Finding the Arclength of an Astroid Curve

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Homework Help Overview

The discussion revolves around finding the arclength of an astroid curve defined by the parametric equations f(θ) = 6cos³(θ) and g(θ) = 6sin³(θ), with θ ranging from 0 to 2π. Participants are exploring the integral setup for calculating the total length of the curve.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are discussing the correctness of the integral for arclength and questioning the results obtained from their calculations. There are attempts to simplify the integrand and suggestions to adjust the integration limits.

Discussion Status

There are multiple interpretations of the integral and varying results from different methods of integration. Some participants have provided guidance on simplifying the integrand and adjusting the integration limits, but there is no explicit consensus on the correct answer.

Contextual Notes

Participants mention the possibility of errors related to the settings of their calculation programs, specifically regarding the use of radians versus degrees. There is also a reference to the need for multiplying results from integrals over specific intervals.

ILoveBaseball
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If [tex]f(\theta)[/tex] is given by:[tex]f(\theta) = 6cos^3(\theta)[/tex] and [tex]g(\theta)[/tex] is given by:[tex]g(\theta) = 6sin^3(\theta)[/tex]
Find the total length of the astroid described by [tex]f(\theta)[/tex] and [tex]g(\theta)[/tex].
(The astroid is the curve swept out by ([tex]f(\theta)[/tex],[tex]g(\theta)[/tex]) as [tex]\theta[/tex] ranges from 0 to 2pi )

[tex]f/d(\theta) = -18*cos(x)^2*sin(x)[/tex]
[tex]g/d(\theta) = 18*sin(x)^2*cos(x)[/tex]

this is asking for arclength right?
my integral:

[tex]\int_{0}^{2\pi}\sqrt{(-18*cos(\theta)^2*sin(\theta))^2+(18*sin(\theta)^2*cos(\theta))^2}[/tex]

anyone what's wrong with my integral? cause i keep getting the wrong answer.
 
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I suppose it's all right with your integral. What a result do You receive? In which way You integrated it?
 
ILoveBaseball said:
my integral:

[tex]\int_{0}^{2\pi}\sqrt{(-18*cos(\theta)^2*sin(\theta))^2+(18*sin(\theta)^2*cos(\theta))^2}[/tex]

anyone what's wrong with my integral? cause i keep getting the wrong answer.

Do not forget that

[tex]\sqrt{(\sin\theta)^2(\cos\theta)^2}=|\sin\theta\cos\theta|[/tex]

Integral from 0 to pi/2 and multiple the result by 4.

ehild
 
i used my calculator to integrate my function. I also used another math program on my computer to verify it. i integrated from 0 to pi/2 and got 9.558*4 = 38.232 but it's incorrect and i don't understand why. Also tried to integrate from 0 to 2pi and got 50.253217, but it won't take that either.
 
From 0 to pi/2 i and "Mathematica" got 9. Thus the total length is 9*4=36. Agree?
 
awesome thanks
 
ILoveBaseball said:
i used my calculator to integrate my function. I also used another math program on my computer to verify it. i integrated from 0 to pi/2 and got 9.558*4 = 38.232 but it's incorrect and i don't understand why. Also tried to integrate from 0 to 2pi and got 50.253217, but it won't take that either.

Simplify your integrand. It becomes

[tex]\int_{0}^{2\pi}18|\sin\theta\cos\theta|d\theta[/tex]

Integral from 0 to pi/2, multiply by 4. The result should be 36.

You could have made the mistake with your programs that you did not set to radians and the program calculated with degrees.

ehild
 

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