Parametric and Cartesian Equations

[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 into t in y.

if so how so i simplify

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

 

[courtrigrad]

$$\frac{x}{4} = \cos t$$$$\frac{y}{5} = \sin t$$$$\sin^{2} t + \cos^{2}t = 1$$

 

[lemurs]

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

 

[Hootenanny]

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

Don't forget they equal one.

 

[Office_Shredder]

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.

 

[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

 

[Office_Shredder]

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)