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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
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...
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
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.
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
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.
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.
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