# Calculus with Parametric Curves

• Mar 30th 2009, 06:38 PM
aaronb
Calculus with Parametric Curves
x=2cos(theta)
y=sin(2theta)

I have to find the coordinates of the horizontal and vertical tangents to this problem. I know that the top of the derivative must be 0 for horizontal and 0 on the bottom for the vertical but how do I go about finding the cartesian coordinates?
• Mar 30th 2009, 07:42 PM
TheEmptySet
Quote:

Originally Posted by aaronb
x=2cos(theta)
y=sin(2theta)

I have to find the coordinates of the horizontal and vertical tangents to this problem. I know that the top of the derivative must be 0 for horizontal and 0 on the bottom for the vertical but how do I go about finding the cartesian coordinates?

Here is how you would find the horizontal tangents...

$\displaystyle \frac{d y}{d \theta}=2\cos(2\theta)$

So we set it equal to zero to get

$\displaystyle \cos(2\theta)=0 \iff 2\theta=\frac{\pi}{2}+n \pi$ where n is an integer

$\displaystyle \theta=\frac{\pi+2n\pi}{4}=\frac{(2n+1)\pi}{4}$

Now to get the cartesian coordinates plug the values of theta into your parametric equations so when n=0 we get pi/4 so

$\displaystyle x=2\cos(\frac{\pi}{4})=\sqrt{2}$

$\displaystyle y=\sin(\frac{\pi}{2})=1$

so ONE of the ordered pairs is $\displaystyle (\sqrt{2},1)$

This should get you started you will need to use more values of n to find the rest.

Hint there are 4 horizontal tangents and 2 vertical good luck.
• Mar 31st 2009, 02:49 AM
aaronb
I think I got this.

I keep on plugging in integers for n until it exceeds the period of 2pi then I would be done.
• Mar 31st 2009, 03:18 AM
simplependulum
x = 2cosu
y = sin2u = 2sinu cosu = sinu (2cosu)=x sinu
sinu = y/x
cosu = | sqrt[ 1- (sinu)^2 ] | = | sqrt[ 1 - y^2 / x^2 ] |

x = 2| sqrt[ 1 - y^2 / x^2 ] | => x^2 = 2| sqrt(x^2 - y^2) |