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anonymous

  • 5 years ago

find ∫G(x,y)ds ??? , on the indicated curve G(X,Y)=2XY, C: X=5*COST, Y=5*SINT, 0≤t≤π÷4

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  1. Akshay_Budhkar
    • 5 years ago
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    ok i will give u a hint, u should solve it yourself :D

  2. anonymous
    • 5 years ago
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    ok no problem

  3. Akshay_Budhkar
    • 5 years ago
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    \[2 \sin(x) \cos(x)=\sin(2x)\]

  4. Akshay_Budhkar
    • 5 years ago
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    so your G(x,y) =

  5. Akshay_Budhkar
    • 5 years ago
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    \[2 \times 5 \times \cos(x) \times 5 \times \sin(x)\]

  6. Akshay_Budhkar
    • 5 years ago
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    = \[25 \sin (2x)\]

  7. Akshay_Budhkar
    • 5 years ago
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    you can intergerate that :D

  8. anonymous
    • 5 years ago
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    i know how to find \[\int\limits_{C}^{} G(x,y) dx \] and \[\int\limits_{C}^{} G(x,y) dy \] what i want is \[\int\limits_{C}^{} G(x,y) ds \]

  9. Akshay_Budhkar
    • 5 years ago
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    one min i need to get a better help 1 second

  10. anonymous
    • 5 years ago
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    i am thinking that i should add \[\int\limits_{C}^{} G(x,y) dx + \int\limits_{C}^{} G(x,y) dy \] but i am not sure

  11. Akshay_Budhkar
    • 5 years ago
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    @turing ds???

  12. anonymous
    • 5 years ago
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    what do you think ?

  13. TuringTest
    • 5 years ago
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    I'm sorry I'm multitasking...

  14. anonymous
    • 5 years ago
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    this is how the question written exactly ... what does he mean by ds?

  15. Akshay_Budhkar
    • 5 years ago
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    i don't know that is why i called turing

  16. Akshay_Budhkar
    • 5 years ago
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    ds? are u sure that is what is asking for?

  17. anonymous
    • 5 years ago
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    yes i am sure

  18. anonymous
    • 5 years ago
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    the answer will be 125/2 .. but i do not know how to get it

  19. Akshay_Budhkar
    • 5 years ago
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    why do you think you need to add?

  20. Akshay_Budhkar
    • 5 years ago
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    ds generally in my questions is the area

  21. anonymous
    • 5 years ago
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    really i am not sure i tried everything but i could not reach the answer

  22. Akshay_Budhkar
    • 5 years ago
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    i am waiting for turing to respond lol.. he will surely help.. i m thinking of possibilities till then

  23. anonymous
    • 5 years ago
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    ok I appreciate your help bro .. thanks.

  24. Akshay_Budhkar
    • 5 years ago
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    no no its not done! i wont sleep till u get ur answer :P

  25. anonymous
    • 5 years ago
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    Lol thanks so much.. me too cuz i have to submit this assignment tomorrow :D

  26. TuringTest
    • 5 years ago
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    ok ds=sqrt[(dx/dt)^2+(dy/dt)^2]=5 2xy=50cost*sint got it from there?

  27. anonymous
    • 5 years ago
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    Integrating with respect to ds means that you are integrating with respect to the arc length (which is often denoted s) of the curve C. Note that \[ds=\sqrt{dx^2+dy^2}=\sqrt{(-5\cos (t) dt)^2+(5\sin (t) dt)^2}=5dt\] Thus, you have that \[\int_C G(x,y)ds=\int_0^{\pi/4}25\sin(2t)\cdot 5dt = \int_0^{\pi/4}125\sin(2t)dt = \frac{125}{2}\]

  28. Akshay_Budhkar
    • 5 years ago
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    ds?

  29. Akshay_Budhkar
    • 5 years ago
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    arc length is what it signifies?

  30. TuringTest
    • 5 years ago
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    yes, you need ds in line integrals, arc length differential

  31. anonymous
    • 5 years ago
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    thanks guys I get it .. I appreciate your help .

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