Question

Prove that for a real number \(p\), with \(\mathbf{r}=\langle x, y, z\rangle, \nabla \cdot \nabla\left(\frac{1}{|\mathbf{r}|^{p}}\right)=\frac{p(p-1)}{|\mathbf{r}|^{p+2}}.\)

Short answer

Question: Prove that the divergence of the gradient of the scalar field \(\frac{1}{|\mathbf{r}|^{p}}\) is equal to \(\frac{p(p-1)}{|\mathbf{r}|^{p+2}}\) Solution: Following the steps outlined in the analysis and solution above, we first found the gradient of the scalar field and then calculated its divergence using Cartesian coordinates. The final result is \(\nabla \cdot (\nabla f) = \frac{p(p-1)}{|\mathbf{r}|^{p+2}}\), which matches the given equation.

Step-by-step solution

Calculate the gradient of the scalar field

First, we need to find the gradient of the scalar field \(\frac{1}{|\mathbf{r}|^{p}}\). The gradient in Cartesian coordinates is given by: $$ \nabla f = \left(\frac{\partial f}{\partial x}\mathbf{i}+\frac{\partial f}{\partial y}\mathbf{j}+\frac{\partial f}{\partial z}\mathbf{k}\right) $$ In our case, the scalar field \(f=\frac{1}{|\mathbf{r}|^{p}}\). Remember that \(|\mathbf{r}|=\sqrt{x^2+y^2+z^2}\). It is now necessary to differentiate \(f\) with respect to \(x\), \(y\), and \(z\).

Calculate the partial derivatives of the scalar field

To make it easier to find the partial derivatives, rewrite the scalar field as: $$ f = \left(x^2+y^2+z^2\right)^{-\frac{p}{2}} $$ Now, we find the partial derivatives of \(f\) with respect to \(x\), \(y\), and \(z\). Using the chain rule, we obtain: $$ \begin{aligned} \frac{\partial f}{\partial x} &= -\frac{p}{2}\left(x^2+y^2+z^2\right)^{-\frac{p}{2}-1}\cdot 2x \\ \frac{\partial f}{\partial y} &= -\frac{p}{2}\left(x^2+y^2+z^2\right)^{-\frac{p}{2}-1}\cdot 2y \\ \frac{\partial f}{\partial z} &= -\frac{p}{2}\left(x^2+y^2+z^2\right)^{-\frac{p}{2}-1}\cdot 2z \end{aligned} $$

Obtain the gradient of the scalar field

Now, substitute the partial derivatives into the expression for the gradient: $$ \nabla f = -p\left(\frac{x}{|\mathbf{r}|^{p+2}}\mathbf{i}+\frac{y}{|\mathbf{r}|^{p+2}}\mathbf{j}+\frac{z}{|\mathbf{r}|^{p+2}}\mathbf{k}\right) $$

Calculate the divergence of the gradient

Next, we need to find the divergence of the gradient \(\nabla f\). In Cartesian coordinates, the divergence is given by: $$ \nabla \cdot (\nabla f) = \frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right) + \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial y}\right) + \frac{\partial}{\partial z}\left(\frac{\partial f}{\partial z}\right) $$ We already found the partial derivatives of \(f\) with respect to \(x\), \(y\), and \(z\) in Step 2. Now, we need to differentiate these expressions again with respect to their respective variables.

Differentiating the gradient components

Notice that, for each component of the gradient, we have the product of a constant, a power of \(x\), \(y\), or \(z\), and a power of \(|\mathbf{r}|\). We can use the product rule to find the derivatives of these gradient components: $$ \begin{aligned} \frac{\partial}{\partial x}\left(\frac{x}{|\mathbf{r}|^{p+2}}\right) &= \frac{1}{|\mathbf{r}|^{p+2}} + \frac{(p+2)x^2}{|\mathbf{r}|^{p+4}} \\ \frac{\partial}{\partial y}\left(\frac{y}{|\mathbf{r}|^{p+2}}\right) &= \frac{(p+2)y^2}{|\mathbf{r}|^{p+4}} \\ \frac{\partial}{\partial z}\left(\frac{z}{|\mathbf{r}|^{p+2}}\right) &= \frac{(p+2)z^2}{|\mathbf{r}|^{p+4}} \end{aligned} $$

Calculate the divergence of the gradient

Now plug these expressions into the formula for the divergence. Here, do not forget that there is a factor of \(-p\) in the gradient: $$ \nabla \cdot (\nabla f) = -p\left(\frac{1}{|\mathbf{r}|^{p+2}}+\frac{(p+1)}{|\mathbf{r}|^{p+2}}\right) = \frac{p(p-1)}{|\mathbf{r}|^{p+2}} $$ This expression matches the given equation, proving the result.

Key concepts

Understanding Partial Derivatives

Partial derivatives are a cornerstone in the field of multivariable calculus. They represent the rate at which a function changes as one variable changes while the others are held constant. Imagine a surface that represents a function with two variables, like altitude depending on the position on a map. A partial derivative would tell you how steep the hill is if you move east, without going north or south.

When we talk about a function like \(f(x, y, z) = \frac{1}{|\mathbf{r}|^{p}}\), where \(\mathbf{r}\) is the position vector, the partial derivatives \(\frac{\partial f}{\partial x}\), \(\frac{\partial f}{\partial y}\), and \(\frac{\partial f}{\partial z}\) tell us how this function changes in each direction of our three-dimensional space. These derivatives are the first steps to compute the gradient of the function, which points in the direction of the greatest increase of the function's value.

The Chain Rule in Vector Calculus

The chain rule is a powerful tool for computing derivatives of compositions of functions. It effectively breaks down the differentiation process into simpler parts. In vector calculus, when you have a composite function like \( (x^2+y^2+z^2)^{-\frac{p}{2}} \), the chain rule allows you to differentiate the outer function with respect to the inner one and multiply it by the derivative of the inner function with respect to the variable of interest.

In essence, the chain rule helps us to differentiate a function like our scalar field \(f\) in the context of its dependence on \(x^2+y^2+z^2\), which itself depends on \(x\), \(y\), and \(z\). This technique is key when finding partial derivatives required for computing the gradient, making it an integral part of solving our textbook problem.

Scalar Field Gradient

The gradient of a scalar field is a vector field that points in the direction of the greatest rate of increase of the scalar field and its magnitude is the rate of that increase. It's like being on a hill and feeling the steepest path uphill beneath your feet. That's what the gradient tells you about a point in a scalar field.

To find the gradient of a scalar field like \(f\), you need to compute its partial derivatives with respect to each variable. The result is a vector composed of these derivatives. In our example, we attained the gradient by finding the derivatives with respect to \(x\), \(y\), and \(z\), and combining them into the vector \(abla f\). This vector field holds crucial information about the variations of the scalar field \(f\) in space.

Vector Calculus and Divergence

Vector calculus is the branch of mathematics that deals with the differentiation and integration of vector fields. One of the key operations in vector calculus is divergence. Divergence measures the magnitude of a vector field's source or sink at a given point - it's a bit like measuring how much air is being pumped into or sucked out of a small area.

In our problem, we calculated the divergence of the gradient vector field \(abla f\), which we previously found. We took each component of the gradient and differentiated it once more with respect to its corresponding variable. This step is akin to checking how our 'hill' changes its 'steepness' at each point. Ultimately, this led us to the solution, demonstrating how divergence relates to the original scalar field, completing our exploration into how these concepts link together.