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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:
\[\hat{r}'(t)=3t^2\hat i2t \hat j+ \hat k\]
 one year ago
 one year 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: \[\hat{r}'(t)=3t^2\hat i2t \hat j+ \hat k\]
 one year ago
 one year ago

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lgbasalloteBest ResponseYou've already chosen the best response.0
you might want to use \cdot rather than \bullet hehe
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.2
I'll change the Function F in terms of t's in just a second
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.2
thanks @lgbasallote. Better? :P
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.2
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?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.2
Hey i think I got it! \[x=t^3;y=t^2;z=t\] YES?
 one year ago

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

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

MathSofiyaBest ResponseYou've already chosen the best response.2
and now I would just do the dot product correct?
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.2
\[\int_0^1(3t^2sin(t^3)2tcos(t^2)+t^4)dt\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.2
wait, I have to take the integral of several products?
 one year ago

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

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

MathSofiyaBest ResponseYou've already chosen the best response.2
oh integration by parts! duh =D
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
no integration by parts you can do it all with usubs
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.2
Oh I see it now!!!!
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.2
u=t^2 du=2t dt for the second integral
 one year ago

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

TuringTestBest ResponseYou've already chosen the best response.1
yeah, it's well setup for the usub thing :)
 one year ago

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

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

MathSofiyaBest ResponseYou've already chosen the best response.2
\[(cos(1)+1)+(sin(1)+1)+\frac 1 5\] \[cos(1)sin(1)+\frac{11}{5}\]
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.2
oops sine of 0 is zero
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
you have an extra +1 in there  yep
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.2
cos(1)sin(1)+ 6/5
 one year ago

TuringTestBest ResponseYou've already chosen the best response.1
looks good to me :)
 one year ago

MathSofiyaBest ResponseYou've already chosen the best response.2
sigh...finally! Thanks guys!
 one year ago
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