anonymous
  • anonymous
Evaluate the line integral \[\int_c F\bullet dr\] where C is given by the vector function r(t). \[F(x,y,z)=(x+y)\hat{i}+(y-z)\hat{j}+z^2\hat{k}\] \[r(t)=t^2\hat{i}+t^3\hat{j}+t^2\hat{k}\] \[0 \le t \le 1\] \[r'(t)=2t\hat{i}+3t^2\hat{j}+2t\hat{k}\] \[\int_c F \bullet dr=\int_a^b F(r(t)) \bullet r'(t)dt\] \[\int_0^1 (t^2\hat{i}+(t^3-t^2)\hat{j} + t^4\hat{k})\bullet (2t\hat{i} +3t^2\hat{j}+2t\hat{k})dt\] how am I doing?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Oops wrong question, it's \[F(x,y,z)=z\hat{i}+y\hat{j}-z\hat{k}\] \[r(t)=t\hat{i}+sint\hat{j}+cost\hat{k}\] \[0 \le t \le pi\]
anonymous
  • anonymous
for your i, x+y
anonymous
  • anonymous
alright guess we'll start over hahha continues

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anonymous
  • anonymous
would that make \[r'(t)=\hat{i}+cost\hat{j}-sint\hat{k}\] and
anonymous
  • anonymous
\[\int_0^{\pi} (t\hat{i}+sint\hat{j}+cost\hat{k})\bullet (\hat{i}+cost\hat{j}-sint\hat{k})dt\]
KingGeorge
  • KingGeorge
So far that looks right to me.
anonymous
  • anonymous
thanks
KingGeorge
  • KingGeorge
You're welcome. Once you do the dot product, it looks like it will be very easy to evaluate.
KingGeorge
  • KingGeorge
So can you show me what you think the dot product will be?
anonymous
  • anonymous
I can't just multiply the i j and k coefficients can I? \[t+sintcost-sintcost\]
KingGeorge
  • KingGeorge
That's exactly what it is. You should end up with \[t+\sin(t)\cos(t)-\sin(t)\cos(t)=t\]
anonymous
  • anonymous
Yaaaaayyyyy!!!!! I just need to build confidence i guess.... sigh
KingGeorge
  • KingGeorge
You'll get used to it.
anonymous
  • anonymous
here is the mistake I see: x=t y=sint z=cost that would give us \[\int_0^{\pi} cost dt+\int_0^{\pi}sintcostdt+\int_0^{\pi}tsint dt\]
TuringTest
  • TuringTest
\[\vec F(x,y,z)=z\hat{i}+y\hat{j}-z\hat{k}\]\[\vec r(t)=t\hat i+\sin t\hat j+\cos t\hat k\]\[\vec r'(t)=\hat i+\cos t\hat j-\sin t\hat k\]\[\int_0^\pi\vec F(\vec r(t))\cdot\vec r'(t)dt=\int_0^\pi\langle\cos t,\sin t,\cos t\rangle\cdot\langle1,\cos t,-\sin t\rangle dt\]do you see a problem with that?
TuringTest
  • TuringTest
\[=\int_0^\pi\cos t+\sin t\cos t-\sin t\cos tdt=\int_0^\pi\cos tdt=-2\]
anonymous
  • anonymous
oh I see what I did! It's z y -z and not z y -x
anonymous
  • anonymous
thanks!
TuringTest
  • TuringTest
welcome!
TuringTest
  • TuringTest
now I go to bed :)
anonymous
  • anonymous
Have a good night's rest!
TuringTest
  • TuringTest
likewise!

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