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I am working on this problem. Even some of the notation has me confused (the vectors $\vec i$ and $\vec j$).

Let $\vec r(t):=ae^{-bt}\cos(t)\vec i +ae^{-bt}\sin(t)\vec j$ where $a$ and $b$ are positive constants. The trace of $\vec r (t)$ is called the logarithmic spiral.

(a) Show that as $t \to +\infty $, $\vec r (t)$ approaches the origin.

(b) Show that $\vec r (t)$ has finite arc length on $[0,\infty)$.

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What is your question? – Aryabhata Feb 14 '12 at 20:47
$\vec i$ and $\vec j$ are common notation for orthogonal vectors. In a more modern notation, you would say that the spiral is a function $f:\mathbb R \to \mathbb{R}^2$ given by $$f(t)=\left(a e^{-bt}\cos(t),a e^{-bt}\sin(t)\right)$$ – yohBS Feb 14 '12 at 20:47
Recall that the arclength of the parametric curve $x=f(t)$, $y=g(t)$ as $t$ goes from $p$ to $q$ is $\int_p^q\sqrt{(f'(t))^2+(g'(t))^2}dt$. Compute the derivatives carefully, square them, add. There will be magic simplification. – André Nicolas Feb 14 '12 at 21:07
Sorry, I should have provided more. Is using the vector notation an alternative to $y=ae^{-bt}cos(t), x=ae^{}$? – J.Borges Feb 14 '12 at 21:25
For part (a), I will be calculating the limit of each $x=f(t)$ and $y=g(t)$ as they approach infinity? – J.Borges Feb 14 '12 at 21:26

The length is given, as the OP wrote, by $$\int_0^\infty\sqrt{a^2b^2e^{-2bt}(\cos^2t+\sin^2t)+a^2e^{-2bt}(\cos^2t+\sin^2t)}\,\,dt=$$ $$=\int_0^\infty ae^{-bt}\sqrt{b^2+1}\,\,dt=\left.-\frac{a\sqrt{b^2+1}}{b}\,e^{-bt}\right|_0^\infty=\frac{a}{b}\sqrt{b^2+1}$$

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