Find-the-length-of-the-arc-of-the-hyperbolic-spiral-r-a-lying-between-r-a-and-r-2a-

Question Number 11413 by agni5 last updated on 24/Mar/17

Find the length of the arc of the hyperbolic

spiral r θ = a lying between r = a and

r = 2 a .

Answered by mrW1 last updated on 26/Mar/17

r = \frac{a}{θ}

\frac{d r}{d θ} = - \frac{a}{θ^{2}}

\sqrt{r^{2} + {(\frac{d r}{d θ})}^{2}} = \frac{a \sqrt{1 + θ^{2}}}{θ^{2}}

L = \int_{θ_{1}}^{θ_{2}} \sqrt{r^{2} + {(\frac{d r}{d θ})}^{2}} d θ = a \int_{θ_{1}}^{θ_{2}} \frac{\sqrt{1 + θ^{2}}}{θ^{2}} d θ

= a {[- \frac{\sqrt{1 + θ^{2}}}{θ} + \ln (θ + \sqrt{1 + θ^{2}})]}_{θ_{1}}^{θ_{2}}

= a [\frac{\sqrt{1 + θ_{1}^{2}}}{θ_{1}} - \frac{\sqrt{1 + θ_{2}^{2}}}{θ_{2}} + \ln \frac{θ_{2} + \sqrt{1 + θ_{2}^{2}}}{θ_{1} + \sqrt{1 + θ_{1}^{2}}}]

w i t h θ_{1} = \frac{a}{r_{1}} = \frac{a}{2 a} = \frac{1}{2} a n d θ_{2} = \frac{a}{r_{2}} = \frac{a}{a} = 1

L = a [\frac{\sqrt{1 + \frac{1}{4}}}{\frac{1}{2}} - \frac{\sqrt{1 + 1}}{1} + \ln \frac{1 + \sqrt{1 + 1}}{\frac{1}{2} + \sqrt{1 + \frac{1}{4}}}]

L = a [\sqrt{5} - \sqrt{2} + \ln \frac{2 (1 + \sqrt{2})}{1 + \sqrt{5}}]

Commented by mrW1 last updated on 26/Mar/17

t h e a n s w e r i s c o r r e c t e d .

p l e a s e s e e a l s o Q 11433 .

Facebook Tweet Pin