1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    [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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    so i think you'll get 4 more for y=0, 4 more for z=0, then 2 for y=z=0
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook