Calculus of Polar Equation
Because polar equation is a parametric equation, the previous discussions on the calculus of parametric equations
are directly applicable.
However, because polar equation relies on a unique set of parameters and it is widely used,
its characteristics are specifically studied and several convenient formulas can be derived. We will begin with finding
the slope.
For parametric equation, remember that the slope is defined as:
dt
dx
dt
dy
dx
dy
=
Polar equation is essentially a type of parametric equation where:
θ
θ
sin
cos
r
y
r
x
=
=
So, for polar equation, the slope becomes:
θ
θ
d
dx
d
dy
dx
dy
=
However, the big difference is that
θ
,
r
are both varying. Therefore:
You must use the product rule and implicit differentiation to find the derivative:
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
θ
d
dr
d
d
r
d
dy
d
dr
d
d
r
d
dx
sin
cos
cos
sin
+
=
+
−
=
Substituting
θ
d
dx
and
θ
d
dy
back into the
dx
dy
equation and after some algebraic arrangement,
the slope of a polar
curve is defined as
:
θ
θ
θ
θ
θ
θ
d
dr
r
d
dr
r
dx
dy
cos
sin
sin
cos
+
−
+
=
The above equation looks complicated but it is actually quite simple to use.
Example: Find
dx
dy
for
4cos
r
θ
=
4cos
4sin
dr
r
d
θ
θ
θ
=
→
=
−
The above formula gives:
(
)
(
)
(
)
(
)
.
4cos
cos
sin
4sin
4cos
sin
cos
4sin
Simplified as needed
dy
dx
θ
θ
θ
θ
θ
θ
θ
θ
+
−
=
=
−
+
−
If the problem asks for a specific tangent point, say at
30
o
θ
=
, just evaluate the above at
30
o
θ
=

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