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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?2012-02-14
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    $\vec i$ 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 $$f(t)=\left(a e^{-bt}\cos(t),a e^{-bt}\sin(t)\right)$$2012-02-14
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    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.2012-02-14
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    Sorry, I should have provided more. Is using the vector notation an alternative to $y=ae^{-bt}cos(t), x=ae^{}$?2012-02-14
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    For part (a), I will be calculating the limit of each $x=f(t)$ and $y=g(t)$ as they approach infinity?2012-02-14
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    @J.Borges: Usually $i$ is the unit vector along the $x$-axis, so in the question two above, you have reversed $x$ and $y$. No big deal, but might as well get it right. For question above, yes, you calculate the limits, which are clearly $0$, since the $e^{-bt}$ term goes to $0$, and the sine and cosine always have absolute value $\le 1$. And it is not as "they" $\to\infty$, it is as $t\to\infty$.2012-02-14
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    I've got it down to $\int_{0}^{\infty}\sqrt{b^{2}a^{2}e^{-2bt}cos^{2}(t)+a^{2}e^{-2bt}sin^{2}(t)+b^{2}a^{2}e^{-2bt}sin^{2}(t)+a^{2}e^{-2bt}cos^{2}(t)}dt$ but now I am at a bit of a loss. I am trying to get rid of the square root.2012-02-15
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    Get rid of the $\cos^2$'s and (+) $\sin^2$'s! You are close..2012-02-16

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