anonymous
  • anonymous
Evaluate the line integral f(x,y)ds along the parametric curve: f(x,y)=3x+4y^2;x=2t-1, y=t+5, -1(
Mathematics
schrodinger
  • schrodinger
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amistre64
  • amistre64
\[ds =\sqrt{(x')^2+(y')^2}\]
amistre64
  • amistre64
well, in this case i spose fx and fy for partials
anonymous
  • anonymous
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?

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amistre64
  • amistre64
yes, but dont forget to replace x and y with their t equations within f f(x,y)=3x+4y^2;x=2t-1, y=t+5 f(t)=3(2t-1)+4(t+5)^2
amistre64
  • amistre64
or am i mixing things up?
anonymous
  • anonymous
ds = |v(t)| dt
amistre64
  • amistre64
im mixing a vector feild in with this i believe
anonymous
  • anonymous
no i believe you are right, let me write it out real quick
anonymous
  • anonymous
you're fine @amistre64
anonymous
  • anonymous
so i would just get a sqrt(5) from the ds, then the function would read integral of (3(2t-1)+4(t+5)^2)dt (times the sqrt(5))?
amistre64
  • amistre64
..... thnx ;)
amistre64
  • amistre64
yep, but put the sqrt5 inside the integral
anonymous
  • anonymous
yes correct, @razroz
anonymous
  • anonymous
thank you so much
amistre64
  • amistre64
good luck ;)

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