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anonymous
 3 years ago
Evaluate the line integral \[\int_c F\cdot dr\] where c is the vector function
\[\hat{r}(t)=t^3\hat it^2 \hat j+ t \hat k\]
and
\[F(x,y,z)=sinx\hat i + cosy \hat j+zx \hat k\]
Here is how far I have gotten so far:
\[\hat{r}'(t)=3t^2\hat i2t \hat j+ \hat k\]
anonymous
 3 years ago
Evaluate the line integral \[\int_c F\cdot dr\] where c is the vector function \[\hat{r}(t)=t^3\hat it^2 \hat j+ t \hat k\] and \[F(x,y,z)=sinx\hat i + cosy \hat j+zx \hat k\] Here is how far I have gotten so far: \[\hat{r}'(t)=3t^2\hat i2t \hat j+ \hat k\]

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you might want to use \cdot rather than \bullet hehe

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'll change the Function F in terms of t's in just a second

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0thanks @lgbasallote. Better? :P

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh 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
 3 years ago
Best ResponseYou've already chosen the best response.0Hey i think I got it! \[x=t^3;y=t^2;z=t\] YES?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That would give me: \[F(r(t))=sin(t^3)\hat i +cos(t^2)\hat j+t^4 \hat k \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int_{t=0}^{t=1}\left(sin(t^3)\hat i +cos(t^2)\hat j+t^4 \hat k\right)\cdot\left(3t^2\hat i 2t\hat j +\hat k \right) dt\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and now I would just do the dot product correct?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int_0^1(3t^2sin(t^3)2tcos(t^2)+t^4)dt\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait, I have to take the integral of several products?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0separate them\[\int_{0}^{1} 3t^2 \sin t^3 dt+...+...\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh ok \[\int_0^1 3t^2sin(t^3)dt\int_0^12tcos(t^2)dt+\int_0^1t^4dt\] like this?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh integration by parts! duh =D

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1no integration by parts you can do it all with usubs

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0u=t^2 du=2t dt for the second integral

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0u=t^3 and du=3t^2 dt for the first integral...LOL that took me a while!

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1yeah, it's well setup for the usub thing :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[cos(1)+sin(1)+2\] My algebra is probably wrong...but Yeah @mukushla that was one long recipe!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\large \cos t^3]_{0}^{1}+\sin t^2]_{0}^{1}+t^5/5]_{0}^{1}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[(cos(1)+1)+(sin(1)+1)+\frac 1 5\] \[cos(1)sin(1)+\frac{11}{5}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oops sine of 0 is zero

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1you have an extra +1 in there  yep

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sigh...finally! Thanks guys!
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