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theEric
 3 years 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
 3 years 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
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.1those two integrals are the same

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

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

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

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

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

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

theEric
 3 years 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
 3 years 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
 3 years 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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