How do you graph #r=8sintheta#?

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Answer 1 · 2016-12-29T14:11:27

Graph is inserted. See explanation.

#r=2a cos(theta-alpha)#

represents the circle through the pole r = 0, with radius a and center

at #(a, alpha)#.

Here, a = 4 and #alpha = pi/2#'

For graphing here, convert to cartesian form using

#r(cos theta, sin theta ) = (x, y)# that gives,

#sin theta= y/r and r=sqrt(x^2+y^2)>=0.#

So, the cartesian form is

#sqrt(x^2+y^2)=8y/sqrt(x^2+y^2)# that gives the standard form

#x^2+(y-4)^2=4^2#

graph{x^2+y^2-6y=0 [-20, 20, -10, 10]}