Partial derivatives.

  • Thread starter Kruum
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1. The problem statement, all variables and given/known data

First problem: Let [tex]f(x,y) = x-y[/tex] and u = vi+wj. In which direction does the function decrease and increase the most? And what u (all of them) satisfies Duf = 0

Second problem: Let [tex]z = f(x,y)[/tex], where [tex]x = 2s+3t[/tex] and [tex]y = 3s-2t[/tex]. Determine [tex]\partial{z^2}/\partial{s^2}[/tex]

2. Relevant equations

Gradient and the chain rule

3. The attempt at a solution

For the first question in the first problem I've gotten using gradient: increase i-j and decrease -i+j. Am I correct? For the second question all I've gotten so far is (nabla)f(dot)u = 0 = (1-y)v+(x-1)w. Where do I get the second equation to solve both v and w?

Second problem gives me 4zxx+12zxy+9zyy. Is that completely wrong?
 
Last edited:

HallsofIvy

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What you have done so far is correct. For the second part, here's any easy hint: in 2 dimensions, a perpendicular to vector <a, b> is <-b, a>.


For the second problem, [tex]f_s= f_x x+_s+ f_y y_s= 2f_x+ 3f_y[/tex] .
That applies to any function of x,y so it applies to [tex]f_s[/tex] itself.

[tex]f_{ss}= (f_s)_s= 2(2f_x+ 3f_y)_x+ 3(2f_x+ 3f_y)_y= 4f_{xx}+ 6f_{yx}+ 6f_{xy}+ 9f_yy[/tex]
and since, for this function, with continuous second derivatives, the mixed derivatives are the same,
[tex]f_ss= 4f_{xx}+ 12f_{xy}+ 9f_{yy}[/tex]
exactly as you say.
 
220
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What you have done so far is correct. For the second part, here's any easy hint: in 2 dimensions, a perpendicular to vector <a, b> is <-b, a>.
So I just mark (1-y)w-(x-1)v = 0?

For the second problem, [tex]f_s= f_x x+_s+ f_y y_s= 2f_x+ 3f_y[/tex] .
That applies to any function of x,y so it applies to [tex]f_s[/tex] itself.
Thank you, I never thought of that and it will make things much easier.
 

HallsofIvy

Science Advisor
41,621
821
So I just mark (1-y)w-(x-1)v = 0?
?? You had already found that the gradient was i- j and you appeared to know that the directional derivative, in the direction of unit vector [itex]\vec{i}[/itex] is the dot product [itex]\nabla f\cdot \vec{v}[/itex]. That will be 0 if and only if the two vectors are orthogonal: the direction in which the derivative is 0 is perpendicular to the gradient vector, i+ j.

Thank you, I never thought of that and it will make things much easier.
 
220
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?? You had already found that the gradient was i- j and you appeared to know that the directional derivative, in the direction of unit vector [itex]\vec{i}[/itex] is the dot product [itex]\nabla f\cdot \vec{v}[/itex]. That will be 0 if and only if the two vectors are orthogonal: the direction in which the derivative is 0 is perpendicular to the gradient vector, i+ j.
Thank you. This is a new subject for me and my thinking was too complicated. I found that the vector is either [itex]\vec{i}+\vec{j}[/itex] or [itex]-\vec{i}-\vec{j}[/itex].

And I just realized that this function is so simple that I could have found the vectors by giving the function a set value and draw it in 2-dimensions. I know there is a fancy name for that, but I've no idea what it is in English.
 

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