# Partial derivatives.

#### Kruum

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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?

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#### HallsofIvy

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, $$f_s= f_x x+_s+ f_y y_s= 2f_x+ 3f_y$$ .
That applies to any function of x,y so it applies to $$f_s$$ itself.

$$f_{ss}= (f_s)_s= 2(2f_x+ 3f_y)_x+ 3(2f_x+ 3f_y)_y= 4f_{xx}+ 6f_{yx}+ 6f_{xy}+ 9f_yy$$
and since, for this function, with continuous second derivatives, the mixed derivatives are the same,
$$f_ss= 4f_{xx}+ 12f_{xy}+ 9f_{yy}$$
exactly as you say.

#### Kruum

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, $$f_s= f_x x+_s+ f_y y_s= 2f_x+ 3f_y$$ .
That applies to any function of x,y so it applies to $$f_s$$ itself.
Thank you, I never thought of that and it will make things much easier.

#### HallsofIvy

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 $\vec{i}$ is the dot product $\nabla f\cdot \vec{v}$. 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.

#### Kruum

?? 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 $\vec{i}$ is the dot product $\nabla f\cdot \vec{v}$. 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 $\vec{i}+\vec{j}$ or $-\vec{i}-\vec{j}$.

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.

"Partial derivatives."

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