# Thread: Determining the length of a logarithmic spiral.

1. ## Determining the length of a logarithmic spiral.

Unfortunately I couldnt convert the pdf into a picture so that I could upload it but i will try to explain my problem as briefly i can
here is a pdf link btw:
<a target="_blank" title="mathhelplength (pdf)" href="http://pdfcast.org/pdf/mathhelplengt...hhelplength</a>
Right so I am faced with the following problem
"Determine the length of the lograrithmic spiral": x = e^-t * cos t y = e^-t * sin t
0<= t <= 2pi
I recal a formula
l(length symbol) = $(integral from a to b) SQRT [x'(t)]^2 + [y'(t)}^ dt I don't know how to evaluate this, could someone show me? EDIT: New link mathhelplength - PDFCast.org 2. Originally Posted by Riazy "Determine the length of the lograrithmic spiral": x = e^-t * cos t y = e^-t * sin t 0<= t <= 2pi$\displaystyle x' = -e^{-t}(\sin{t}+\cos{t})\displaystyle (x')^2 = e^{-2t}(\sin{t}+\cos{t})^2\displaystyle y' = e^{-t}(\cos{t}-\sin{t})\displaystyle (y')^2 = e^{-2t}(\cos{t}-\sin{t})^2\displaystyle (x')^2 + (y')^2 = 2e^{-2t}\displaystyle \displaystyle L = \int_0^{2\pi} \sqrt{2e^{-2t}} \, dt$finish it. 3. Sorry, but the my problem is really how you added together (x')^2+(y')^2 = 2e^-2t I would like to see that step in detail, because thats what I dont understand, i.e how they are added together to 2e^-2t THanks alot 4. Hello, Riazy! This requires some very careful work . . . are you up for it?$\displaystyle \text{Determine the length of the lograrithmic spiral:}$. .$\displaystyle \begin{Bmatrix}x &=& e^{\text{-}t}\cos t \\ y &=& e^{\text{-}t}\sin t \end{Bmatrix}\;\;0\,\le\, t\,\le\,2\pi\displaystyle \displaystyle \text{Formula: }\;L \;=\; \int^b_a \sqrt{[x'(t)]^2 + [y'(t)]^2}\, dt\displaystyle \text{Derivatives: }\;\begin{Bmatrix}x'(t) &=& \text{-}e^{\text{-}t}\sin t - e^{\text{-}t}\cos t &=& \text{-}e^{\text{-}t}(\sin t + \cos t) \\ \\[-3mm] y'(t) &=& e^{\text{-}t}\cos t - e^{\text{-}t}\sin t &=& e^{\text{-}t}(\cos t - \sin t ) \end{Bmatrix}\displaystyle \text{Squares: }\;\begin{Bmatrix}[x'(t)]^2 &=& e^{\text{-}2t}(\sin^2\!t + 2\sin t\cos t + \cos^2\!t) \\ \\[-3mm] [y'(t)]^2 &=& e^{\text{-}2t}(\cos^2\!t - 2\sin t\cos t + \sin^2\!t) \end{Bmatrix}\displaystyle \text{Add: }\;[x'(t)]^2 + [y'(t)]^2 \;=\;e^{\text{-}2t}\bigg[(\underbrace{\sin^2\!t + \cos^2\!t)}_{\text{This is 1}} + \underbrace{(\cos^2\!t + \sin^2\!t)}_{\text{This is 1}}\bigg] $. . . . . . . . . . . . . . . . .$\displaystyle =\;2e^{\text{-}2t}\displaystyle \text{Hence: }\;\sqrt{[x'(t)]^2 + [y'(t)]^2} \;=\;\sqrt{2e^{\text{-}2t}} \;=\;\sqrt{2}\,e^{\text{-}t}\displaystyle \displaystyle \text{Therefore: }\;L \;=\;\sqrt{2}\!\int^{2\pi}_0 \!\!e^{\text{-}t}\,dt $Got it? 5.$\displaystyle (x')^2 = e^{-2t}(\sin{t}+\cos{t})^2\displaystyle (x')^2 = e^{-2t}(\sin^2{t}+2\sin{t}\cos{t}+\cos^2{t})\displaystyle (x')^2 = e^{-2t}(1+2\sin{t}\cos{t})\displaystyle (y')^2 = e^{-2t}(\cos{t}-\sin{t})^2\displaystyle (y')^2 = e^{-2t}(\cos^2{t}-2\cos{t}\sin{t} + \sin2{t})\displaystyle (y')^2 = e^{-2t}(1-2\cos{t}\sin{t})\displaystyle (x')^2 + (y')^2 = e^{-2t}(1+2\sin{t}\cos{t}) + e^{-2t}(1-2\cos{t}\sin{t}) = e^{-2t}[(1+2\sin{t}\cos{t}) + (1-2\sin{t}\cos{t})]\$

can you finish it now?