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razroz

  • 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

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  1. amistre64
    • 3 years ago
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    \[ds =\sqrt{(x')^2+(y')^2}\]

  2. amistre64
    • 3 years ago
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    well, in this case i spose fx and fy for partials

  3. razroz
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    or am i mixing things up?

  6. Algebraic!
    • 3 years ago
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    ds = |v(t)| dt

  7. amistre64
    • 3 years ago
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    im mixing a vector feild in with this i believe

  8. razroz
    • 3 years ago
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    no i believe you are right, let me write it out real quick

  9. Algebraic!
    • 3 years ago
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    you're fine @amistre64

  10. razroz
    • 3 years ago
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    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
    • 3 years ago
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    ..... thnx ;)

  12. amistre64
    • 3 years ago
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    yep, but put the sqrt5 inside the integral

  13. Algebraic!
    • 3 years ago
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    yes correct, @razroz

  14. razroz
    • 3 years ago
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    thank you so much

  15. amistre64
    • 3 years ago
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    good luck ;)

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