How do you convert #r = 5 cos (t)# into cartesian form?

1 Answer
Nov 28, 2015

Answer:

The equation relates #r and theta#, and describes a set of points #(r,theta)#which form a curve in the plane. We want to know the curve's equation in #(x,y)# coordinates.

Explanation:

You used #t# instead of #theta#. We can do that.

Start with knowing where we're going: the variables are related by things like
#x = r cos(t) and y = r sin(t) and r^2 = x^2+y^2.#

You have #r = 5 cos(t),# let's create an #r cos(t#) by multiplying both sides by #r.#

#r^2 = 5 r cos(t)# . . . Now substitute and see that it turns into

#x^2+y^2=5x.#

This is the answer. If you complete the square you get the form

#x^2 - 5x + y^2 = 0# . . . Half of 5 is 5/2, whose square is 25/4:

#x^2 - 5x+25/4 + y^2 = 25/4 #, or in circle form:

#(x-5/2)^2+(y-0)^2=(5/2)^2#,

You answer these ?'s: "It's a circle of center #(?,?)# and radius#?#."

...// dansmath strikes again! \\...