# Polar graph: stuck on sketching a cardioid

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• Oct 17th 2008, 12:56 PM
ssadi
Polar graph: stuck on sketching a cardioid, something about chords
Sketch the cardioid with polar equation r=2(1+cos theta)
Prove that for all the chords POQ of the cardioid which pass through the pole O, the length of PQ is constant.
How do I do the green part?
I have gone as far as to deduce that if P (r1, theta1) and Q is (r2, theta2)
Then theta2=theta1-180
r2=2+2cos (theta1-180)
There isn't given a formula in the book to find the distance between two polar coordinates, I found one on net:
http://www.thestudentroom.co.uk/atta...1&d=1224276450
Here is the sketch of the graph:
http://www.thestudentroom.co.uk/atta...1&d=1224276450
I am stuck here for over two hours. Emergency help...I am drow...(Oops I gulped up some water)
• Oct 17th 2008, 01:13 PM
Laurent
Quote:

Originally Posted by ssadi
There isn't given a formula in the book to find the distance between two polar coordinates, I found one on net:

You don't need any formula from the internet... The distance from $\displaystyle O$ to $\displaystyle P(r,\theta)$ is $\displaystyle r$. So what you have to do is add the values of $\displaystyle r$ for $\displaystyle \theta_1$ and $\displaystyle \theta_2=180-\theta_1$ (in degrees), and check that it does not depend on $\displaystyle \theta_1$.
• Oct 17th 2008, 01:54 PM
ssadi
Quote:

Originally Posted by Laurent
You don't need any formula from the internet... The distance from $\displaystyle O$ to $\displaystyle P(r,\theta)$ is $\displaystyle r$. So what you have to do is add the values of $\displaystyle r$ for $\displaystyle \theta_1$ and $\displaystyle \theta_2=180-\theta_1$ (in degrees), and check that it does not depend on $\displaystyle \theta_1$.

theta2=theta1-180 according to my calculations. I am giving it a try.
• Oct 17th 2008, 01:57 PM
Laurent
Quote:

Originally Posted by ssadi
theta2=theta1-180 according to my calculations. I am giving it a try.

Sorry I made a mistake, in fact $\displaystyle \theta_2=\theta_1+180$. Because $\displaystyle \theta_2-\theta_1$ is a flat angle.
• Oct 17th 2008, 01:59 PM
ssadi
Quote:

Originally Posted by Laurent
You don't need any formula from the internet... The distance from $\displaystyle O$ to $\displaystyle P(r,\theta)$ is $\displaystyle r$. So what you have to do is add the values of $\displaystyle r$ for $\displaystyle \theta_1$ and $\displaystyle \theta_2=180-\theta_1$ (in degrees), and check that it does not depend on $\displaystyle \theta_1$.

It worked. Thanks.
• Oct 17th 2008, 02:00 PM
ssadi
Quote:

Originally Posted by Laurent
Sorry I made a mistake, in fact $\displaystyle \theta_2=\theta_1+180$. Because $\displaystyle \theta_2-\theta_1$ is a flat angle.

Does that mean I am wrong here? I have verified it with calculator, my formula worked for polar -pi<theta<pi
• Oct 17th 2008, 02:07 PM
Laurent
Quote:

Originally Posted by ssadi
Does that mean I am wrong here? I have verified it with calculator, my formula worked for polar -pi<theta<pi

No, indeed, you're right as well... I considered the graph defined for $\displaystyle 0<\theta_2<2\pi$, while you considered $\displaystyle -\pi<\theta_2<\pi$, that's why. Forget my previous post, I was correcting my first one, not yours in fact.
• Oct 17th 2008, 02:10 PM
ssadi
Quote:

Originally Posted by Laurent
No, indeed, you're right as well... I considered the graph defined for $\displaystyle 0<\theta_2<2\pi$, while you considered $\displaystyle -\pi<\theta_2<\pi$, that's why. Forget my previous post, I was correcting my first one.

I guessed so, that your limits might be different. But my problem is solved anyway. :)