Question

Find the length of the parametric curve curve x = et cos t, y = et sin t
for 0 <= t <= 2.

Answer 1

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

Answer 2

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,\[\frac{ds}{dt}=\sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2} \rightarrow ds =\sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2}dt\]

Answer 3

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

Answer 4

yup :)

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

you've got some errors

Answer 10

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

Answer 11

otherwise its great

Answer 12

yeah i got that :D pretty neat calculus

Answer 13

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

Answer 14

:)