Determining the length of a logarithmic spiral.

**Riazy** · February 13th 2011, 07:35 AM

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 <= 2piI 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

**skeeter** · February 13th 2011, 07:48 AM

Originally Posted by Riazy

"Determine the length of the lograrithmic spiral":
x = e^-t * cos t
y = e^-t * sin t 0<= t <= 2pi

$x' = -e^{-t}(\sin{t}+\cos{t})$

$(x')^2 = e^{-2t}(\sin{t}+\cos{t})^2$

$y' = e^{-t}(\cos{t}-\sin{t})$

$(y')^2 = e^{-2t}(\cos{t}-\sin{t})^2$

$(x')^2 + (y')^2 = 2e^{-2t}$

$\displaystyle L = \int_0^{2\pi} \sqrt{2e^{-2t}} \, dt$

finish it.

**Riazy** · February 13th 2011, 08:39 AM

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

**Soroban** · February 13th 2011, 08:45 AM

Hello, Riazy!

This requires some very careful work . . . are you up for it?

$\text{Determine the length of the lograrithmic spiral:}$

. . $\begin{Bmatrix}x &=& e^{\text{-}t}\cos t \\ y &=& e^{\text{-}t}\sin t \end{Bmatrix}\;\;0\,\le\, t\,\le\,2\pi$

$\displaystyle \text{Formula: }\;L \;=\; \int^b_a \sqrt{[x'(t)]^2 + [y'(t)]^2}\, dt$

$\text{Derivatives: }\;\begin{Bmatrix}x'(t) &=& \text{-}e^{\text{-}t}\sin t - e^{\text{-}t}\cos t &=& \text{-}e^{\text{-}t}(\sin t + \cos t) \\ \\[-3mm] y'(t) &=& e^{\text{-}t}\cos t - e^{\text{-}t}\sin t &=& e^{\text{-}t}(\cos t - \sin t ) \end{Bmatrix}$

$\text{Squares: }\;\begin{Bmatrix}[x'(t)]^2 &=& e^{\text{-}2t}(\sin^2\!t + 2\sin t\cos t + \cos^2\!t) \\ \\[-3mm] [y'(t)]^2 &=& e^{\text{-}2t}(\cos^2\!t - 2\sin t\cos t + \sin^2\!t) \end{Bmatrix}$

$\text{Add: }\;[x'(t)]^2 + [y'(t)]^2 \;=\;e^{\text{-}2t}\bigg[(\underbrace{\sin^2\!t + \cos^2\!t)}_{\text{This is 1}} + \underbrace{(\cos^2\!t + \sin^2\!t)}_{\text{This is 1}}\bigg]$
. . . . . . . . . . . . . . . . . $=\;2e^{\text{-}2t}$

$\text{Hence: }\;\sqrt{[x'(t)]^2 + [y'(t)]^2} \;=\;\sqrt{2e^{\text{-}2t}} \;=\;\sqrt{2}\,e^{\text{-}t}$

$\displaystyle \text{Therefore: }\;L \;=\;\sqrt{2}\!\int^{2\pi}_0 \!\!e^{\text{-}t}\,dt$

Got it?

**skeeter** · February 13th 2011, 08:49 AM

$(x')^2 = e^{-2t}(\sin{t}+\cos{t})^2$

$(x')^2 = e^{-2t}(\sin^2{t}+2\sin{t}\cos{t}+\cos^2{t})$

$(x')^2 = e^{-2t}(1+2\sin{t}\cos{t})$

$(y')^2 = e^{-2t}(\cos{t}-\sin{t})^2$

$(y')^2 = e^{-2t}(\cos^2{t}-2\cos{t}\sin{t} + \sin2{t})$

$(y')^2 = e^{-2t}(1-2\cos{t}\sin{t})$

$(x')^2 + (y')^2 = e^{-2t}(1+2\sin{t}\cos{t}) + e^{-2t}(1-2\cos{t}\sin{t}) = e^{-2t}[(1+2\sin{t}\cos{t}) + (1-2\sin{t}\cos{t})]$

can you finish it now?

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