Question

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( 13 years ago

Answer 1

\[ds =\sqrt{(x')^2+(y')^2}\]

Answer 2

well, in this case i spose fx and fy for partials

Answer 3

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?

Answer 4

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

Answer 5

or am i mixing things up?

Answer 6

ds = |v(t)| dt

Answer 7

im mixing a vector feild in with this i believe

Answer 8

no i believe you are right, let me write it out real quick

Answer 9

you're fine @amistre64

Answer 10

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))?

Answer 11

..... thnx ;)

Answer 12

yep, but put the sqrt5 inside the integral

Answer 13

yes correct, @razroz

Answer 14

thank you so much

Answer 15

good luck ;)