Calculating Arc Length in Polar Coordinates

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SUMMARY

The discussion focuses on calculating the arc length of the polar curve defined by r = sin³(x/3) over the interval 0 < x < 3π/2. The derivative r' was computed as r' = cos(x/3)sin²(x/3), and the expression for r² was established as r² = cos²(x/3)sin⁴(x/3). The integral formula for arc length, which involves the square root of the sum of the squares of r' and r, was introduced but left unresolved, prompting requests for further assistance.

PREREQUISITES
  • Understanding of polar coordinates and their properties
  • Knowledge of calculus, specifically integration techniques
  • Familiarity with trigonometric identities, such as sin²(x) = (1 - cos(2x))/2
  • Ability to differentiate polar functions
NEXT STEPS
  • Study the arc length formula for polar coordinates in detail
  • Practice integration of functions involving square roots
  • Explore trigonometric identities and their applications in calculus
  • Learn about numerical methods for evaluating integrals when analytical solutions are complex
USEFUL FOR

Students studying calculus, particularly those focusing on polar coordinates and arc length calculations, as well as educators seeking to enhance their teaching methods in these topics.

JosephR
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Homework Statement



Find The length of r=sin³(x/3) 0<x<3pi/2

2. The attempt at a solution

well first i found r'=3.cos(x/3).1/3.sin²(x/3)=cos(x/3)sin²(x/3)
r²=cos²(x/3)sin^4(x/3)

then i put the formula

integral of radical (r'²+r²)dx and I'm stuck here

any help?
 
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JosephR said:
Find The length of r=sin³(x/3) 0<x<3pi/2

well first i found r'=3.cos(x/3).1/3.sin²(x/3)=cos(x/3)sin²(x/3)
r²=cos²(x/3)sin^4(x/3)

then i put the formula

integral of radical (r'²+r²)dx and I'm stuck here

any help?

Hi JosephR! :smile:

Hint: sin6(x/3) = sin²(x/3)sin^4(x/3) :wink:
 
hey tiny-tim:)

i knew this but it would take some time to be solved !

because sin²(x/3)=[1-cos(2x/3)]/2

anyway thanks buddy !
 

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