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Find the length of the following curve:

$r(t) = e^{-t} \sin(t)+e^{-t}\sin(t) i$ for $0 \leq t\leq 1$.

Any ideas?

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you're missing a "j" I think. –  James S. Cook Aug 28 '12 at 2:41
    
No the question specifcally doesnt have a j –  Nick Aug 28 '12 at 2:42
    
Is $i$ the imaginary unit? –  EuYu Aug 28 '12 at 2:44
    
No it is the x component –  Nick Aug 28 '12 at 2:44
1  
Am puzzled by the notation. If it is the parametric "curve" $x(t)=e^{-t}\sin t$, $y(t)=e^{-t}\sin t$ then it is really easy. –  André Nicolas Aug 28 '12 at 2:53

4 Answers 4

Once you correct what appears to be an error, can you follow the definition and examples here?

http://www.mathwords.com/a/arc_length_of_a_curve.htm

HTH

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My interpretation: $x = e^{-t}\sin(t)$ and $y =e^{-t}\sin(t)$ where $r = x+iy$ apparently. For $0 \leq t \leq 1$ we have a rather silly parameterization of a line from $(0,0)$ to $(\sin(1)/e,\sin(1)/e)$ the length of which by the distance formula is simply $\sqrt{2}\sin(1)/e$

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This is indeed true, however, I think the OP may possibly have made a mistake. I think it should have been $t \mapsto (e^{-t}\sin t, e^{-t}\cos t).$ If the OP's question was correct then we need to consider the turning points of $e^{-t}\sin t$. It's possible that the curve could retrace itself, e.g. $t \mapsto (\sin t, \sin t).$ The image is a line segment of length $\sqrt{2}$, but the curve goes backwards and forwards in that segment forever. In fact $e^{-t}\sin t$ has a turning point at $t = \pi/4.$ –  Fly by Night Aug 28 '12 at 17:06

As far as I understand we have $$ x = e^{-t} \sin(t)\\ y = e^{-t} \sin(t)$$ and $\dfrac{dx}{dt} = \dfrac{dy}{dt} = 2e^{-t}(\cos t - \sin t)$. Recall the length is given by $$ \int_0^{1} \sqrt{\Big(\frac{dx}{dt}\Big)^2 + \Big(\frac{dy}{dt}\Big)^2} \ dt$$ which this simplifies to $$ \sqrt{2} \int_0^{1} \frac{dx}{dt}\ dt = \sqrt{2} \int_0^{1} 2e^{-t}(\cos t - \sin t)\ dt $$ So it boils down to evaluating the integral $$ \int_0^{1} e^{-t}\cos t \ dt - \int_0^{1} e^{-t} \sin t\ dt $$ which not hard to do (I guess by parts).

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If we interpret the problem as meaning $x(t)=y(t)=e^{-t}\sin t$, then there are two possible answers, depending on whether we think of what is described as a set of points or as a path.

We are travelling along the line $y=x$. It is easy to verify that the distance from the origin reaches a maximum at $t=\pi/4$, and then we travel some distance back towards the origin to the point $(e^{-1}\sin 1, e^{-1}\sin 1)$. So the length of the set traced out is $\sqrt{2}f(\pi/4)$ where $f(t)=e^{-t}\sin t$. This is simply $e^{-\pi/4}$.

If we think in terms of the length of the path travelled, we must add in the length of the line segment from $f(\pi/4,\pi/4)$ to $f(1,1)$. That gives length of the path equal to $2e^{-\pi/4}-\sqrt{2}\,e^{-1}\sin 1$.

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