57,370 --> 00:08:00,380
Well, we can just go back
to our original formula.

161
00:08:00,380 --> 00:08:01,840
We just go back to our original

162
00:08:01,840 --> 00:08:02,990
formula that we derived.

163
00:08:02,990 --> 00:08:06,500
If we're shooting it off at 45
degrees, what is the sine of

164
00:08:06,500 --> 00:08:07,500
45 degrees?

165
00:08:07,500 --> 00:08:11,330
The sine of 45 degrees is
equal to the square

166
00:08:11,330 --> 00:08:13,030
root of 2 over 2.

167
00:08:13,030 --> 00:08:15,390
You could use a calculator for
that or maybe you know it from

168
00:08:15,390 --> 00:08:16,500
the unit circle.

169
00:08:16,500 --> 00:08:21,410
The cosine of 45 degrees is also
square root of 2 over 2.

170
00:08:21,410 --> 00:08:23,700
And if you'd actually just taken
the principal roots at

171
00:08:23,700 --> 00:08:26,300
this stage of the equation,
you'd have gotten that the

172
00:08:26,300 --> 00:08:29,130
cosine of theta has to equal
sine of theta over this

173
00:08:29,130 --> 00:08:31,940
interval, and that only
happens at 45 degrees.

174
00:08:31,940 --> 00:08:32,710
But given this.

175
00:08:32,710 --> 00:08:37,760
We can put this back into the
original expression right up

176
00:08:37,760 --> 00:08:40,020
here, our original function.

177
00:08:40,020 --> 00:08:43,240
So the optimal distance that
we are going to travel, so

178
00:08:43,240 --> 00:08:47,440
distance as a function-- the
distance we travel at 45

179
00:08:47,440 --> 00:08:54,020
degrees is going to be equal
to 2s squared over g times

180
00:08:54,020 --> 00:08:57,690
cosine of theta, which is
square root of 2 over 2.

181
00:08:57,690 --> 00:09:01,570
Cosine of 45 is square root of
2 over 2 time sine of theta,

182
00:09:01,570 --> 00:09:04,590
which is square root
of 2 over 2.

183
00:09:04,590 --> 00:09:07,440
Well what's the square root of
2 times the square root of 2?

184
00:09:07,440 --> 00:09:08,880
Well that's just 2.

185
00:09:08,880 --> 00:09:10,770
Let me simplify this.

186
00:09:10,770 --> 00:09:12,240
So the square root of 2
times the square root

187
00:09:12,240 --> 00:09:14,010
of 2, that is 2.

188
00:09:14,010 --> 00:09:16,330
This 2 cancels out
with that 2.

189
00:09:16,330 --> 00:09:19,600
And then, this 2 cancels
out with this 2.

190
00:09:19,600 --> 00:09:22,850
So then the optimal distance you
travel at 45 degrees, all

191
00:09:22,850 --> 00:09:28,220
we're left with is the
s squared over g.

192
00:09:28,220 --> 00:09:31,030
Assuming no air resistance, kind
of an ideal circumstance.

193
00:09:31,030 --> 00:09:34,410
No matter what planet you're
on, how fast you do it, the

194
00:09:34,410 --> 00:09:37,250
best angle is always 45 degrees
assuming no air

195
00:09:37,250 --> 00:09:37,980
resistance.

196
00:09:37,980 --> 00:09:41,010
And if you do it on that best
angle, you're going to travel

197
00:09:41,010 --> 00:09:44,080
s squared over g.

198
00:09:44,080 --> 00:09:47,880
Going back to the original
problem, if s is

199
00:09:47,880 --> 00:09:49,150
10 meters per second.

200
00:09:49,150 --> 00:09:51,130
Let's say s was 10 meters
per second.

201
00:09:51,130 --> 00:09:54,690


202
00:09:54,690 --> 00:09:58,270
And let's say we're dealing with
a world where let's say,

203
00:09:58,270 --> 00:10:03,640
gravity is equal to 10 meters
per second squared, then

204
00:10:03,640 --> 00:10:13,630
according to what we've derived,
your optimal distance

205
00:10:13,630 --> 00:10:20,410
is going to be s squared-- so
it's going to be 100-- divided

206
00:10:20,410 --> 00:10:21,050
by gravity.

207
00:10:21,050 --> 00:10:22,530
It's going to be 10.

208
00:10:22,530 --> 00:10:25,080
And if you square meters per
second, you're going to get

209
00:10:25,080 --> 00:10:28,420
meters squared per second
squared divided by the

210
00:10:28,420 --> 00:10:32,200
acceleration of gravity, meters
per second squared.

211
00:10:32,200 --> 00:10:33,550
Second squared's cancel out.

212
00:10:33,550 --> 00:10:36,290
You have a meter squared
divided by meters.

213
00:10:36,290 --> 00:10:39,080
Your optimal distance
would be 10 meters.

214
00:10:39,080 --> 00:00:00,000
Pretty neat.