Lecture 4
January 10, 2012
Reference: Miller Analytic Geometry and Calculus
1
Calculus in Polar coordinates
Example 1.1.
Find an expression for the angle of the radius vector to the tangent
vector of a curve in polar coordinates.
ADD PICTURE
The slope
OP
is given by
y
x
and the slope of
PT
is given by
d
y
d
x
. Thus
tan
θ
=
y
x
and
tan
φ
=
d
y
d
x
. Using geometric properties of triangles we deduce that
180

φ
+
θ
+
ψ
= 180
and hence that
φ
=
θ
+
ψ
and so
tan
ψ
= tan(
φ

θ
)
=
sin(
φ

θ
cos(
φ

θ
)
=
sin
φ
cos
θ

sin
θ
cos
φ
cos
φ
cos
θ

sin
φ
sin
θ
=
tan
φ

tan
θ
1

tan
φ
tan
θ
hence
tan
ψ
=
d
y
d
x

y
x
1 +
y
x
d
y
d
x
=
x
d
y

y
d
x
x
d
x
+
y
d
y
1
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Now
x
=
r
cos
θ
,
y
=
r
sin
θ
so
d
x
=

r
sin
θ
d
θ
+ cos
θ
d
r
and
d
y
=
r
cos
θ
d
θ
+ sin
θ
d
r
so
tan
φ
=
r
cos
θ
(
r
cos
θ
d
θ
+ sin
θ
d
r
)

r
sin
θ
(

r
sin
θ
d
θ
+ cos
θ
d
r
)
r
cos
θ
(

r
sin
θ
d
θ
+ cos
θ
d
r
) +
r
sin
θ
(
r
cos
θ
d
θ
+ sin
θ
d
r
)
=
r
2
d
θ
r
d
r
=
r
d
θ
d
r
Theorem 1.1.
In polar coordinates
r
d
θ
d
r
gives the angle between the tangent to a curve and the radius vector.
Corollary 1.1.
Suppose that two curves intersect at a point
P
. Let
ψ
1
and
ψ
2
be
the angles from the radii to the tangent vectors.
ADD PICTURE
Then the angle
α
between the curves can be found by means of the formula
tan
α
= tan(
ψ
2

ψ
1
) =
tan
ψ
2

tan
ψ
1
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 Winter '08
 JackWaddell
 Calculus, Geometry, Polar Coordinates, Sin, Cos, cos cos, Miller Analytic Geometry and Calculus

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