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find the exact length of the polar curve |
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| Sep22-12, 11:56 PM | #1 |
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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 |
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| Sep23-12, 12:59 AM | #3 |
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multivariable calculus 1....the way you did it looks really complicated lol....would it be the same if i used this for length?
(sqrt((dx/dtheta)^2+(dy/dtheta)^2)dtheta)? 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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