## anonymous 3 years ago 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(<or=) t (<or=) 2

1. amistre64

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

2. amistre64

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

3. 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?

4. 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

5. amistre64

or am i mixing things up?

6. anonymous

ds = |v(t)| dt

7. amistre64

im mixing a vector feild in with this i believe

8. anonymous

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

9. anonymous

you're fine @amistre64

10. 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))?

11. amistre64

..... thnx ;)

12. amistre64

yep, but put the sqrt5 inside the integral

13. anonymous

yes correct, @razroz

14. anonymous

thank you so much

15. amistre64

good luck ;)