Find the Cartesian equation given the parametric equations

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Homework Help Overview

The discussion revolves around finding the Cartesian equation corresponding to the given parametric equations \(x = (\cos t)^3\) and \(y = (\sin t)^3\). Participants explore the implications of these equations, particularly in relation to trigonometric identities and the behavior of the functions involved.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the transformation of the parametric equations into a Cartesian form, leading to the equation \(x^{\frac{2}{3}} + y^{\frac{2}{3}} = 1\). Some participants question the treatment of negative values for \(x\) and \(y\) and how these affect the validity of the derived equation. There is also a debate about the conventionality of expressions involving negative bases and fractional exponents.

Discussion Status

The discussion is ongoing, with participants offering different perspectives on the implications of negative values in the context of the equations. Some guidance has been provided regarding the treatment of negative numbers in exponentiation, but no consensus has been reached on the best approach to handle these cases.

Contextual Notes

Participants are considering the implications of the parametric equations on the Cartesian form, particularly focusing on the restrictions that arise when \(x\) or \(y\) are negative. There is an awareness of the potential complications that arise from manipulating fractional exponents with negative bases.

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...
 
Last edited:
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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.
 
Last edited:
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.
 
Last edited:
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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