Partial derivative with respect to a function, rather than variable?

In summary, the person is struggling with partial derivatives in their math/physics masters program and is seeking help on a forum. They are trying to solve a question involving partial derivatives and are unsure of how to proceed. Another forum member explains the solution, which involves expressing z as an explicit function of the independent variables y and theta. The person realizes their mistake and thanks the forum member for their help.
  • #1
la6ki
53
0
Hi all. I've recently started working a lot on my background in math and physics, since this year I began a new masters program which is quite math/physics heavy and I don't have a formal background in either field. I will try to get active on this forum, since I've been luring for some time and I find it really useful.

Right now I'm going through partial derivatives and some of the questions in the book I'm working with are quite confusing and I'm stuck. Here is one few examples:

z=x[itex]^{2}[/itex]+2y[itex]^{2}[/itex]
x=rcos(θ)
y=rsin(θ)

Question: [itex]\frac{∂^{2}z}{∂y∂θ}[/itex]

Here's what I've done so far:

I took the partial derivative with respect to θ = 2r[itex]^{2}[/itex]cosθsinθ

Now, from here I'm supposed to take the partial of the above with respect to y. I could either express it in terms of x and y:

[itex]\frac{∂2xy}{∂y}[/itex]

or in terms of r and θ:

[itex]\frac{∂2r^{2}sinθcosθ}{∂rsinθ}[/itex]

The thing is that I have no idea how to solve any of the above. The first I don't know how to solve because x and y aren't independent, so I can't just treat x as a constant. I got stuck when I attempted using the chain rule as well. The second I can't solve, since I don't know how to take a partial derivative with respect to a function, rather than a variable.

I would appreciate it if somebody explained how I can solve the problem using both methods (if both are possible, that is).
 
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  • #2
You have to express z as an explicit function of the two independent variables (y,theta).

From your second transformation rule, you have r=y/sin(theta).
Inserting this in your first, you get x=y*cot(theta), where cot(theta)=cos(theta)/sin(theta)

Thus, you have:
z(y,theta)=y^2(cot^2(theta)+2)

Now, proceed as usual.
----------------------------------------------------------------------------------------
Visually, the coordinate grid you are using are lines parallell to the x-axis (fixed values of y), and rays emanating from the origin (fixed values of theta)
 
  • #3
Oh... That was easy! :) I can't believe I was struggling so much with this when the obvious solution was in front of me the whole time.

Well, thanks a lot, you saved me a lot of stress!

Now, since I already attempted to solve it in a strange way, could you tell me if I made a mistake anywhere? Is it possible to continue the solution from any (or both) of the ways I went to? Or is it not doable analytically?
 
  • #4
"Now, since I already attempted to solve it in a strange way, could you tell me if I made a mistake anywhere? "
Yes.
When you made your partial with respect to the angular variable, you deliberately held "r" constant.
Thus, you actually "moved along" a circular arc (the only possible way to change the value of the angle with "r" constant). But, to move like that necessarily invokes a change in "y", which you are not allowed to!

HOW can you change the value of the angle if you are not allowed to change the "y"-value?
Answer: To follow a line parallell to the x-axis!
 
  • #5
Yeah, I think that makes sense. Thanks again!
 
  • #6
:smile:
 

1. What is a partial derivative with respect to a function?

A partial derivative with respect to a function is a mathematical concept used in multivariable calculus to calculate the rate of change of a function with respect to one of its input variables while holding all other variables constant.

2. How is a partial derivative with respect to a function different from a regular derivative?

A regular derivative calculates the rate of change of a function with respect to a single variable, while a partial derivative with respect to a function calculates the rate of change with respect to one variable while holding all other variables constant.

3. Can a partial derivative with respect to a function be negative?

Yes, a partial derivative with respect to a function can be negative. This indicates that the function is decreasing with respect to that particular variable.

4. How is a partial derivative with respect to a function denoted?

A partial derivative with respect to a function is denoted by using the symbol "∂" followed by the name of the function and the variable with respect to which the derivative is being taken. For example, "∂f(x,y)" would represent the partial derivative of the function f with respect to the variable x.

5. What is the practical application of partial derivatives with respect to a function?

Partial derivatives with respect to a function have many practical applications in fields such as physics, economics, and engineering. They are used to analyze the behavior of complex systems and to optimize functions with multiple input variables. For example, in economics, partial derivatives are used to analyze the impact of changing one variable on the overall output of a production function.

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