Finding 1st and 2nd total partial differential

Then differentiate with respect to y and z similarly to find the other two partials. Their sum will be [itex]\partial^2 r/\partial x^2+ \partial^2 r/\partial y^2+ \partial^2 r/\partial z^2[/itex].
  • #1
ScottO
1
0
I've been struggling a bit with the following problem...

Given:
[tex]r^2 = (x - a)^2 + (y - b)^2 + (z - c)^2[/tex]

Find:
1. [tex]\frac{\partial r}{\partial x} + \frac{\partial r}{\partial y} + \frac{\partial r}{\partial z}[/tex]

and

2. [tex]\frac{\partial^2 r}{\partial x^2} + \frac{\partial^2 r}{\partial y^2} + \frac{\partial^2 r}{\partial z^2}[/tex]

I've found the following for 1.

[tex]2rdr = 2(x - a)dx + 2(y - b)dy + 2(z - c)dz[/tex]

[tex]dr = \frac{(x - a) + (y - b) + (z - c)}{r}[/tex]

This matches the answer in the book. So I think I'm OK to this point.

Figuring 2 is where I'm having trouble. How do I handle the r in the denominator? Do I substitute [tex]\sqrt{(x - a)^2 + (y - b)^2 + (z - c)^2}[/tex] from the original given? Or start from here:

[tex]2rdr = 2(x - a)dx + 2(y - b)dy + 2(z - c)dz[/tex]

which leads to

[tex]2d^2r = 2dx^2 + 2dy^2 + 2dz^2[/tex]

[tex]d^2r = 3[/tex]

I don't think latter is correct, as it seems like it doesn't consider r. Plus, the answer in the book is

[tex]\frac{2}{r}[/tex]

I've gotten close using the substitution, but things are a bit messy, and I'm missing some simplification or something.

Here's what I get for one partial using the substitution...

[tex]dr = \frac{(x - a) + (y - b) + (z - c)}{r}[/tex]

[tex]\frac{\partial^2 r}{\partial x^2} = \frac{r - \frac{[(x - a) + (y - b) + (z - c)](x - a)}{r}}{r^2}[/tex]

-Scott
 
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  • #2
ScottO said:
I've been struggling a bit with the following problem...
I would consider the phrase "total partial differential" a contradiction in terms! Perhaps you mean what I would call just the "total differential".

Given:
[tex]r^2 = (x - a)^2 + (y - b)^2 + (z - c)^2[/tex]

Find:
1. [tex]\frac{\partial r}{\partial x} + \frac{\partial r}{\partial y} + \frac{\partial r}{\partial z}[/tex]
This is neither a partial derivative nor a differential. In fact, since it would change with any change in coordinate system, I'm not sure it has any "real" meaning!

and

2. [tex]\frac{\partial^2 r}{\partial x^2} + \frac{\partial^2 r}{\partial y^2} + \frac{\partial^2 r}{\partial z^2}[/tex]

I've found the following for 1.

[tex]2rdr = 2(x - a)dx + 2(y - b)dy + 2(z - c)dz[/tex]
Okay, now that's a total differential.

[tex]dr = \frac{(x - a) + (y - b) + (z - c)}{r}[/tex]
What happened to the "dx", "dy", and "dz"?

This matches the answer in the book. So I think I'm OK to this point.[/quote]
If the problem really was to find
[tex]\frac{\partial r}{\partial x} + \frac{\partial r}{\partial y} + \frac{\partial r}{\partial z}[/tex]
then there was no reason to find the total differential. Just the three partial derivatives would suffice.

Figuring 2 is where I'm having trouble. How do I handle the r in the denominator? Do I substitute [tex]\sqrt{(x - a)^2 + (y - b)^2 + (z - c)^2}[/tex] from the original given? Or start from here:

[tex]2rdr = 2(x - a)dx + 2(y - b)dy + 2(z - c)dz[/tex]

which leads to

[tex]2d^2r = 2dx^2 + 2dy^2 + 2dz^2[/tex]

[tex]d^2r = 3[/tex]

I don't think latter is correct, as it seems like it doesn't consider r. Plus, the answer in the book is

[tex]\frac{2}{r}[/tex]

I've gotten close using the substitution, but things are a bit messy, and I'm missing some simplification or something.

