anonymous
  • anonymous
Evaluate the line integral \[\int_c F\cdot dr\] where c is the vector function \[\hat{r}(t)=t^3\hat i-t^2 \hat j+ t \hat k\] and \[F(x,y,z)=sinx\hat i + cosy \hat j+zx \hat k\] Here is how far I have gotten so far: \[\hat{r}'(t)=3t^2\hat i-2t \hat j+ \hat k\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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lgbasallote
  • lgbasallote
you might want to use \cdot rather than \bullet hehe
anonymous
  • anonymous
I'll change the Function F in terms of t's in just a second
anonymous
  • anonymous
thanks @lgbasallote. Better? :P

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lgbasallote
  • lgbasallote
yes. yes it is.
anonymous
  • anonymous
oh my, I did I similar problem 28 days ago http://openstudy.com/study#/updates/50060b97e4b06241806745b8 and I'm also looking at http://tutorial.math.lamar.edu/Classes/CalcIII/LineIntegralsVectorFields.aspx but I can't seem to remember... What do i substitute in for x and y, do I pick which one I want t to be?
anonymous
  • anonymous
Hey i think I got it! \[x=t^3;y=-t^2;z=t\] YES?
anonymous
  • anonymous
That would give me: \[F(r(t))=sin(t^3)\hat i +cos(-t^2)\hat j+t^4 \hat k \]
anonymous
  • anonymous
\[\int_{t=0}^{t=1}\left(sin(t^3)\hat i +cos(-t^2)\hat j+t^4 \hat k\right)\cdot\left(3t^2\hat i -2t\hat j +\hat k \right) dt\]
anonymous
  • anonymous
and now I would just do the dot product correct?
anonymous
  • anonymous
exactly
anonymous
  • anonymous
\[\int_0^1(3t^2sin(t^3)-2tcos(-t^2)+t^4)dt\]
anonymous
  • anonymous
well done
anonymous
  • anonymous
wait, I have to take the integral of several products?
anonymous
  • anonymous
separate them\[\int_{0}^{1} 3t^2 \sin t^3 dt+...+...\]
anonymous
  • anonymous
oh ok \[\int_0^1 3t^2sin(t^3)dt-\int_0^12tcos(-t^2)dt+\int_0^1t^4dt\] like this?
anonymous
  • anonymous
yep
anonymous
  • anonymous
oh integration by parts! duh =D
anonymous
  • anonymous
\[-\cos t^3\] :-)
TuringTest
  • TuringTest
no integration by parts you can do it all with u-subs
TuringTest
  • TuringTest
do you get that?
anonymous
  • anonymous
Oh I see it now!!!!
TuringTest
  • TuringTest
cool :)
anonymous
  • anonymous
u=t^2 du=2t dt for the second integral
TuringTest
  • TuringTest
yep
anonymous
  • anonymous
u=t^3 and du=3t^2 dt for the first integral...LOL that took me a while!
TuringTest
  • TuringTest
yeah, it's well set-up for the u-sub thing :)
anonymous
  • anonymous
u cooked the problem
anonymous
  • anonymous
\[-cos(1)+sin(1)+2\] My algebra is probably wrong...but Yeah @mukushla that was one long recipe!
anonymous
  • anonymous
\[\large -\cos t^3]_{0}^{1}+\sin -t^2]_{0}^{1}+t^5/5]_{0}^{1}\]
anonymous
  • anonymous
\[(-cos(1)+1)+(-sin(1)+1)+\frac 1 5\] \[-cos(1)-sin(1)+\frac{11}{5}\]
anonymous
  • anonymous
oops sine of 0 is zero
TuringTest
  • TuringTest
you have an extra +1 in there - -yep
anonymous
  • anonymous
-cos(1)-sin(1)+ 6/5
TuringTest
  • TuringTest
looks good to me :)
anonymous
  • anonymous
sigh...finally! Thanks guys!
TuringTest
  • TuringTest
welcome !
anonymous
  • anonymous
:)

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