parametric to cartesian equations

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May 23rd 2010, 04:35 PM
kevin11

parametric to cartesian equations

Find the equivalent Cartesian equation for the parametric equations given below. Which curve does this represent?

$ x = 2 + 4\cos t $

$ y = 4 + 6\sin t $

So far all I've done is isolate x and y by itself (not sure if this is the write course of action).
$ \cos t = \frac {x}{8} $

$ \sin t = \frac {y}{24} $

Any help would be appreciated!
May 23rd 2010, 04:47 PM
sa-ri-ga-ma

Quote:

Originally Posted by kevin11

Find the equivalent Cartesian equation for the parametric equations given below. Which curve does this represent?

$ x = 2 + 4\cos t $

$ y = 4 + 6\sin t $

So far all I've done is isolate x and y by itself (not sure if this is the write course of action).
$ \cos t = \frac {x}{8} $

$ \sin t = \frac {y}{24} $

Any help would be appreciated!

No. Your simplification is not correct. It should be

$\cos(t) = \frac{x-2}{4}$

$\sin(t) = \frac{y-4}{6}$

Now use identity $\cos^2(t) + \sin^2(t) = 1$

.
May 23rd 2010, 04:48 PM
Sudharaka

Quote:

Originally Posted by kevin11

Find the equivalent Cartesian equation for the parametric equations given below. Which curve does this represent?

$ x = 2 + 4\cos t $

$ y = 4 + 6\sin t $

So far all I've done is isolate x and y by itself (not sure if this is the write course of action).
$ \cos t = \frac {x}{8} $

$ \sin t = \frac {y}{24} $

Any help would be appreciated!

Dear kevin11,

Your method of isolation of sine and cosine terms was incorrect.

You should get, $sint=\frac{y-4}{6}$ and $cost=\frac{x-2}{4}$

Using the trignometric identity $sin^{2}t+cos^{2}t=1$ you can eliminate t to obtain the cartesian equation.

Hope this will help you.