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Find the exact length of the polar curve

  • #1

Homework Statement


r=5^theta theta goes from 0 to 2Pi


Homework Equations



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

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?
 

Answers and Replies

  • #2
986
8

Homework Statement


r=5^theta theta goes from 0 to 2Pi


Homework Equations



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

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?
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.
 
  • #3
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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