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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Martoman on April 06, 2010, 09:49:00 pm

Title: fun parametric equation
Post by: Martoman on April 06, 2010, 09:49:00 pm: I think this is in the domain (no pun intended) of Spech maths, so I will ask it here.

Parametric curves defined by:
$x= 3(cos(\theta) - sin(\theta))$
$y = 2(sin(\theta)+cos(\theta))$

Find the cartesian equation.

Ok so this requires slightly more creativity than the normal ones. I recognised that if you square both sides, they will give you a $2sin(\theta)cos(\theta)$ term which will no doubt cancel.

So I did thus, let $\theta = x$ for simplicity of typing (no, i will not Copy paste thank you very much)
$<br />2x = 6(cos(x) - sin(x))$
$3y= 6((sin(x)+cos(x)))$

Squaring both sides yields:

$4x^2 = 36(cos^2(x) -2cos(x)sin(x) + sin^2(x))$
$9y^2 = 36(sin^2(x) + 2cos(x)(sin(x) + cos^2(x))$

Undoubtably, adding the two equations we get:

$4x^2 + 9y^2 = 36(1-2cos(x)sin(x)) + 36(1 + 2cos(x)sin(x))$
so $4x^2 + 9y^2 = 72$

Then cartesian equation is:

$\boxed{\frac{x^2}{18}+\frac{y^2}{8} = 1}$

Is this valid?

*edit*: Thanks brightsky
Title: Re: fun parametric equation
Post by: m@tty on April 06, 2010, 09:59:57 pm: It appears so. Why shouldn't it be?
Title: Re: fun parametric equation
Post by: Martoman on April 06, 2010, 10:04:27 pm: Specialist book has this insanely contrived answer, this is q12 dii from ext-res chapter 5 btw.
Title: Re: fun parametric equation
Post by: brightsky on April 06, 2010, 10:24:40 pm: Soz for hi-jacking your thread Martoman...I got a related question...

Show that the rational points on the circle $x^2 + y^2 = 1$ can all be found by the parametrization:

$\boxed{x = \frac{2t}{1+t^2}, y = \frac{1-t^2}{1+t^2}}$

By constructing a line segment of gradient $t$ from the known rational point (-1,0).

Do you do this by solving $y = tx + t$ and $x^2 + y^2 = 1$ simultaneously?
Title: Re: fun parametric equation
Post by: Martoman on April 06, 2010, 10:35:34 pm: Quote from: brightsky on April 06, 2010, 10:24:40 pm
Soz for hi-jacking your thread Martoman...I got a related question...

Show that the rational points on the circle $x^2 + y^2 = 1$ can all be found by the parametrization:

$\boxed{x = \frac{2t}{1+t^2}, y = \frac{1-t^2}{1+t^2}}$

By constructing a line segment of gradient $t$ from the known rational point (-1,0).

Do you do this by solving $y = tx + t$ and $x^2 + y^2 = 1$ simultaneously?

In short? Yes. Just reject the x=-1, y = 0 fail solution.
Title: Re: fun parametric equation
Post by: kamil9876 on April 06, 2010, 10:39:53 pm: that's a historical classic :) I think Diophantus was the first to do this. Check here.
Title: Re: fun parametric equation
Post by: /0 on April 06, 2010, 10:43:12 pm: That is (also) known as the weierstrass substitution

It is useful for turning integration problems involving sines and cosines into simple rational functions.
Title: Re: fun parametric equation
Post by: Martoman on April 06, 2010, 10:55:21 pm: Quote from: /0 on April 06, 2010, 10:43:12 pm
That is (also) known as the weierstrass substitution

It is useful for turning integration problems involving sines and cosines into simple rational functions.

HOW DO YOU KNOW THIS? ahahaah you are getting karma for it.
Title: Re: fun parametric equation
Post by: TrueTears on April 06, 2010, 10:58:03 pm: coz /0 is insane at maths thats why
Title: Re: fun parametric equation
Post by: Martoman on April 06, 2010, 11:02:59 pm: Quote from: TrueTears on April 06, 2010, 10:58:03 pm
coz /0 is insane at maths thats why

I always thought maths wasn't much about knowledge, then again, now i can integrate trig functions without thinking!
Title: Re: fun parametric equation
Post by: brightsky on April 06, 2010, 11:06:45 pm: Quote from: Martoman on April 06, 2010, 09:49:00 pm
I think this is in the domain (no pun intended) of Spech maths, so I will ask it here.

Parametric curves defined by:
$x= 3(cos(\theta) - sin(\theta))$
$y = 2(sin(\theta)+cos(\theta))$

Find the cartesian equation.

Ok so this requires slightly more creativity than the normal ones. I recognised that if you square both sides, they will give you a $2sin(\theta)cos(\theta)$ term which will no doubt cancel.

So I did thus, let $\theta = x$ for simplicity of typing (no, i will not Copy paste thank you very much)
$<br />2x = 6(cos(x) - sin(x))$
$3y= 6((sin(x)+cos(x)))$

Squaring both sides yields:

$4x^2 = 36(cos^2(x) -2cos(x)sin(x) + sin^2(x))$
$9y^2 = 36(sin^2(x) + 2cos(x)(sin(x) + cos^2(x))$

Undoubtably, adding the two equations we get:

$4x^2 + 9y^2 = 36(1-2cos(x)sin(x)) + 36(1 + 2cos(x)sin(x))$
so $4x^2 + 9y^2 = 72$

Then cartesian equation is:

$\boxed{\frac{x^2}{8}+\frac{y^2}{18} = 1}$

Is this valid?

Is it me or is this supposed to be $\frac{x^2}{18} + \frac{y^2}{8} = 1$ instead..
Title: Re: fun parametric equation
Post by: Martoman on April 06, 2010, 11:08:20 pm: Um yes, switch em.
Title: Re: fun parametric equation
Post by: /0 on April 06, 2010, 11:18:26 pm: Quote from: Martoman on April 06, 2010, 11:02:59 pm
Quote from: TrueTears on April 06, 2010, 10:58:03 pm
coz /0 is insane at maths thats why

I always thought maths wasn't much about knowledge, then again, now i can integrate trig functions without thinking!

lol TT also knew this, he studied from Stewart Calculus like I did