1. Not finding help here? Sign up for a free 30min 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!

Functiosn with multiple variables

  1. Oct 15, 2007 #1
    1. The problem statement, all variables and given/known data

    1) I have a C1-function f(u,v), and f(0,0) = 1, df/du(0,0) = 3 and df/dv(0,0)=5. I have to find k'(1), when k(x) = f(x^2-1;x-1).

    2) A function g(x,y) = 3x^2-x-y+y^2. I have to find the minimum and maximum of the function on D_1 = [0,1] x [0,1] and D_2 = [1,2] x [0,1].

    3. The attempt at a solution

    1) g(x,y) = x^2-1 and h(x,y) = x-1. dk/dx = (df/du)*(dg/dx) + (df/dv)*(dh/dx). I know g and h, but I have to find f - how do I do that?

    2) I know how to find maximum and minimum of a function, but I don't know what they mean by "on D_1 = [0,1] x [0,1] and D_2 = [1,2] x [0,1]."?

    EDIT: Sorry for the spelling-error in the title.
     
    Last edited: Oct 15, 2007
  2. jcsd
  3. Oct 15, 2007 #2
    I solved #1 - simply by using the chain rule:

    dk/dx = (df/du)(g(x),h(x))*(dg/dx) + (df/dv)(g(x),h(x))*(dh/dx)

    In my book, the arguments weren't included.. that confused me.

    For #2, I have found the critical points as usual, and the minimum is (1/6;1/2). But still, the D_1 and D_2 confuses me - what do they mean by that?
     
    Last edited: Oct 15, 2007
  4. Oct 15, 2007 #3

    D H

    User Avatar
    Staff Emeritus
    Science Advisor


    These refer to two domains. D1 is the set [itex]\{(x,y):x\in[0,1]\ \text{and}\ y\in[0,1]\}[/itex]. D2 is the set [itex]\{(x,y):x\in[1,2]\ \text{and}\ y\in[0,1]\}[/itex].
     
  5. Oct 15, 2007 #4
    Ok, thanks. I have another question, but this is about the gradient of a function.

    Take a look at the attached level curve of a function. At which point is the gradient 0?
    I believe it is C, since the gradient there is 0 - but apparently it is A - why is that?

    And btw, to find the critical points in the two sets, I can look at the function and the interval, and see what f(x,y) = 3x^2-x-y+y^2 has to be less than or equal and what is has to be bigger than or equal, and solve it from there?

    I get that the minimum is -1/3 - I get that there is no maximum, which I don't understand. I just find the gradient, and equal it to zero - I get one result, and by processing it, I get the minimum-point. How come there is no maximum?

    Thank you for all your help so far.
     

    Attached Files:

    Last edited: Oct 15, 2007
  6. Oct 15, 2007 #5
    Or perhaps I should "extend" the question and ask, what the link between the gradient and the level curve is.

    + the other question about the minimum/maximum.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Functiosn with multiple variables
Loading...