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2 Gradients

  1. May 15, 2009 #1
    1. The problem statement, all variables and given/known data

    grad f = (3x2 + y2)i + (2xy - 3)j

    a) find f

    b) evalue the integral of grad f dot dr where c is any path from 1,2 to 2,1

    c)evaluate the integral of gradf dot dr where c is the curve given by 4x2 + 9y2 = 36

    2. Relevant equations



    3. The attempt at a solution

    a) f = x3 + xy2 - 3j

    b) integral grad f dot dr = f(Q) - f(P) = (1 + 4 - 12) - (8 + 2 - 3) = -14

    c) integral grad f dot dr = integral F(r(t)) dot r'(t) is this the correct method

    r(t) = 4cost i + 9sint j
    r'(t) = -4sint i + 9cost j

    F(r(t)) = (3(4cost)2 + (9sint)2)i + (2(4cost)(9sint) - 3)j

    F(r(t)) dot r'(t) = (3(-4sint)(4cost)2 + (-4sint)(9sint)2) + (2(9cost)(4cost)(9sint) - 3(9cost)

    = (3(-4sint)(4cost)2 + (-4sint)(9sint)2) + (2(9cost)(4cost)(9sint) - 3(9cost)

    =[tex]\int[/tex] -192(cost)2sint + 324(sint)3 + 648(cost)2(sint)2 - 27cost dt from 0 to pi

    is this correct thus far
     
  2. jcsd
  3. May 15, 2009 #2

    gabbagabbahey

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    Be more mindful of typos; I'm sure you meant to write " f = x3 + xy2 - 3y", right?

    I see two problems here:

    (1) [itex] 3*2\neq 12[/itex]

    (2) You seem to be going from P=(2,1) to Q=(1,2), but the problem statement you posted suggests that the curve goes from P=(1,2) to Q=(2,1)

    Careful, [itex]\vec{r}'(t)=\frac{d\vec{r}}{dt}\neq d\vec{r}[/itex]. Intstead you should have written [itex]d\vec{r}=\vec{r}'(t)dt[/itex]

    This parameterization does not describe the correct curve. Using your parameterization, x(t)=4cos(t) and y(t)=9sin(t)....That would mean that [itex]4x^2+9y^2=64\cos^2 t+ 729\sin^2 t=64+665 sin^2 t\neq 36[/itex]
     
  4. May 17, 2009 #3
    b) f(Q) - f(P) = (8+2-3) - (1+4-6) = 5

    c) dont i have to keep them as vectors

    r(t) = 64cos2t i + 729sin2t j

    r'(t) = -64sin2t i - 729sin2t j
     
  5. May 17, 2009 #4

    gabbagabbahey

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    [itex](8+2-3)-(1+4-6)=(7)-(-1)=8\neq 5[/itex]

    I'm not exactly sure what you are trying to ask here, but you do understand that any 2D curve can be parameterized as [itex]\vec{r}(t)=x(t)\hat{i}+y(t)\hat{j}[/itex] right?

    So when you write [itex]\vec{r}(t)=4\cos(t)\hat{i}+9\sin(t)\hat{j}[/itex], you are essentially saying that at every point along your curve, [itex]x(t)=4\cos(t)[/itex] and [itex]y(t)=9\sin(t)[/itex]. This isn't true though, because you are told that [itex]4x^2+9y^2=36[/itex] at every point on your curve; and [itex]x(t)=4\cos(t)[/itex] and [itex]y(t)=9\sin(t)[/itex] don't satisfy that equation.

    You need to choose an [itex]x(t)[/itex] and [itex]y(t)[/itex] that do satisfy the equation [itex]4x(t)^2+9y(t)^2=36[/itex] for all values of t.
     
  6. May 17, 2009 #5
  7. May 17, 2009 #6

    gabbagabbahey

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  8. May 18, 2009 #7
    r(t) = 3cost i + 2sint j
    r'(t) = -3sint i + 2cost j

    f = x3 + xy2 - 3y

    f(r(t)) = (3cost)3 + (3cost)(2sint)2 - 3(2sint)

    I then have to dot it with r'(t), but my f(r(t)) is not in vector form, there are no i,j,z compnents, how do i proceed
     
  9. May 18, 2009 #8

    gabbagabbahey

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    I thought the question wanted you to compute Grad(f) dot dr....
     
  10. May 18, 2009 #9
    you right

    [tex]\int[/tex] (27cos2 + 4sin2)i + (12sincos - 3)j dot (-3sin i + 2cos j) dt =

    [tex]\int[/tex] -81cos2sin - 12sin3 + 24sincos2 - 6cos dt

    how do i integrate that
     
  11. May 18, 2009 #10

    gabbagabbahey

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    The easiest way is to write [itex]\sin^3(t)=\sin(t)(1-\cos^2(t))=\sin(t)-\sin(t)\cos^2(t)[/itex], then separate everything into three integrals of the form

    [tex]c_1\int \sin(t)\cos^2(t)dt+c_2\int \sin(t)dt+c_3\int\cos(t)dt[/tex]

    Which I'm sure you know how to integrate.

    By the way, are you supposed to calculate this over the entire ellipse, or just some section of it?
     
  12. May 18, 2009 #11
    evaluate grad f dot dr, where c is the curve given by 4x^2 + 9y^2 = 36
     
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