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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 \[\int_0^{\pi} cost dt+\int_0^{\pi}sintcostdt+\int_0^{\pi}tsint dt\]
 2 years 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?
 2 years ago

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

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

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

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

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