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Homework Help: Adv Multi V Calc questions

  1. Apr 27, 2010 #1
    1. The problem statement, all variables and given/known data

    .pdf file of assignment is here: http://www.math.washington.edu/~sullivan/3264_sp10.pdf

    So, I have a few questions about the assignment.

    Problem #1 I need to know how to find the 14 points.

    Problem #2 When it tells me to computer ∂f/∂x amd ∂f/dy at (0,0), does that mean I just I take the limit as (x,y)-->(0,0), since obviously the two are undefined at (0,0)?

    Problem #3 The two conditions for differentiability at a point are that f(x,y) is continuous at that point and that lim (h,k)-->(0,0) |f(x+h,y+k)-f(x,y)-(Ah+Bh)| / √(x2+y2) = 0, where A= [f(x+h,y)-f(x,y)]/h, B=[f(x,y+k)-f(x,y)]/k, according to my textbook. Right? How am I supposed to show that such approaches 0 is h,k approach zero?

    2. Relevant equations

    They're scattered around the assignment

    3. The attempt at a solution

    The only one worth showing my work is Problem 1. I used the theorem on the front.

    grad(F)=<1, 3y2, z3>
    grad(G)=<x2, y2, z2>

    grad(F), then, is never the zero vector.
    grad(G) is at the point (0,0,0).

    If I make grad(F) a scalar multiple of grad(G), my system of equations looks like


    ------> 1/ß + y2 + z3/ß = 1 -----> ß2 + 3y2 + ßz2=1.

    Now, how do I solve for the 13 other points with that equation????

    Thanks in advance.
  2. jcsd
  3. Apr 27, 2010 #2


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    so the way i read the question, the function & constraint are:
    [tex] F(x,y,z) = x + y^3 + z^3[/tex]
    [tex] G(x,y,z) = x^1 + y^2 + z^2 = 1[/tex]

    these gradients don't look right
  4. Apr 27, 2010 #3


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    [tex] \nabla F = <\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}> [/tex]
  5. Apr 27, 2010 #4
    I have no idea what I was thinking!

    Let me redo that part.



    Now what?
  6. Apr 27, 2010 #5


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    ok so you have 4 equations, try & solve them...
  7. Apr 27, 2010 #6

    To simplify,



    (2ß)2 + (2ß/3)2 + (2ß/3)2 = 1.

    ----> 4ß2 + 4ß2/9 + 4ß2/9 = 1

    ----> 36ß2 + 8ß2 = 1

    ----> 44ß2 = 1

    ----> ß= 1/√44

    How am I supposed to get 13 points? That would only give me one.
  8. Apr 27, 2010 #7


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    shouldnt this be
    [tex] \beta = \frac{1}{2x} [/tex]
    to get these you have divided through by y & z respectively, which is ok if tehy are noth non-negative, but what about the cases when one or both are zero?

    if last step was correct, which i don't think it is based on the first comment, when you take a squareroot you get a positive & negative solution
  9. Apr 28, 2010 #8
    Attempt 3:


    into the 4th equation gives me

    1/(4ß2) + 4ß2/9 + 4ß2/9=1.

    Multiply everything by 36ß2 and I get

    9 + 16ß4 + 16ß4 = 36ß2.

    ----> I don't know how to solve that, but I assume there would be 4 solutions, (By the way, how would I solve that?) which would then given me 4 different points (x,y,z).

    I wouldn't matter is y or z were negative, because they'd just become positive when I plug them into the equation of a sphere, right?

    As for working with zeros, that would give me 6 more solutions......

    Am I getting closer?

    Explain, please.
  10. Apr 28, 2010 #9


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    solve for another variable [itex] \alpha = \beta^2 [/itex] then solve for beta. also not that beta cannot be zero
    right, so if it doesn't matter if y or z are negtive, they are both separate solutions to the problem
    maybe, you'll have to try and see, try the cases:
    [tex] y = 0, z \neq 0 [/tex]
    [tex] y \neq 0, z = 0 [/tex]
    [tex] y = 0, z = 0 [/tex]

    14 solutions is a lot, so you'll have to try & be systematic in working through
  11. Apr 28, 2010 #10
    So, I got ß= +/- ( (9 +/- √153) / 16). That's four different values which can be plugged in for four points (x,y,z).

    If I set y=0, z=z, that's the same as z=0, y=y, ..... Okay, I get it.
  12. Apr 28, 2010 #11


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    so i think you'll get 4 more for y=0, 4 more for z=0, then 2 for y=z=0
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