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Functiosn with multiple variables

 
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Oct15-07, 07:46 AM   #1
 

Functiosn with multiple variables


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.
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Oct15-07, 09:16 AM   #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?
Oct15-07, 10:09 AM   #3
D H
 
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Quote by Niles View Post
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]."?

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].
Oct15-07, 10:35 AM   #4
 

Functiosn with multiple variables


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 Thumbnails
level.GIF  
Oct15-07, 01:01 PM   #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.
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