find the arc length of a curve
\[r(t)=tcosti+tsintj+\frac{ 2\sqrt{2} }{ 3}t ^{3/2}k\]
r(t) such that t>0 and t<pi
\[\large L=\int\limits_{a}^{b}\sqrt{r\prime(t)}dt\] just compute the derivative and sub un the integral
correction: \[\large L= \int\limits_{a}^{b}\sqrt{\left| r\prime(t) \right|}dt\] compute the derivative and sub in the integral
-sinti+costj
how do i do the other derivation
isn't it \[\large L= \int\limits_{a}^{b}|r\prime(t)|dt\]
it is
because it is the magnitude of r(t) where the derivative is squared and square rooted
i don't know how about going next
find the derivative of your 3 functions
i get -sin(t)+cos(t)+sqrt(2t)
but i dont think its right
you get that for what exactly
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
\[\sqrt{(-sint)^2+(cost)^2+(\sqrt{2t})^2}\] this was my derivative
did you use the product rule on the first two functions?
no
maybe you should :)
haha yeah
what about the third one
3rd is fine
cos(t)-tsin(t)+sin(t)+tcos(t)
what is that
thatas the first 2
remember ..you take the derivative of each of the 3 functions...square each of them...then add...then take square root
i dont know messing up right now
your two derivatives are fine...you just don't add them like you did
I need to go to my office...I'll check back in later on your progress.
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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