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Homework Help: Find the exact length of the polar curve

  1. Sep 22, 2012 #1
    1. The problem statement, all variables and given/known data
    r=5^theta theta goes from 0 to 2Pi

    2. Relevant equations

    Length= integral between a and b of sqrt(r^2+(dr/dtheta)^2)dtheta

    3. The attempt at a solution

    r^2=25^theta or 5^(2theta) dr/dtheta=5^theta (ln 5) (dr/dtheta)^2=25^theta+10^theta (ln 5)+ (ln 5)^2

    so the integral would be= sqrt(25^theta+25^theta+10^theta (ln 5)+ (ln 5)^2) dtheta

    ive been working on this problem all day only to find no clue how to solve it,wolfram alpha keeps timing out and cant solve it,maple cant solve it

    i also tried the length= integral(sqrt(dx/dtheta)^2+(dy/dtheta)^2) but both of these turn out to be ridiculously hard integrals to solve...can someone help please?
  2. jcsd
  3. Sep 23, 2012 #2
    It would be helpful to know which class this is for, since there are fancy ways to solve it through Complex Analysis. However, just use the fact that 5θ = eln5θ = eθ ln(5) = (eθ)ln (5). Make a u-substitution with u = eθ; then you have ∫ uln (5) du = uln(5) + 1 / (ln(5) + 1) + C = (eθ)ln(5)+1 / (ln(5) + 1) + C.
  4. Sep 23, 2012 #3
    multivariable calculus 1....the way you did it looks really complicated lol....would it be the same if i used this for length?


    i get an integral of sqrt(25^x(1+(ln5)^2)

    however when i try to compute the exact integral, the results i get from wolfram alpha are slighty different....

    (sqrt((dx/dtheta)^2+(dy/dtheta)^2)dtheta) this way gives a length of 24649.1

    and the first way (sqrt(r^2+(dr/dtheta)^2) gives 21719.3

    and when i calculate it on my own using this way,(sqrt((dx/dtheta)^2+(dy/dtheta)^2)dtheta) i get29015.56297....

    this question has got me soo frustrated,how can they expect us to do this without computers?
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