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MathSofiya
Group Title
Evaluate the line integral \[\int_c F\bullet dr\]
where C is given by the vector function r(t).
\[F(x,y,z)=(x+y)\hat{i}+(yz)\hat{j}+z^2\hat{k}\]
\[r(t)=t^2\hat{i}+t^3\hat{j}+t^2\hat{k}\]
\[0 \le t \le 1\]
\[r'(t)=2t\hat{i}+3t^2\hat{j}+2t\hat{k}\]
\[\int_c F \bullet dr=\int_a^b F(r(t)) \bullet r'(t)dt\]
\[\int_0^1 (t^2\hat{i}+(t^3t^2)\hat{j} + t^4\hat{k})\bullet (2t\hat{i} +3t^2\hat{j}+2t\hat{k})dt\]
how am I doing?
 2 years ago
 2 years ago
MathSofiya Group Title
Evaluate the line integral \[\int_c F\bullet dr\] where C is given by the vector function r(t). \[F(x,y,z)=(x+y)\hat{i}+(yz)\hat{j}+z^2\hat{k}\] \[r(t)=t^2\hat{i}+t^3\hat{j}+t^2\hat{k}\] \[0 \le t \le 1\] \[r'(t)=2t\hat{i}+3t^2\hat{j}+2t\hat{k}\] \[\int_c F \bullet dr=\int_a^b F(r(t)) \bullet r'(t)dt\] \[\int_0^1 (t^2\hat{i}+(t^3t^2)\hat{j} + t^4\hat{k})\bullet (2t\hat{i} +3t^2\hat{j}+2t\hat{k})dt\] how am I doing?
 2 years ago
 2 years ago

This Question is Closed

MathSofiya Group TitleBest ResponseYou've already chosen the best response.3
Oops wrong question, it's \[F(x,y,z)=z\hat{i}+y\hat{j}z\hat{k}\] \[r(t)=t\hat{i}+sint\hat{j}+cost\hat{k}\] \[0 \le t \le pi\]
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
for your i, x+y
 2 years ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
alright guess we'll start over hahha continues
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.3
would that make \[r'(t)=\hat{i}+cost\hat{j}sint\hat{k}\] and
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.3
\[\int_0^{\pi} (t\hat{i}+sint\hat{j}+cost\hat{k})\bullet (\hat{i}+cost\hat{j}sint\hat{k})dt\]
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
So far that looks right to me.
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
You're welcome. Once you do the dot product, it looks like it will be very easy to evaluate.
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
So can you show me what you think the dot product will be?
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.3
I can't just multiply the i j and k coefficients can I? \[t+sintcostsintcost\]
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
That's exactly what it is. You should end up with \[t+\sin(t)\cos(t)\sin(t)\cos(t)=t\]
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.3
Yaaaaayyyyy!!!!! I just need to build confidence i guess.... sigh
 2 years ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
You'll get used to it.
 2 years ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.3
here is the mistake I see: x=t y=sint z=cost that would give us \[\int_0^{\pi} cost dt+\int_0^{\pi}sintcostdt+\int_0^{\pi}tsint dt\]
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
\[\vec F(x,y,z)=z\hat{i}+y\hat{j}z\hat{k}\]\[\vec r(t)=t\hat i+\sin t\hat j+\cos t\hat k\]\[\vec r'(t)=\hat i+\cos t\hat j\sin t\hat k\]\[\int_0^\pi\vec F(\vec r(t))\cdot\vec r'(t)dt=\int_0^\pi\langle\cos t,\sin t,\cos t\rangle\cdot\langle1,\cos t,\sin t\rangle dt\]do you see a problem with that?
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
\[=\int_0^\pi\cos t+\sin t\cos t\sin t\cos tdt=\int_0^\pi\cos tdt=2\]
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.3
oh I see what I did! It's z y z and not z y x
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.3
thanks!
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
welcome!
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
now I go to bed :)
 one year ago

MathSofiya Group TitleBest ResponseYou've already chosen the best response.3
Have a good night's rest!
 one year ago

TuringTest Group TitleBest ResponseYou've already chosen the best response.0
likewise!
 one year ago
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