Polar coordinates (trig question)

[Oxygenate]

Homework Statement

Given r = 2tan(theta)sec(theta)

Find cos(theta) then use inverse key to find sec(theta)

The answer given in the solution guide is y = 1/2 x^2

Attempt at solution
Since tan = sin/cos and sec = 1/cos
We have r = 2sin/cos * 1/cos
So rcos^2 = 2sin
rcos^2 is defined as x^2 so x^2 = 2sin

Unless 2 sin is defined as 2 y, I don't see how we can get to y = 1/2x^2??

 

[jamesrc]

I do not understand where you got this part from:

rcos^2 is defined as x^2 so x^2 = 2sin

In polar coordinates,
<br /> x=r\cos\theta<br />
and
<br /> y=r\sin\theta<br />

Substitute your expression for r into these two equations. Solve the first one in such a way that you can plug your result into the second one to solve for y in terms of x.

 

[Defennder]

So rcos^2 = 2sin
rcos^2 is defined as x^2 so x^2 = 2sin

r cos^2 isn't the same as x^2. r^2 \cos^2 \theta = x^2.

For this problem just use the equations for converting between rectangular and polar coordinates, using the expression for cos theta and sine theta as hinted and substituting them into the problem equation.

 

[Oxygenate]

Okay, so:

r = 2sin/cos * 1/cos

rcos = 2sin * 1/cos

x = 2sin * 1/cos
x = 2tan

Where does the x^2 come from and where does the y come from?

 

[jamesrc]

Don't worry about where the x^2 comes from, that is part of the answer and is a result of you solving the problem correctly.

You know that y=r\sin\theta by definition. Find y as a function of theta only.

Then you can eliminate theta when you find y as a function of x only.

 

[Oxygenate]

r = 2tan(theta)sec(theta)
r = 2sin(theta)/cos(theta) * 1/cos(theta)
rcos^2 = 2sin(theta)
r^2cos^2 = r2sin(theta)
x^2 = 2y
y = 1/2(x^2)

It seems like this works too. Thanks! I haven't done trig in like forever...