# Parametric and Cartesian Equations

1. Oct 29, 2006

### lemurs

ok I am give a parametric equations of

x= 4 cos t and y=5 sin t

I know that i have to solve the x equation for t then stick it in the y equation but i getting stuck or not rembering some simple stuff i should be.

I believe i get t= cos(inv) (x/4) and substiute it in to t in y.

if so how so i simplify

5 sin(cos(inv)(x/4))

2. Oct 29, 2006

$$\frac{x}{4} = \cos t$$

$$\frac{y}{5} = \sin t$$

$$\sin^{2} t + \cos^{2}t = 1$$

3. Oct 29, 2006

### lemurs

so then i would justs substiute in those in so it would be
x^2/16 +y^2/25

4. Oct 29, 2006

### Hootenanny

Staff Emeritus
Don't forget they equal one.

5. Oct 29, 2006

### Office_Shredder

Staff Emeritus
in general, sin(arcos(x)) can be solved by drawing a right triangle. arcos(x) is an angle theta whose cosine is x. So pick one of the angles that's not 90 degrees. Label that theta. Since theta is arcos(x), the adjacent side is x, and the hypotenuse is 1. So sin(theta) is opposite over hypotenuse. You can get the opposite side by pythagoras, and you're done.

6. Oct 29, 2006

### lemurs

ok i understnad that now but how would you do the reverse.. go form given a cartesian equation to a cause given X^2-y^2=1 how would you solve that

7. Oct 29, 2006

### Office_Shredder

Staff Emeritus
You're trying to make that into a parametric form? We know cosh2t - sinh2t = 1. so if you let x=cosht, and y=sinht, it works

In general, if you're completely at a loss, you can try to solve for y=f(x), let x=t, and y=f(t)