# cartesian equation

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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?

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?

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#2

(Original post by

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?

**velaris08**)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?

The graph is

https://www.desmos.com/calculator/w61rkgxifh

which is typical for a parametric equation.

Last edited by mqb2766; 1 month ago

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#3

**velaris08**)

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?

described by x = cos t, y = sin t, but you can't describe a circle by a single function y = f(x).

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(Original post by

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).

**davros**)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).

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(Original post by

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.

**mqb2766**)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.

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#7

(Original post by

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?

**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?

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(Original post by

Is there a question?

**mqb2766**)Is there a question?

x = cos t and y = sin2t between 0 < t < 2pi

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#9

(Original post by

question is find the cartesian curves given by the para eqs.

x = cos t and y = sin2t between 0 < t < 2pi

**velaris08**)question is find the cartesian curves given by the para eqs.

x = cos t and y = sin2t between 0 < t < 2pi

y^2 = 4x^2 * sin^2(t) = 4x^2(1-x^2)

Problem solved.

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(Original post by

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

**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.

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(Original post by

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.

**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.

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#12

(Original post by

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

**velaris08**)Hmm, but for general reference - I'm not wrong in thinking you can't define sin t to be positive here?

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.

Last edited by mqb2766; 1 month ago

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(Original post by

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.

**mqb2766**)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.

Last edited by velaris08; 1 month ago

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#14

(Original post by

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 .

**velaris08**)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 .

Last edited by mqb2766; 1 month ago

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