Question

Find the length traced out along the parametric curve x=cos(e^(2t)), y=sin(e^(2t)) and has endpoints [0,1].

so far my answer is....
sin^2(e^(2t))*4e^(4t)+cos^2(e^(2t))*4e^(4t)

Answer 1

and then i know you put together like terms...so....

=4e^(4t)+4e^(4t) * (sin^2(e^(2t))+cos^2(e^(2t))

Answer 2

@agent0smith

Answer 3

it's basically asking for arc length... which for parametric curves is \[\Large \int\limits_{a}^{b} \sqrt{\left( \frac{ dx }{dt } \right)^2 +\left( \frac{ dy }{dt } \right)^2} dt\]

Answer 4

yup thats how i got my answer but idk where to go from where im at

Answer 5

Assuming your derivatives are correct, now plug that into the square root... but first, sin^2x+cos^2x = 1... so what's sin^2(e^(2t))+cos^2(e^(2t)) = ??? (note they both have e^2t so it's just like sin^2x+cos^2x=1

Answer 6

Your derivatives look right. So 4e^(4t)+4e^(4t) * (sin^2(e^(2t))+cos^2(e^(2t)) is really just 4e^(4t)+4e^(4t) since the whole sin^2 + cos^2 is equal to 1.

\[\Large \int\limits\limits_{0}^{1} \sqrt{4e^{4t}} dt\]

Answer 7

Oh you had an extra 4e^4t. it's just 4e^(4t), you factored it out twice at the start.

Answer 8

After simplifying the square root... \[\Large \int\limits\limits\limits_{0}^{1}2e^{2t} dt\]

Answer 9

so it can equal to 1 even though the e^(2t) is there opposed to just t?

Answer 10

Yep. sin^2(e^2t) + cos^2(e^2t) is still 1. As long as it's the same thing in both sine and cosine (the e^2t).

Answer 11

awesome so its 6.38906!!!

Answer 12

well my online homework says it is.

Answer 13

http://tutorial.math.lamar.edu/Classes/CalcII/ParaArcLength.aspx

Answer 14

6.38

Answer 15

Oh, you meant \[\Large \int\limits r d \theta\] i'm guessing...