find the arc length and the length of the part of the tangen

[Smily]

sketch the exponential spiral r=ae^(bx) for the case in which a is positive and b is negative, and show that the arc length form (theta)= 0 to (theta)=infinity to the length of the part of the tangent at (theta)=0 that is cut off by the x- and y- axis.
Basically, find arc length and length of the tangent.
i did
L= (from 0 to infinity)sqrt( (ae^(b(theta)) )^2 + (abe^(b(theta)))^2 ) d(theta) =
lim(t-->infinity) (from 0 to t)sqrt ( ae^(b(theta)) (1+ b^2) ) d(theta)=
and here i don`t know what to do.
:roll:

 

[galactus]

\(\displaystyle \L\\\sqrt{(ae^{-b{\theta}})^{2}+(-abe^{-b{\theta}})^{2}}\)

\(\displaystyle =\L\\\sqrt{\L\\a^{2}(b^{2}+1)e^{-2b{\theta}}}\)

\(\displaystyle =\L\\\sqrt{a^{2}(b^{2}+1)}e^{-b{\theta}}\)

\(\displaystyle =\L\\\int\sqrt{a^{2}(b^{2}+1)}e^{-b{\theta}}d{\theta}\)

\(\displaystyle =\L\\\frac{-e^{-b{\theta}}a\sqrt{b^{2}+1}}{b}\)

Does this help?. Is this something like you were aiming at?.

 

[Smily]

im sorry, i missed few words.
sketch the exponential spiral r=ae^(bx) for the case in which a is positive and b is negative, and show that the arc length form (theta)= 0 to (theta)=infinity IS EQUAL to the length of the part of the tangent at (theta)=0 that is cut off by the x- and y- axis.
I have problem to solve to the length of the part of the tangent at (theta)=0 that is cut off by the x- and y- axis.

 

[galactus]

Perhaps I am going off on a tangent here, but try the limit:

\(\displaystyle \L\\\lim_{{\theta}\to\0}\frac{-e^{-b{\theta}}a\sqrt{b^{2}+1}}

{b}=\frac{-a\sqrt{b^{2}+1}}{b}\)

 

[Smily]

but how the last relate to the length of the part of the tangent at (theta)=0? ho wcan i find it?

 

[galactus]

As theta aproaches 0, the length approaches \(\displaystyle \frac{a\sqrt{b^{2}+1}}{b}\)