cartesian equation

x = cos t y = sin2t

I basically used the double angle formulae to get y =2sin(theta)cos(theta) so y = 2sin(theta)x

cos^2(t )+ sin^2(t) is always equal to 1 so sin^2(t) = 1 - x^2
sin t = + or - sqrt (1-x^2)

the thing is - the range is 0 < t < 2pi so how can you see which one it is i.e. + or - the sqrt? pmt solution bank just takes the positive as a given and continues, but as the range is a whole cycle, how can you say it's either?

Reply 1

mqb2766

Original post by velaris08

Can you upload the actual question/pmt solution? Not too sure why you're writing it in that form.

The graph is
https://www.desmos.com/calculator/w61rkgxifh
which is typical for a parametric equation.

(edited 2 years ago)

Reply 2

davros

Original post by velaris08

What do you mean by "which one it is"? As t varies from 0 to 2pi, so x and y vary accordingly taking on positive and negative values. Note that you cannot write y = f(x) where f is a function in this case because you will not get a 1-1 mapping (true function). It's similar to how a circle can be
described by x = cos t, y = sin t, but you can't describe a circle by a single function y = f(x).

Reply 3

velaris08

Original post by davros

What I mean is - in the worked solutions, it takes the value of sin t to be the positive - when I don't get how they can determine that since the domain means that it could be positive or negative?

Reply 4

velaris08

Original post by mqb2766

Sorry - attached

Screenshot 2021-09-14 211408.png34.7KB

Reply 5

mqb2766

Original post by velaris08

Sorry - attached

Is there a question?

Reply 6

davros

Original post by velaris08

What I mean is - in the worked solutions, it takes the value of sin t to be the positive - when I don't get how they can determine that since the domain means that it could be positive or negative?

Unless I'm missing something, they should not have taken the positive square root - they could just square everything to get the final expression.

Reply 7

velaris08

Original post by mqb2766

Is there a question?

question is find the cartesian curves given by the para eqs.
x = cos t and y = sin2t between 0 < t < 2pi

Reply 8

mqb2766

Original post by velaris08

question is find the cartesian curves given by the para eqs.
x = cos t and y = sin2t between 0 < t < 2pi

Tbh, I wouldn't have bothered with all that. You know its going to be a quadratic type relationship and simply squared (1) to get
y^2 = 4x^2 * sin^2(t) = 4x^2(1-x^2)
Problem solved.

Reply 9

velaris08

Original post by davros

Unless I'm missing something, they should not have taken the positive square root - they could just square everything to get the final expression.

So basically in a situation I'd have to give the final expression in a way that is squared? Otherwise it doesn't make sense, right?

Reply 10

velaris08

Original post by mqb2766

Tbh, I wouldn't have bothered with all that. You know its going to be a quadratic type relationship and simply squared (1) to get
y^2 = 4x^2 * sin^2(t) = 4x^2(1-x^2)
Problem solved.

Hmm, but for general reference - I'm not wrong in thinking you can't define sin t to be positive here?

Reply 11

mqb2766

Original post by velaris08

Hmm, but for general reference - I'm not wrong in thinking you can't define sin t to be positive here?

So with the proviso that its a rubbish way to do it (square it up, square root, then square back up), then
y = +/-2*x*sqrt(1-x^2)
However, they square it back up which reintroduces the missing negative solutions. Whether this is deliberate or not, ...

https://www.desmos.com/calculator/wojlznvwc2
Supports your original question, 1/2 the curve is missing if you dont consider -sqrt(). However, its not missing when you square it up.

Its easier to simply ignore this and square (1) directly and plug the x-y variables directly into the pythagorean identity, as mentioned previously.

(edited 2 years ago)

Reply 12

velaris08

Original post by mqb2766

Got it! Thanks so much for that - I did think that perhaps the squaring was why they ignored it but it vexed me how they had the intermediate bit. Also - I know I've asked a lot of questions that you've answered and I've not gotten back to - I appreciate all those responses too and sorry I haven't thanked you directly on those - I've just realised that I have a lot to catch up on so conveniently, the only times I get on here is when I need a question answered :smile:

. Your help is genuinely, ridiculously invaluable to me - and I know I ask a lot of stupid questions and so it's much appreciated. To everyone else who helps me with answers also - but I know you stumble upon mine quite a lot!

(edited 2 years ago)

Reply 13

mqb2766

Original post by velaris08

No problem. My guess is you're right, they missed out 1/2 the solutions and were "lucky" when they reintroduced them by squaring up. I guess they knew the Cartesian answer, so didnt notice they were a bit sloppy with the intermediate steps. However, it really is easier to simply avoid all the hassle and siply work with quadratic / quartic x and y.