Determining the Length of a Polar Curve

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M9-6: length of polar curves

Tip: This isn't the place to ask a question because the teacher can't reply.

Key Questions

  • We can find the arc length #L# of a polar curve #r=r(theta)# from #theta=a# to #theta=b# by

    #L=int_a^bsqrt{r^2+({dr}/{d theta})^2}d theta#

  • The Arc Length in Polar Coordinates is given bu:

    # L = int \ dS # where # dS=sqrt(r^2+((dr)/(d theta))^2) \ d theta #

  • Answer:

    Use the chain rule.

    Explanation:

    By Chain Rule,

    #{dr}/{d theta}=3cos^2(theta/3)cdot[-sin(theta/3)]cdot1/3#

    by cleaning up a bit,

    #=-cos^2(theta/3)sin(theta/3)#

    Let us first look at the curve #r=cos^3(theta/3)#, which looks like this:

    enter image source here

    Note that #theta# goes from #0# to #3pi# to complete the loop once.

    Let us now find the length #L# of the curve.

    #L=int_0^{3pi}sqrt{r^2+({dr}/{d theta})^2} d theta#

    #=int_0^{3pi}sqrt{cos^6(theta/3)+cos^4(theta/3)sin^2(theta/3)}d theta#

    by pulling #cos^2(theta/3)# out of the square-root,

    #=int_0^{3pi}cos^2(theta/3)sqrt{cos^2(theta/3)+sin^2(theta/3)}d theta#

    by #cos^2theta=1/2(1+cos2theta)# and #cos^2theta+sin^2theta=1#,

    #=1/2int_0^{3pi}[1+cos({2theta}/3)]d theta#

    #=1/2[theta+3/2sin({2theta}/3)]_0^{3pi}#

    #=1/2[3pi+0-(0+0)]={3pi}/2#

    I hope that this was helpful.

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