Find the Cartesian equation given the parametric equations

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The discussion centers on deriving the Cartesian equation from the parametric equations x = (cos t)³ and y = (sin t)³, leading to the equation x^(2/3) + y^(2/3) = 1. Participants debate the implications of negative values for x and y, noting that the equation holds true only for positive values, creating inconsistencies when negative values are introduced. The conversation also touches on the interpretation of cube roots of negative numbers, with some expressing uncertainty about conventional expressions. Ultimately, it is agreed that while the parametric equations work for positive values, complications arise with negative inputs. The need for caution in manipulating fractional exponents with negative bases is emphasized.
chwala
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Homework Statement
Find the cartesian equation given the parametric equations;

##x=\cos ^3t, y=\sin^3 t ##
Relevant Equations
parametric equations
hmmmmm nice one...boggled me a bit; was trying to figure out which trig identity and then alas it clicked :wink:

My take;

##x=(\cos t)^3 ## and ##y=(\sin t)^3##

##\sqrt[3] x=\cos t## and ##\sqrt[3] y=\sin t##

we know that

##\cos^2 t + \sin^2t=1##

therefore we shall have,

##x^{\frac{2}{3}} + y^{\frac{2}{3}}=1##

Any other way is highly appreciated...
 
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You should consider negative x y case.
 
anuttarasammyak said:
You should consider negative x y case.
I seem not to get you. Consider negative cases in what way?
 
I interpreted the equation as well as Wolfram do as below shown.

1674349062272.png
1674349135976.png

As for your
\sqrt[3] x=\cos t ,\sqrt[3] y=\sin t
I am not sure whether \sqrt[3]{-1}=-1 is a conventional expression. Say it is so,
(-1)^{2/6}=1^{1/6}=1 \neq (-1)^{1/3}
That seems inconvenient.
 
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anuttarasammyak said:
I am not sure whether \sqrt[3]{-1}=-1 is a conventional expression.
Sure it is. A negative real number has a real and negative root, plus a couple complex roots.
 
If we allow negative x for ##x^{m/n}##
x^{m/n}=x^{2m/2n}=(x^2)^{m/2n}=(|x|^2)^{m/2n}=|x|^{m/n}
for an example
\sqrt[n]{-1}=\sqrt[n]{1}=1
which is a false statement. So I think we had better use ##x^{m/n}## for only positive x.

[EDIT] In order the defnition of imaginary number unit,
\sqrt{-1}=i, remains sound, we should just watch as it is and never apply the procedures above.
 
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anuttarasammyak said:
I am not sure whether
\sqrt[3]{-1}=-1 is a conventional expression. Say it is so,
(-1)^{2/6}=1^{1/6}=1 \neq (-1)^{1/3}
That seems inconvenient.
I don't believe there is any problem with ##\sqrt[3]{-1} = (-1)^{1/3}##, both of which are equal to -1.
However, when you start manipulating the exponent, using the properties of fractional exponents, then I agree that you run into trouble when the base is negative.

Having thought about things a bit longer, @anuttarasammyak, I agree with what you are saying. The parametric equations ##x = \cos^3(t)## and ##y = \sin^3(t)## agree with the resulting equation that @chwala got: ##x^{2/3} + y^{2/3} = 1## only when both x and y are positive. For negative values of x or y, which map to values of t in the parametric equations, you run into problems of inconsistency in the non-parametric equation.
 
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