Partial derivatives with Gradient and the chain rule

In summary: 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.
  • #1
Kruum
220
0

Homework Statement



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]

Homework Equations



Gradient and the chain rule

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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  • #2
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.
 
  • #3
HallsofIvy said:
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.
 
  • #4
Kruum said:
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.
 
  • #5
HallsofIvy said:
?? 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.
 

Related to Partial derivatives with Gradient and the chain rule

1. What is a partial derivative?

A partial derivative is a mathematical concept that measures the rate of change of a function with respect to one of its variables, while holding all other variables constant.

2. What is the gradient of a function?

The gradient of a function is a vector that represents the direction and magnitude of the steepest slope of the function at a given point. It is calculated by taking the partial derivatives of the function with respect to each of its variables.

3. How is the chain rule used in partial derivatives?

The chain rule is used in partial derivatives to calculate the change in a function due to changes in one or more of its independent variables. It involves taking the partial derivatives of each variable and multiplying them together.

4. Why are partial derivatives important in science?

Partial derivatives are important in science because they allow us to analyze and understand how a system or function changes in response to changes in its variables. This is essential in fields such as physics, engineering, and economics.

5. Can the chain rule be applied to functions with more than two variables?

Yes, the chain rule can be applied to functions with any number of variables. The general formula involves taking the partial derivatives of the function with respect to each variable and multiplying them together.

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