Parametric -> Implicit Equations

JC3187
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Hi guys,

I have done what I can with the following:

Given a parametric curve x = xsint, y = sin(2t) where t is in R.

Find an implicit equation of this curve.

MY ANSWER:
y = 2costsint = costx

Therefore sint = x/2, cost = y/x

sin^2(t) + cos^2(t) = x^2 / 4 + y^2 / x^2 = 1

Would this be right?
I understand that x != 0 but why can't I just multiply every value underlined by 4x^2?

Thank you, any input is appreciated!
 
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JC3187 said:
Hi guys,

I have done what I can with the following:

Given a parametric curve x = xsint, y - sin(2t) where t is in R.



Please correct the typo's so we know what the actual equations are.
 
JC3187 said:
Hi guys,

I have done what I can with the following:

Given a parametric curve x = xsint, y - sin(2t) where t is in R.

I will assume, from your work below, that the equations are ##x=2\sin t,~y=\sin(2t)##

Find an implicit equation of this curve.
MY ANSWER:
y = 2costsint = costx

Therefore sint = x/2, cost = y/x

sin^2(t) + cos^2(t) = x^2 / 4 + y^2 / x^2 = 1

Would this be right?
I understand that x != 0 but why can't I just multiply every value underlined by 4x^2?

Thank you, any input is appreciated!

Yes, you can multiply through by ##4x^2##. The original equations certainly allow ##x=0## and the only problem is that you divided by ##x## in your solution. You could have derived the same equation as you get when you multiply through by ##4x^2## without ever dividing by ##x##. So you should give the answer as $$
4x^2 = 4y^2+x^4$$or some variation of that.
 
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JC3187 said:
Hi guys,

I have done what I can with the following:

Given a parametric curve x = xsint, y - sin(2t) where t is in R.

Find an implicit equation of this curve.

MY ANSWER:
y = 2costsint = costx

Therefore sint = x/2, cost = y/x

sin^2(t) + cos^2(t) = x^2 / 4 + y^2 / x^2 = 1

Would this be right?
I understand that x != 0 but why can't I just multiply every value underlined by 4x^2?

Thank you, any input is appreciated!
Do you mean
x = sin(t)

and

y = sin(2t)

?​
 
Yes sorry That is what I meant.

How do you derive it from the original equation?
x=2sint, y=sin(2t) without dividing by x?
 
JC3187 said:
Yes sorry That is what I meant.

How do you derive it from the original equation?
x=2sint, y=sin(2t) without dividing by x?

$$x^2+y^2=4\sin^2 t + (2\sin t\cos t)^2=4\sin^2t(1+\cos^2 t)
=4\sin^2 t(2-\sin^2 t) = x^2(2-\frac{x^2} 4) = \frac{x^2(8-x^2)}{4}$$
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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