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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).
 11 months ago
 11 months ago
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).
 11 months ago
 11 months ago

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theEricBest ResponseYou've already chosen the best response.0
dw:1367800601926:dw
 11 months ago

theEricBest ResponseYou've already chosen the best response.0
dw:1367800740264:dw
 11 months ago

theEricBest 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\[\frac{dy(t)}{dt}=cos(t)\]
 11 months ago

theEricBest 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?
 11 months ago

ZarkonBest ResponseYou've already chosen the best response.1
those two integrals are the same
 11 months ago

ZarkonBest ResponseYou've already chosen the best response.1
you might want to look at your parametric equations again
 11 months ago

theEricBest 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?
 11 months ago

ZarkonBest ResponseYou've already chosen the best response.1
"circle with radius 3"
 11 months ago

theEricBest 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...
 11 months ago

theEricBest 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\[x(t)=3cos(t)\]\[y(t)=3sin(t)\]\[\text{where }t\text{ goes from }0 \text{ to } \pi \text{.}\]
 11 months ago

theEricBest ResponseYou've already chosen the best response.0
The first one is from \[\pi\] to 0, I mean.
 11 months ago

theEricBest ResponseYou've already chosen the best response.0
And so they're the same?
 11 months ago

ZarkonBest ResponseYou've already chosen the best response.1
What is your force field \(F\)?
 11 months ago

ZarkonBest ResponseYou've already chosen the best response.1
\[W=\int\limits_{a}^{b}F(c(t))\cdot c'(t) dt\]
 11 months ago

ZarkonBest ResponseYou've already chosen the best response.1
\(c(t)\) is the particular path
 11 months ago

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

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

theEricBest 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!
 11 months ago
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