Find the length of a curve 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?
 A: 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
A: 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 $\sqrt{2}\sin(1)/e$
A: 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).
A: 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 $2e^{-\pi/4}-\sqrt{2}\,e^{-1}\sin 1$.
