Convert into cartesian form?

Question

the following

r(1 + sinANGLE) = a

anyone know?

Answer 1

To go from cartesian to polar, all you need to know are:

x = r cos θ
y = r sin θ

From polar to cartesian, it's:

r = √(x² + y²)
θ = tan ˉ¹ y/x if x > 0,
      tan ˉ¹ y/x + π if x < 0,
      π/2 if x = 0 and y > 0,
      3π/2 if x = 0 and y < 0.

So starting with:

r(1 + sin θ) = a

Distribute the r first:

r + r sin θ = a

Then substitute for r and r sin θ:

√(x² + y²) + y = a

The rest is just making it pretty, if you want. Subtract y from both sides:

√(x² + y²) = a - y

Square both sides:

x² + y² = a² - 2ay + y²

Notice that the y²s drop out...and it's "traditional" to solve for y:

x² - a² = -2ay

y = -(x² - a²)/2a = (a² - x²)/2a = a/2 - x²/(2a)

Answer 2

enable's take the second one question first. 2) The unit circle is the circle interior the complicated plane with radius a million established on the inspiration. by technique of definition, e^(ix) is the point that is a distance x alongside the unit circle, starting up at (a million,0) and shifting counterclockwise. So e^(ipi/2) is (0,a million), e^(ipi) is (-a million,0), e^(ipi/4) = (2^0.5,2^0.5), etc. a million) (e^x)^2 = e^(2x), so (e^(ix))^2 = e^(2ix). this signifies that once you sq. a complicated quantity, you double the attitude from the x (genuine) axis. So i^2 = (0,a million)^2 = e(ipi/2)^2 = e(ipi) = (-a million,0) = -a million. once you're taking the sq. root of a complicated quantity that is on the unit circle, you get the quantity which, at the same time as the attitude from the genuine axis is doubled, resources the unique quantity. There are 2 such numbers: one with 0.5 the attitude from the genuine axis because the unique quantity, and one on the alternative fringe of the unit circle from the first. So the first sq. root of e^(ipi/2) is e(ipi/4). the second one is e^(i*(pi+pi/4)) or e^(5ipi/4), or equivalently, e^(i*(-pi+pi/4)) or e^(-3pi/4).