Parametric Equations of an ellipse

[ProBasket]

The ellipse \frac{x^2}{3^2} + \frac{y^2}{4^2} = 1
can be drawn with parametric equations. Assume the curve is traced clockwise as the parameter increases.

If x=3cos(t)

then y = ___________________________


wouldnt i just sub x into the ellipse equation and solve for y?

well i did that and got \sqrt{(-1/16*((3*cos(t))^2/9)+1)}

but there's a negative sign inside the sqrt function, so it's not possible

 

[whozum]

\sqrt{(-1/16*((3*cos(t))^2/9)+1)}

\sqrt{(-1/16*(9cos^2(t)/9)+1)}

\sqrt{(-cos^2(t)/16+16/16)}

\sqrt{\frac{(16-cos^2(t))}{16}}

\frac{\sqrt{16-cos^2(t)}}{4}

\frac{\sqrt{(4-cos(t))(4+cos(t))}}{4}


Im sure that can simplify more, but I'm out of ideas.

 

[whozum]

Also consider that a circle is an ellipse with a = b = 1, in which case the parametric equations are:

x(t) = a cos(t) = cos(t)
y(t) = b sin(t) = sin(t)

 

[dextercioby]

Okay.I think it's not too difficult to show that
y=4\sin t

Daniel.

 

[ProBasket]

\frac{\sqrt{(4-cos(t))(4+cos(t))}}{4}

and y = 4*sin(t) is incorrect. I really get and understand how you got 4*sin(t). but anyone know why these answers are incorrect?

 

[dextercioby]

\frac{y^{2}}{16}=1-\cos^{2}t=\sin^{2}t\Rightarrow y^{2}=(4\sin t)^{2}\Rightarrow y=\pm 4\sin t...U can choose the "-" sign (y\searrow \ \mbox{when} \ t\nearrow)...

Daniel.

 

[whozum]

The answer would be y = -4sin(t) because the particle moves clockwise, and as t \nearrow, sin(t) \mbox { travels counter clockwise.}

For sin(t) \mbox{ to travel clockwise you would need to multiply the parameter by -1}

y(t) = 4sin(-t) \mbox{ which equals } y(t) = -4sin(t) \mbox{ by properties of the sin function}

 

[dextercioby]

Well,what do you know,it's the same thing with what I've written...

Daniel.

 

[whozum]

I was explaining to him why :)

 

[dextercioby]

Im sure that can simplify more, but I'm out of ideas.

Sure you were... However,i still think the OP needs to do some thinking on this problem.

Daniel.