Parametric Equations of an ellipse

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

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

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

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

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

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

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

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

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

Daniel.

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

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

Daniel.

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
Homework Helper
Well,what do you know,it's the same thing with what i've written...:tongue2:

Daniel.

I was explaining to him why :)

dextercioby