Could somebody explain this problem and solution(about partial derivatives)

In summary, the conversation is discussing a problem involving a function with variables x, y, and z, and a solution provided for finding fxx. The question raised is whether there is a contradiction in the solution, as x is stated to be a function of r and \theta, but the solution seems to assume that r and \theta are functions of x. The expert explains that as long as a function is invertible, you can work with it either way, and in this case, the problem involves polar coordinates.
  • #1
oahsen
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Homework Statement



there is a question and a solution in this page;
http://www.fen.bilkent.edu.tr/~otekman/math102/s03/m2q5.html"

Please firstly examine the question and solution. (5b)...
There it says f(x,y)=z and x=g(r,teta) and y=h(r,teta) and asks fxx.
He solves this problem by starting so;
fx=zr*rx+zteta*tetax ...
My question is that by writing fx such as above aren't you assume that r and teta are a function of x? However in the question it says x=f(r,teta); this means x is a function of r and teta then how can you write fx=zr*rx+zteta*tetax? Isn't there a contradiction or am i understanding something false?:confused:
 
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  • #2
You are told that [itex]x= r cos(\theta)[/itex] so obviously x is a function of r and [itex]\theta[/itex] and therefore r and [itex]\theta[/itex] are functions of x. (It does not however anywhere say x= f(r,[/itex]\theta[/itex]).) As long as a function is invertible, you can work either way: If x= f(r), and r is invertible, then r= f-1(x) and dr/dx= 1/(dx/dr).\

Here, you really have "polar coordinates": [itex]x= r cos(\theta)[/itex] and [itex]y= r sin(\theta)[/itex] so that [itex]r= \sqrt{x^2+ y^2}[/itex] and [itex]\theta= arctan(\frac{y}{x})[/itex]. You can do it either way.
 

FAQ: Could somebody explain this problem and solution(about partial derivatives)

1. What are partial derivatives?

Partial derivatives are a type of derivative used in multivariable calculus to calculate the instantaneous rate of change of a function with respect to one of its variables, while holding all other variables constant. This is useful in solving optimization problems and understanding the behavior of a function in multiple dimensions.

2. How are partial derivatives calculated?

To calculate a partial derivative, you take the derivative of a multivariable function with respect to one of its variables, treating all other variables as constants. This means that you only differentiate the part of the function that contains the variable you are interested in, and leave the other parts unchanged. The resulting derivative is called a partial derivative.

3. What are partial derivatives used for?

Partial derivatives are used in many areas of mathematics and science, including economics, physics, engineering, and statistics. They are particularly useful in optimization problems, where the goal is to find the maximum or minimum value of a function with multiple variables. They also help us understand the behavior of a function in multiple dimensions.

4. Can you give an example of a problem involving partial derivatives?

Sure, one example is finding the maximum or minimum volume of a box with a fixed surface area. In this problem, the volume of the box is a function of two variables - length and width. We can use partial derivatives to find the values of length and width that will give us the maximum or minimum volume, while keeping the surface area constant.

5. Are partial derivatives difficult to understand?

Partial derivatives can be challenging to grasp at first, especially if you are new to multivariable calculus. However, with practice and a solid understanding of basic calculus concepts, you can master the concept of partial derivatives. It's important to remember that they are just another tool to help us solve problems and understand functions in multiple dimensions.

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