Math help plz

[canis]

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[Cogman]

To convert to rectangular, solve for t (preferably in the x equation) and then replace the t's in the y equation with the t in terms of x equation.

To tell if it is conic or not, once it is into rectangular form, see if you can't fiddle around with it to turn it into a standard conic equation. (i'm sure there is a better way, its just been a while since I've dealt with conic sections... Not very useful to Computer engineers.)

 

[Lemon law]

Since we have no idea how x and y are related, all we have is a mismash. But as we look at the variable t, it would in theory be a a conic section if t is simply an real number.

But since t could also stand for some variable like Z cubed -17, we can't say anything without knowing how x and y are related.

But if we assume x=y, then we get 0=-9 t squared which is an imaginary number unless t is zero.

 

[Cogman]

Since we have no idea how x and y are related, all we have is a mismash. But as we look at the variable t, it would in theory be a a conic section if t is simply an real number.

But since t could also stand for some variable like Z cubed -17, we can't say anything without knowing how x and y are related.

But if we assume x=y, then we get 0=-9 t squared which is an imaginary number unless t is zero.

What are you smoking? This a parametric equation problem.

 

[canis]

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[Cogman]

For this particular curve, solving for t is the hard part. I cannot see a way to use simple substitution.

Treat x (or y for that matter) like an unknown constant when solving for t. Completing the square should get you a solution for t in terms of x (or y).

You should notice a + or - sneak in. That means that this is not going to be a linear solution. Remember, you can easily remove the +- by squaring.

 

[CitanUzuki]

lemon law you just blew my mind!

Listen to cogman, complete the square and substitute.

 

[Mr. Pedantic]

No, I don't think it's a parabola. But then my maths skills are sorely unused, so you might want to check that. Complete the square, and then substitute for t in the y equation.

 

[canis]

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[Mr. Pedantic]

Erm...that's not an equation. That's a statement.

And I got:

y = 10(22(x+100)^0.5-220-x)

How did you get what you got?

 

[Cogman]

Thanks Cogman. Did you see the completing the square technique in a reference about parametric equations?

Simplified the equation and got:

Does not appear to be a conic. Am I correct? What can this curve be classified as?

non-conic . I don't know that it is a defined shape.

 

[Cogman]

Erm...that's not an equation. That's a statement.

And I got:

y = 10(22(x+100)^0.5-220-x)

How did you get what you got?

Pst, when you take the sqrt of a square, the answer on the other side of the equation is always + or -. He squared y to get rid of the + or -. Otherwise, you should end up with two equations.

The graph seems like a slanted parabolic curve.

 

[CitanUzuki]

Im pretty sure conics refers to a general set of shapes/lines including parabolas, hyperbolas, and ellipses. Punching the above parametric equation into my calculator yields a giant parabolic like curve. But because there is no axis of symmetry it is not a parabola.

 

[canis]

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[Mr. Pedantic]

Pst, when you take the sqrt of a square, the answer on the other side of the equation is always + or -. He squared y to get rid of the + or -. Otherwise, you should end up with two equations.

The graph seems like a slanted parabolic curve.

Damn. I keep forgetting about that. Like bloody c.

And I don't think it's a parabola. The original parametric equation only gives values for +ve x, but if you take negative values by using my equation, the curve does some pretty un-parabola-like stuff.

 

[Cogman]

Damn. I keep forgetting about that. Like bloody c.

And I don't think it's a parabola. The original parametric equation only gives values for +ve x, but if you take negative values by using my equation, the curve does some pretty un-parabola-like stuff.

I didn't mean to infer that it was parabolic, only somewhat parabolic-like It isn't anything. Graphing it shows it to have this weird slanted look.

