Find the Cartesian equation given the parametric equations

In summary, the equations for x and y can be solved for when both x and y are positive, but when x or y are negative, the equations become inconsistent.
  • #1
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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  • #2
You should consider negative x y case.
 
  • #3
anuttarasammyak said:
You should consider negative x y case.
I seem not to get you. Consider negative cases in what way?
 
  • #4
I interpreted the equation as well as Wolfram do as below shown.

1674349062272.png
1674349135976.png

As for your
[tex]\sqrt[3] x=\cos t ,\sqrt[3] y=\sin t[/tex]
I am not sure whether [tex]\sqrt[3]{-1}=-1[/tex] is a conventional expression. Say it is so,
[tex](-1)^{2/6}=1^{1/6}=1 \neq (-1)^{1/3}[/tex]
That seems inconvenient.
 
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  • #5
anuttarasammyak said:
I am not sure whether [tex]\sqrt[3]{-1}=-1[/tex] is a conventional expression.
Sure it is. A negative real number has a real and negative root, plus a couple complex roots.
 
  • #6
If we allow negative x for ##x^{m/n}##
[tex]x^{m/n}=x^{2m/2n}=(x^2)^{m/2n}=(|x|^2)^{m/2n}=|x|^{m/n}[/tex]
for an example
[tex]\sqrt[n]{-1}=\sqrt[n]{1}=1[/tex]
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,
[tex]\sqrt{-1}=i[/tex], remains sound, we should just watch as it is and never apply the procedures above.
 
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  • #7
anuttarasammyak said:
I am not sure whether
[tex]\sqrt[3]{-1}=-1[/tex] is a conventional expression. Say it is so,
[tex](-1)^{2/6}=1^{1/6}=1 \neq (-1)^{1/3}[/tex]
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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FAQ: Find the Cartesian equation given the parametric equations

What are parametric equations?

Parametric equations represent a set of related quantities as explicit functions of an independent parameter. For example, for a curve in the plane, the coordinates x and y might be given as functions of a parameter t: x = f(t) and y = g(t).

What is a Cartesian equation?

A Cartesian equation is an equation that specifies a curve, surface, or other geometric object in terms of Cartesian coordinates (x, y, z, etc.). For example, the equation of a circle in Cartesian coordinates is x^2 + y^2 = r^2.

How do you convert parametric equations to a Cartesian equation?

To convert parametric equations to a Cartesian equation, eliminate the parameter by solving one of the parametric equations for the parameter and then substituting this expression into the other parametric equation. This process results in an equation that relates x and y directly.

Can you provide an example of converting parametric equations to a Cartesian equation?

Sure, consider the parametric equations x = 3t + 2 and y = 2t - 1. To eliminate the parameter t, solve the first equation for t: t = (x - 2) / 3. Substitute this into the second equation: y = 2((x - 2) / 3) - 1. Simplify to get the Cartesian equation: y = (2/3)x - 5/3.

Are there any special cases or difficulties to be aware of when converting parametric to Cartesian equations?

Yes, some parametric equations may describe curves that are difficult to express in Cartesian form, such as loops or curves that intersect themselves. Additionally, if the parametric equations involve trigonometric functions, the conversion might require using trigonometric identities, which can complicate the process.

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