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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
 one year ago
 one year 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
 one year ago
 one year ago

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amistre64Best ResponseYou've already chosen the best response.1
\[ds =\sqrt{(x')^2+(y')^2}\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
well, in this case i spose fx and fy for partials
 one year ago

razrozBest ResponseYou've already chosen the best response.0
so 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?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
yes, 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
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
or am i mixing things up?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
im mixing a vector feild in with this i believe
 one year ago

razrozBest ResponseYou've already chosen the best response.0
no i believe you are right, let me write it out real quick
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
you're fine @amistre64
 one year ago

razrozBest ResponseYou've already chosen the best response.0
so 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))?
 one year ago

amistre64Best ResponseYou've already chosen the best response.1
yep, but put the sqrt5 inside the integral
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.1
yes correct, @razroz
 one year ago
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