Download Lecture 19: Arclength and trigonometric functions. We begin this

Lecture 19: Arclength and trigonometric functions.
We begin this lecture by defining the arclength of the graph of a differentiable function
y = f (x) between x = a and x = b. We motivate our definition by calculating the length
of the part of the line y = mx between x = a and x = b. This is the hypotenuse of a right
triangle whose legs have lengths b − a and m(b − a) respectively.
√ Thus by the Pythagorean
theorem, the length of the hypotenuse is given by (b − a) 1 + m2 . This motivates the
following definition of arclength. (We view arclength as the limit of lengths of splines
along the curve.)
Definition Let f be a differentiable function on the interval [a, b]. The arclength of
the graph of f is
Z bp
1 + (f 0 (x))2 dx.
a
We immediate apply this definition to our favorite curve from plane geometry: the
unit circle. The part of the unit circle in the upper half plane is the curve
p
1 − x2 .
y=
With f (x) =
√
1 − x2 , we calculate
f 0 (x) = √
−x
,
1 − x2
so that
p
1 + (f 0 (x))2 = √
1
.
1 − x2
Thus the arclength A of the circle between x = 0 and x = a (actually the negative of the
arclength if a is negative) is given by
Z
A=
a
√
0
dx
.
1 − x2
We have no especially nice way of computing this integral, so we just give it a name.
We say
Z a
dx
√
arcsin a =
.
1 − x2
0
This entirely corresponds to our intuition from plane geometry. When studying geometry, we keep going about the arclengths of parts of circles, even before we can define
what arclength actually means. We associate arcs on circles with the angles that subtend
them, and the arc we are describing corresponds to the angle whose sine is a. The reason
that inverse function to sin has such an odd name is that it is computing the length of an
arc. For our purposes, we look at things a little differently. We have no way of describing
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the function sin x without its inverse, because we don’t know what an angle means without
the notion of arclength. However clearly arcsin is increasing as a goes from −1 to 1 (and
it is odd). Thus it has an inverse. We define sin to be the inverse of arcsin. We have not
yet named the domain of sin. We make the definition that
π
=
2
1
Z
√
0
dx
.
1 − x2
This is really the usual definition for π. It is the arclength of the unit semicircle. We have
π
defined sin x on the interval [ −π
2 , 2 ]. On the same interval, we may define cos x by
cos x =
p
1 − sin2 x.
It is not hard to see that for x ∈ [0, π2 ], we have that
cos(
π
− x) = sin(x),
2
because this is just the symmetry between x and y in the definition of the unit circle.
It is definitely interesting to extend sin and cos to be defined on the whole real line.
We are already in a position to do this by symmetry as well, but for the moment we refrain.
We will have a much clearer way of defining this extension later when we introduce complex
numbers.
But, for now, as long as we stay in the interval [− π2 , π2 ], we are in a position to obtain
all the basic facts of calculus for trigonometric functions. Thus, for instance,
x = arcsin(sinx).
Differentiating in x, we get
1= p
1
d
sin x),
1 − sin2 x dx
(
and solving for the second factor, we obtain the famous formula that
d
sin x = cos x.
dx
Applying the symmetry
cos(
π
− x) = sin(x),
2
we immediately obtain that
d
cos x = − sin x.
dx
2
Using these two results, we can easily build up all the famous formulae in the calculus of
trigonometric functions.
sin x
For instance, we define sec x = cos1 x and tan x = cos
x . We readily use the quotient
rule to calculate
d
sec x = sec x tan x,
dx
and
d
tan x = sec2 x.
dx
Then we are free to observe
sec x
sec x(sec x + tan x)
sec x + tan x
d
(sec x + tan x)
= dx
sec x + tan x
d
=
(log(sec x + tan x)).
dx
=
In short, all the identities of calculus just come to life.
The final thing I wanted to bring up today is the dual role of π. Perhaps, we all
remember π as the arclength of the unit semi-circle, but we might also remember it as
the area of the unit circle. The first can be a definition, but then the second should be a
consequence. Here is how we see it:
We calculate
Z
1
p
1 − x2 dx.
0
This is just the area of one quarter of the unit circle. We will do this using the (quite
natural) trigonometric substitution x = sin u. (Wasn’t x already the sin of something!?!)
We obtain
dx = cos udu.
The integral now runs from 0 to
π
2
and becomes
Z
π
2
cos2 udu.
0
We calculate this integral without any fancy double angle identities. We just use again the
symmetry
π
cos( − u) = sin(u),
2
to obtain
Z π
Z π
2
2
cos2 udu =
0
0
3
sin2 udu.
Thus
Z
0
π
2
1
cos udu =
2
2
π
2
Z
(cos2 u + sin2 u)du,
0
and since it is easy to integrate 1, we get
Z
1
p
1 − x2 dx =
0
π
.
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How is this related to the usual Euclidean proof of the same fact?
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