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theEric
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Calculus 3  Find work done along curve in force field. How do I set up the integral? The curve is the upper half of a circle with radius 3, centered at the origin, from (3,0) to (3,0).
 one year ago
 one year ago
theEric Group Title
Calculus 3  Find work done along curve in force field. How do I set up the integral? The curve is the upper half of a circle with radius 3, centered at the origin, from (3,0) to (3,0).
 one year ago
 one year ago

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theEric Group TitleBest ResponseYou've already chosen the best response.0
dw:1367800601926:dw
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
dw:1367800740264:dw
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
So I think it works nicely if I chose the parametric equations for the curve as\[x(t)=cos(t)\]\[y(t)=sin(t)\]and so\[\frac{dx(t)}{dt}=sin(t)\]and\[\frac{dy(t)}{dt}=cos(t)\]
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
\[\text {So then the parameter, }t\text{, goes from }\pi \text{ to }0 \text{.}\] \[\text{Is it okay to write:} \int_\pi ^0 F(t)dt\text{, or must it be }\int_0 ^ \pi F(t)dt \text{?}\] Don't they equal the same thing?
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
those two integrals are the same
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
you might want to look at your parametric equations again
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
\[\text{Thanks.. I've considered rewriting them to go from }0\text{ to }\pi \text{. In which case:}\] \[x(t)=cos(t)\]\[y(t)=sin(t)\] Is this what you mean? Or were my original equations incorrect?
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
neither are correct
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
"circle with radius 3"
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Oh... Yes.. Thank you! I tend to not carry all of the math into the next step. I tend to carry over my immediate thoughts and stop there...
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
\[x(t)=3cos(t)\]\[y(t)=3sin(t)\]\[\text{where }t \text{ goes from }0\text{ to }\pi \text{.}\] or\[x(t)=3cos(t)\]\[y(t)=3sin(t)\]\[\text{where }t\text{ goes from }0 \text{ to } \pi \text{.}\]
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
The first one is from \[\pi\] to 0, I mean.
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Those work, then?
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
And so they're the same?
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
What is your force field \(F\)?
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
\[W=\int\limits_{a}^{b}F(c(t))\cdot c'(t) dt\]
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
\(c(t)\) is the particular path
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
To continue the problem I'm given,\[F(x,y)=(2y+x^2)i + (x^2  2x)j\]
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
yielding... \[W=\int _0 ^\pi ((6sin(t)+9cos(t))(3sint)+(9cos^2(t)6cos(t))(3cos(t)))dt\]
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Which is ugly.. I've simplified it a lot, and I think I've solved it, but I'm not sure. I'll just check with Wolfram Alpha. Thank you for clearing up the integration!
 one year ago

theEric Group TitleBest ResponseYou've already chosen the best response.0
Take care!
 one year ago
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