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theEric
 one year 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).
theEric
 one year 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).

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theEric
 one year ago
Best ResponseYou've already chosen the best response.0So 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\[\frac{dy(t)}{dt}=cos(t)\]

theEric
 one year ago
Best 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?

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

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

theEric
 one year ago
Best 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?

theEric
 one year ago
Best ResponseYou've already chosen the best response.0Oh... 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...

theEric
 one year ago
Best 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\[x(t)=3cos(t)\]\[y(t)=3sin(t)\]\[\text{where }t\text{ goes from }0 \text{ to } \pi \text{.}\]

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

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

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

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

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

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

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

theEric
 one year ago
Best ResponseYou've already chosen the best response.0Which 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!
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