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anonymous
 4 years ago
Evaluate the line integral f(x,y)ds along the parametric curve: f(x,y)=3x+4y^2;x=2t1, y=t+5, 1(<or=) t (<or=) 2
anonymous
 4 years ago
Evaluate the line integral f(x,y)ds along the parametric curve: f(x,y)=3x+4y^2;x=2t1, y=t+5, 1(<or=) t (<or=) 2

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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1\[ds =\sqrt{(x')^2+(y')^2}\]

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1well, in this case i spose fx and fy for partials

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so i take the partial of both x and y with respect to t and plus that under a square root next to the function f(x,y) and integrate?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1yes, but dont forget to replace x and y with their t equations within f f(x,y)=3x+4y^2;x=2t1, y=t+5 f(t)=3(2t1)+4(t+5)^2

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1or am i mixing things up?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1im mixing a vector feild in with this i believe

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no i believe you are right, let me write it out real quick

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you're fine @amistre64

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so i would just get a sqrt(5) from the ds, then the function would read integral of (3(2t1)+4(t+5)^2)dt (times the sqrt(5))?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.1yep, but put the sqrt5 inside the integral
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