Calculus 2 Rectangular to polar?

x = r * cos(t)

y = r * sin(t)

x^2 + y^2 = r^2

x^2 + y^2 - 5px = 0

r^2 - 5p * r * cos(t) = 0

r - 5p * cos(t) = 0

r = 5 * p * cos(t)

Since

x = r * cos(t)

y = r * sin(t)

Then it follows that

(y/x) =>

(r * sin(t)) / (r * cos(t)) =>

sin(t)/cos(t) =>

tan(t)

Therefore

t = arctan(y/x)

r = 7t

r^2 = 49t^2

x^2 + y^2 = 49 * arctan(y/x)^2

That's about as nice as it's going to get.

Taking your polar coords as {r, a} (Ican't type theta} x = rcosa, t = rsina; thatbchanges your equation to r^2cos^2a + r^2sin^2a - 5prcosa = 0; take out a common factor of r^2 to get r^2(cos^2 a + sin^2a) - 5prcos a = 0 or r^2 - 5prcos a = 0 - there are two or three variations on that you could write

e.g. r^2= 5rpcos a etc. For the second one, x = rcos a, y = rsina so x =- 7acos a, y = 7asin a , but I don't see any way of rewriting that. In general, to convert either way, r = sqrt(x^2 + y^2), tan a = y/x .

Multiply the two components via r,it follows that r^2=3rcosO replace r^2 via x^2 + y^2 and rcosO via x Then x^2+y^2=3x , so x^2-3x=-y^2 fixing for x: (x-3/2)^2=-y^2 via winding up the sq. (a-b)^2=a^2-2ab+b^2 and x-3/2=iy that's a complicated huge type, i being sqrt(-a million) x=3/2+iy, sturdy success