Polar Curve Functions
(Differential Calc)
Published in
4 min readJan 21, 2019
Refer to Khan academy: Polar functions derivatives
In the
Polar World
,
instead of the relationship betweeny & x
,
the function is now representing the relationship betweenRadius & Angle
,
which could be presented as:
Finding the right boundaries
The most tricky part in Polar system, is finding the right boundaries for
θ
, and it will be the first step for polar integral as well.
Differentiate Polar Functions
Taking derivative of Polar function is actually DIFFERENTIATING PARAMETRIC FUNCTION.
To take the derivative we need to:
- Convert the
Polar function
in terms ofx & y
:
- Take derivative of the parametric function.
Example
Solve:
- Since it’s asking for the
Rate of change of y-coordinate
, so we convert the polar function torectangular function
:
- And we take the derivative
dy/dΘ
:
- Plug in the point
Θ=π
and get:
Tangents to Polar curves
Steps:
- Find the slope
dy/dx
- Convert the polar function to get the
x(θ)
andy(θ)
parametric equations - Solve
dy/dx
and get the slope - Plug in the point’s information to solve for
x & y
- Get the equation of the line (tangent).
Example
Solve:
- To find the tangent line, we need to get the slope first, which is
dy/dx
. - And
dy/dx
would be aparametric problem
:
- Plug in the Θ value, to evaluate the slope:
- Find the
x & y
value according to the Θ:
- Now we got everything to form the equation for the tangent line:
Example
Solve:
- First, we need to convert the polar function to
x(θ) & y(θ)
:
- And we need to find the slope
dy/dx
:
- Since it’s a horizontal tangent, so
Slope =0
, which meansdy/dx =0
. Butdx
is dominator can't be zero, so we can setdy = 0
and solve for θ:
- So the answer is:
Example
Solve:
- To find a vertical tangent, we have to set the dominator of the slope as 0, that’s the only thing makes it undefined.
- The slope is
dy/dx
, so we setdx = 0
. - The equation for
x
is: