Parametric Equations

[boneill3]

HI Guys,
I have been asked to sketch the function x^2+2y=4.

I'm not sure if I need to treat this as a Parametric curve and use a value of (t) for both functions to solve x = a cos (t) and y = b sin (t), or just solve for x and Y. ie say when x = 0 y = 2 and plot the and graph where x and y cut the Axis.

Any pushes in the right direction would be greatly appreciated
regards
Brendan

[NickThissen]

I don't see any need for a parametric curve..?

x^2 + 2y = 4
2y = 4 - x^2
y = 2 - 1/2 x^2

This is just the function of a parabola and you can use all the usual things to find the x and y intercepts etc...

[boneill3]

Thanks for your reply.
If the equation was x^2 + 2y^2 =4 would you then use .

x = cos (t) and y = 2 sin (t)

and plot for t ?

Brendan

[NickThissen]

Those parameters are invalid:

x^2 + 2y^2 = 4

cost^2 + 4sint^2 != 4

This is because x^2 + 2y^2 = 4 is not a circle, it's an ellipse.

[boneill3]

Yeah sorry I stuffed UP.

The equation should be.

- 2cos(t) + 4sin(t) = -4 when t = (0 < t < 2*pi)
for 1 loop and

2cos(t) + 4sin(t) = 4 when t = (0 < t < 2*pi)
for the other
regards
Brendan

[NickThissen]

I still don't think that's correct.

If you look at the parametrization of a circle:
x^2 + y^2 = r^2

x = r cos(t)
y = r sin(t)

Substituting x and y back in the first equation:
r^2 cos^2(t) + r^2 sin^2(t) = r^2 ( cos^2(t) + sin^2(t) ) = r^2 (1) = r^2.

So the equation is satisfied for every t.

Yours are not... I don't think you can use a simple cos(t) and sin(t) to parametrize an ellipse.

[jemidiah]

No... you can use sin and cos to parametrize an ellipse. The wrong constants got used above, which is why it didn't work out.

Given:
x^2 + 2y^2 = 4

Use these instead:
x=2cos(t), so x^2 = 4cos^2(t)
y=2/Sqrt(2)*sin(t), so y^2 = 4/2*sin^2(t), and2y^2 = 4sin^2(t).

Then x^2 + 2y^2 = 4cos^2(t) + 4sin^2(t) = 4(1) = 4 as needed.


In the original question, like Nick said there's really no use in using a parametric curve. I mean I guess you could do x = Sqrt(t) so x^2 = t, so that 2y needs to be 4-t so y = 2-t/2. But that's silly--it's more work than just solving the equation for y and graphing like you usually do, since you don't have to bother with the range of t.

[boneill3]

Thanks for all your replys.
I'll think I take your advice and just solve for Y in the original question.
It was good to see the different ways to solve the ellipse. No doubt I will need these soon enough.
regards
Brendan