Question

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

Answer 1

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! \\...