## anonymous 5 years ago Find the length of the parametric curve curve x = et cos t, y = et sin t for 0 <= t <= 2.

1. anonymous

you take the (dx/dt)^2+ (dy/dt)^2 amd then take the ontegral ove rthe inerval is this idea correct

2. anonymous

A differential element of a curve in the x-y plane, ds, is given in terms of the differentials of x and y by Pythagoras' Theorem. $(ds)^2=(dx)^2+(dy)^2$Dividing both sides of the equation by (dt)^2, you have$\left( \frac{ds}{dt} \right)^2=\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2$and from this,$\frac{ds}{dt}=\sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2} \rightarrow ds =\sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2}dt$

3. anonymous

Integrating,$s=\int\limits_{t_1}^{t_2}\sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2}dt$

4. anonymous

yup :)

5. anonymous

$x=e^t \sin t \rightarrow \frac{dx}{dt}=e^t(\cos t - \sin t )$$y=e^t \sin t \rightarrow x=e^t \sin t \rightarrow \frac{dy}{dt}=e^t(\sin t + \cos t )$

6. anonymous

Squaring and adding each, you have$\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2=2e^2t$

7. anonymous

So your arc length is$s=\int\limits_{0}^{2}\sqrt{2}e^tdt=\sqrt{2}(e^2-1)$

8. anonymous

*should be 2e^(2t) in second-last part of derivation.

9. anonymous

you've got some errors

10. anonymous

x= e^t cos t , not e^t sin t

11. anonymous

otherwise its great

12. anonymous

yeah i got that :D pretty neat calculus

13. anonymous

well, transcribing from notes to this site hardly ever goes to plan.

14. anonymous

:)