How do you change an equation from polar coordinate to cartesian?

Question

I am in Calc 2 and we are working on polar coodinate graphs and I can't seem to handle them. The question that I need help with is how to change the equation, r=4sec(theta)(1+ tan(theta)), into the cartesian equation. Thank you so much for your help!

Answer 1

x = r cos (theta)

y = r sin (theta)

x = 4 sec (theta)(1+tan(theta)) cos (theta)
y = 4 sec(theta)(1+tan(theta)) sin(theta)

I'm going to use T instead fo "theta"

x = 4 sec(T)(1+tan(T)) cos (T)
x = 4 (1+tan(T)) since sec = 1/cos, and cos/cos = 1

y = 4 sec(T)(1+tan(T)) sin(T)
y = 4 (1+tan(T)) tan(T) since sin/cos = tan
y = 4 (tan(T) + tan^2(T))
you might be able to simplify it a bit more since tan^2(T) + 1 = sec^2(T), but I think that just makes it messier.

I hope this helps (but double check its accuracy!).

Answer 2

Your desired equation is x^2 - 4 x - 4 y = 0, the equation of a parabola. It can also be written:

4 y = (x - 2)^2 - 4.

From the defining relationships:

x = r cos(theta), and
y = r sin(theta)

one sees that sec(theta) = 1/cos(theta) = r/x, and tan(theta) = y/x.

Substituting those terms into your original equation, it becomes:

r = 4 (r/x) (1 + y/x) or x = 4 (1 + y/x), that is x^2 = 4 x + 4 y , or :

x^2 - 4 x - 4 y = 0, or 4 y = (x - 2)^2 - 4. QED

This is actually the equation of a parabola with vertex (x, y) = (2, - 1), downwards, or in other words open to ( y --> + infinity).

Live long and prosper.

LATER ADDITIONS:

1.) I'm afraid that the first responder, who initially gave you a perfectly reasonable formal prescription, became hopelessly entangled when he edited some of his original answer out and tried instead to apply those formal principles to a relatively simple example.

There's a moral here for all those voters who vote for a "described method" that is not actually applied to the problem in hand, rather than a properly executed result, as "Best Answer." (However, I sadly doubt that such voters are capable of learning from this moral.)

Where I was taught my mathematics, the ability to describe "method" counted for only 30-40%, while "execution" of an illustrative problem counted for the rest. I've found it distressing in my teaching in an American university that students seem to believe that describing "method" alone should merit full credit. Only very bad and/or indulgent teaching could have convinced them of the truth of this fallacy.

2.) The second responder dropped the ' 4 ' in the original equation. The resulting cartesian equation is therefore wrong by the consequences of dropping that factor.

Answer 3

Remember the identity:

x = rcosθ
x/r = cosθ

Now write the equation.

r = 4secθ(1 + tanθ) = (4/cosθ)(1 + tanθ) = 4/(x/r)(1 + y/x)

r = (4r/x)(1 + y/x)

1 = (4/x)(1 + y/x) = 4/x + 4y/x²

x² = 4x + 4y

x² - 4x = 4y

x² - 4x + 4 = 4y + 4

(x - 2)² = 4(y + 1)

This is the equation of a parabola with vertex (2,-1) that opens upward.

Answer 4

you change to polars with
r = sqrt( x^2 + y^2)
tan@ = y/x

you can switch back with
x = r cos@, y = rsin@

your equation is
r = (r/x)(1+y/x)
so x = 1+y/x
so y = x(x-1)