 

[canis]

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[iCyborg]

For the conic, the general equation from wiki is
Ax2 + Bxy + Cy2 + Dx + Ey + F = 0 for all x,y

One straightforward approach is to express all terms via t. You will end up with 5 linear equations (t will have at most 4th power, and a polynomial is equal to 0 if all of its coefficients are equal to 0) and 6 unknowns. Based on that, looks like there should be a solution (with even one free param), but it's possible to obtain some kind of degenerate set of equations, e.g. if two equations are something like A+B=0 and A+B=2, obviously there are no solutions to this. I'd do it, but it kind of looks tedious to write it all out, and it's late here...

 

[Cogman]

Can the part of the curve y>=0 be classified as parabolic? Is "parabolic" a definition or a vague expression?

No. A curve is either parabolic or not. Only in a piecewise situation could you claim that some part of the curve is parabolic (which this is not... Ok, it kind of is, but not really).

x in terms of t is parabolic and y in terms of t is parabolic. But putting them together does not result in a parabolic curve.

 

[Wizlem]

if you see that y=20t+t^2-11t^2=x-11t^2 and then solve the x equation for t^2=200+x+2sqrt(100+x) and plug that back in you can then solve for +sqrt(100+x). Squaring both sides should eventually end up with y^2+4x^2-4xy+4400y-57200x=0 which is a conic section

 

[iCyborg]

@Wizlem
You got something wrong there. For t=1 you get (x,y)=(21,10) which should be on this curve, and when you plug in x=21,y=10 into your equation, you end up with -1.1M instead of 0.

I followed up on my own suggestion, as you don't have to deal with square roots (and is easier to avoid mistakes like above ), and it's mostly about solving a quite simple set of linear equations. I noted before that one param will even be free (6 unknowns, 5 equations) like an oddity - I failed to see that this is required, all zeros is always a solution, so it'd be bad if we had a unique solution, as that would be it. And for a conic, one of A,B,C needs to be non-zero. Furthermore, if you have some equation, you can multiply it by any non-zero real to get another set of valid coeffs, so it's a further reason that we need a free param.

Anyway, I get 5X^2 + xy + (1/20)y^2 -220x + 220y = 0 (if you multiply by 20, you can get all-integer coeffs, but I like it this way...)
Since B^2 - 4AC = 0, this turns out to be a parabola.

 

[Wizlem]

@iCyborg
Thanks for the correction. I attempted to see if your method was easier but it seems much more difficult. I think the main issue in posts previous to mine is before squaring any x or y equations you need to make sure you have only 1 square root term and you place it by itself on 1 side of the equation. When you square both sides, it just disappears.

Thanks Cogman. Did you see the completing the square technique in a reference about parametric equations?

Simplified the equation and got:

This image has been resized. Click this bar to view the full image. The original image is sized 1228x63 and weights 6KB.

Does not appear to be a conic. Am I correct? What can this curve be classified as?

Erm...that's not an equation. That's a statement.

And I got:

y = 10(22(x+100)^0.5-220-x)

How did you get what you got?

These 2 are obviously using the same technique as me. Canis even attempted to square both sides but didn't move the square root to its own side to get rid of it.

 

[iCyborg]

It's not difficult, the tedious part is expanding Ax2 + Bxy + ... with t. Then you group coeffs around powers of t so you have
t^4(A-10B+100C) + t^3(40A-180B-400C) + t^2 etc.

Each of these has to be 0, so you have:
A-10B+100C = 0
40A-180B-400C = 0
400A+400B+400C+D-10E = 0
20D+20E = 0
F=0

The second can be reduced to 2A-9B-20C=0, from here 20C=2A-9B and you can plug this in into the first and get A=5B. After that you get 220B+D=0 from the third, so if you fix D=-E=-220, everything else just unrolls. Computing y^2, xy and x^2 is really the most time consuming, but I feel is still easier than getting rid of square roots, which could be very tricky in the general case.

 

[canis]

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[Wizlem]

My answer had a mistake in it. When I went back through I got what iCyborg has. If you have to graph something with y as a funciton of x, you will have to seperate the equation into 2 parts where y=-x-220+22sqrt(100+x) and y=-x-220-22sqrt(100+x) with x>-100.