Polar functions are graphed in the (`color(black)(r)`, `color(black)(theta)`) plane. r is the distance from the origin to the point, and the `color(black)(theta)` is the angle. They can have parameters which determine how much is graphed. Here is an example of the equation `color(black)(r=3+2cos(theta))` with a parameter of `color(black)(0 <= theta <= 2pi)`
You can switch between polar and rectangular forms by using the following equations:
`color(black)(x^2+y^2=r^2)`
`color(black)(x=rcos(theta))`
`color(black)(y=rsin(theta))`
`color(black)(tan(theta)=y/x)`
The derivative of a polar function is similar to parametric functions.
The first derivative is: `color(black)((dy)/(dx)=((dy)/(d theta))/((dx)/(d theta)))`
The second derivative is: `color(black)((d/(d theta)[(dy)/(dx)])/((dx)/(d theta)))`. Knowing this is not nearly as important as knowing the first derivative of a polar function.
Here's some practice.
Question: What does `color(black)((dy)/(dx))` equal if `color(black)(r=2+3sin(theta))`?
Click on the up carrot to get the answer.
Answer: `color(black)((dy)/(dx)=(2cos(theta)+6sin(theta)cos(theta))/(3cos^2(theta)-3sin^2(theta)-2sin(theta)))`
In order to find the intersecting points between polar functions just make the functions equal each other and then solve for `color(black)(theta)`
For example, the intersection points between `color(black)(r=cos(theta))` and `color(black)(r=sin(theta))` would be found through `color(black)(sin(theta)=cos(theta))` which would be `color(black)(theta = pi/4, (5pi)/4, pi/4+kpi)`
In order to find the area of one polar equation from [a, b], plug in the polar equation into this formula: `color(black)(A=1/2int_a^br^2d theta)`
Set the equations equal to each other and solve for θ, which are the points of intersection. Set up the integral between both points and subtract the bottom equation from the top equation. Make sure to square each equation individually and remember the ½ on the outside.
`color(black)(A=1/2int_a^b(r_1^2-r_2^2)d theta)`
Here's some practice.
Question: What is the area of one petal of `color(black)(r=3cos(2theta))`?
Click on the up carrot to get the answer.
Answer: `color(black)(1/2int_(pi/4)^((3pi)/4)(3cos(2theta))^2d theta=3.5343)`