Here's what I get for one partial using the substitution...

[tex]dr = \frac{(x - a) + (y - b) + (z - c)}{r}[/tex]

[tex]\frac{\partial^2 r}{\partial x^2} = \frac{r - \frac{[(x - a) + (y - b) + (z - c)](x - a)}{r}}{r^2}[/tex]

-Scott
If you are really asked to find just the sum of the second partials (which I would consider to have no real significance, as, unlike the gradient, etc., it depends on the coordinate system.) then just find the partial derivatives.
Since [itex]r^2= (x-a)^2+ (y-b)^2+ (z-c)^2[/itex] then [itex]2r\partial r/\partial x= 2(x-a)[/itex] and [itex]r\partial r/\partial x= x-a. Now differentiate both sides again with respect to x: [itex]\left(\partial r/\partial x\right)^2+ 2r \partial^2 r/\partial x^2= 1[/itex]. Solve that for [itex]\partial^2 r/\partial x^2[/itex].
 
  • #3


Hi Scott,

First of all, great job on finding the first partial derivative! You are correct in using the substitution for r in finding the second partial derivatives. However, you are missing a few steps in your calculations.

Here is how you can find the second partial derivatives:

1. Begin with the equation 2rdr = 2(x - a)dx + 2(y - b)dy + 2(z - c)dz.

2. Substitute r = \sqrt{(x - a)^2 + (y - b)^2 + (z - c)^2} from the original given equation.

3. Expand the equation to get:

2rdr = 2(x - a)dx + 2(y - b)dy + 2(z - c)dz

2\sqrt{(x - a)^2 + (y - b)^2 + (z - c)^2}dr = 2(x - a)dx + 2(y - b)dy + 2(z - c)dz

4. Now, take the first partial derivative with respect to x, keeping y and z as constants:

\frac{\partial}{\partial x}(2\sqrt{(x - a)^2 + (y - b)^2 + (z - c)^2}dr) = \frac{\partial}{\partial x}(2(x - a)dx + 2(y - b)dy + 2(z - c)dz)

2\frac{\partial}{\partial x}(\sqrt{(x - a)^2 + (y - b)^2 + (z - c)^2}dr) = 2dx

5. Simplify and solve for \frac{\partial^2 r}{\partial x^2}:

\frac{\partial^2 r}{\partial x^2} = \frac{1}{\sqrt{(x - a)^2 + (y - b)^2 + (z - c)^2}} \left(1 - \frac{(x - a)^2}{\sqrt{(x - a)^2 + (y - b)^2 + (z - c)^2}} \right)

6. Follow the same steps for the other variables y and z to get the second partial derivatives with respect to y and z.

In the end, you should get the following answers for the second partial derivatives:

\frac{\partial^2 r}{\partial x^2
 

1. What is the definition of a total partial differential?

A total partial differential is a mathematical tool used in multivariate calculus to calculate the rate at which a function changes with respect to multiple variables simultaneously. It is a combination of partial derivatives, which measure the rate of change of a function with respect to one variable while holding all other variables constant.

2. How is the first total partial differential calculated?

The first total partial differential is calculated by taking the partial derivative of a function with respect to one variable, and then multiplying it by the corresponding partial derivative of another variable. This is repeated for each variable in the function, and the results are added together to get the total partial differential.

3. What is the difference between first and second total partial differentials?

The first total partial differential measures the instantaneous rate of change of a function with respect to multiple variables, while the second total partial differential measures the rate of change of the first total partial differential with respect to one of the variables. In other words, the second total partial differential is the partial derivative of the first total partial differential.

4. Why are total partial differentials important in science?

Total partial differentials are important in science because they allow us to analyze the behavior of complex functions with multiple variables. They are used in fields such as physics, chemistry, and engineering to model and understand the relationships between different quantities and how they change over time.

5. Can total partial differentials be applied to real-world problems?

Yes, total partial differentials can be applied to real-world problems in various scientific and engineering fields. They are used to solve optimization problems, model physical systems, and analyze how different variables affect the behavior of a system. Examples of real-world applications include predicting weather patterns, designing efficient engines, and optimizing chemical reactions.

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