|Sep22-12, 11:56 PM||#1|
find the exact length of the polar curve
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?
|Sep23-12, 12:25 AM||#2|
|Sep23-12, 12:59 AM||#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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