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Mathematics 18 Online
OpenStudy (anonymous):

find the arc length of a curve

OpenStudy (anonymous):

\[r(t)=tcosti+tsintj+\frac{ 2\sqrt{2} }{ 3}t ^{3/2}k\]

OpenStudy (anonymous):

r(t) such that t>0 and t<pi

OpenStudy (sirm3d):

\[\large L=\int\limits_{a}^{b}\sqrt{r\prime(t)}dt\] just compute the derivative and sub un the integral

OpenStudy (sirm3d):

correction: \[\large L= \int\limits_{a}^{b}\sqrt{\left| r\prime(t) \right|}dt\] compute the derivative and sub in the integral

OpenStudy (anonymous):

-sinti+costj

OpenStudy (anonymous):

how do i do the other derivation

OpenStudy (zarkon):

isn't it \[\large L= \int\limits_{a}^{b}|r\prime(t)|dt\]

OpenStudy (anonymous):

it is

OpenStudy (anonymous):

because it is the magnitude of r(t) where the derivative is squared and square rooted

OpenStudy (anonymous):

i don't know how about going next

OpenStudy (zarkon):

find the derivative of your 3 functions

OpenStudy (anonymous):

i get -sin(t)+cos(t)+sqrt(2t)

OpenStudy (anonymous):

but i dont think its right

OpenStudy (zarkon):

you get that for what exactly

OpenStudy (zarkon):

as is typical for these problems, the functions above are chosen just right so that the integration is trivial. if you end up with a difficult integral then you did something wrong

OpenStudy (anonymous):

\[\sqrt{(-sint)^2+(cost)^2+(\sqrt{2t})^2}\] this was my derivative

OpenStudy (zarkon):

did you use the product rule on the first two functions?

OpenStudy (anonymous):

no

OpenStudy (zarkon):

maybe you should :)

OpenStudy (anonymous):

haha yeah

OpenStudy (anonymous):

what about the third one

OpenStudy (zarkon):

3rd is fine

OpenStudy (anonymous):

cos(t)-tsin(t)+sin(t)+tcos(t)

OpenStudy (zarkon):

what is that

OpenStudy (anonymous):

thatas the first 2

OpenStudy (zarkon):

remember ..you take the derivative of each of the 3 functions...square each of them...then add...then take square root

OpenStudy (anonymous):

i dont know messing up right now

OpenStudy (zarkon):

your two derivatives are fine...you just don't add them like you did

OpenStudy (zarkon):

I need to go to my office...I'll check back in later on your progress.

OpenStudy (phi):

remember that those i,j,k mean you have a vector with three components your derivative is (cos(t)-tsin(t)) *i + (sin(t)+tcos(t))*j + sqrt(2t)* k or written as three components of a vector v= [ (cos(t)-tsin(t)) (sin(t)+tcos(t)) sqrt(2t)] the magnitude squared is v dot v or v^T v (v transpose times v)

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