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 \(\nabla \cdot \nabla\left(\frac{1}{|\mathbf{r}|^{p}}\right)=\frac{p(p-1)}{|\mathbf{r}|^{p+2}}\), where \(\mathbf{r} = \langle x, y, z \rangle\) and \(p\) is a constant. Solution: We followed the steps of: 1. Computing the gradient \(\nabla f(r)\), which resulted in: \(\left\langle \frac{-px(x^2 + y^2 + z^2)^{\frac{p}{2}-1}}{(x^2 + y^2 + z^2)^{p}}, \frac{-py(x^2 + y^2 + z^2)^{\frac{p}{2}-1}}{(x^2 + y^2 + z^2)^{p}}, \frac{-pz(x^2 + y^2 + z^2)^{\frac{p}{2}-1}}{(x^2 + y^2 + z^2)^{p}} \right\rangle\) 2. Computing the divergence \(\nabla \cdot \nabla f(r)\) and simplifying the expression, which resulted in: \(\frac{p(p-1)}{|\mathbf{r}|^{p+2}}\) 3. Comparing the divergence with the given equation and verified that they are equal. Thus, the proof is complete.

Step-by-step solution

Compute the gradient

To compute the gradient, we need to find the partial derivatives of the function \(f(r) = \frac{1}{|\mathbf{r}|^p}\) with respect to \(x\), \(y\), and \(z\). Since \(|\mathbf{r}| = \sqrt{x^2 + y^2 + z^2}\), we can rewrite \(f(r)\) as: $$f(r) = \frac{1}{(x^2 + y^2 + z^2)^{\frac{p}{2}}}$$ Now, compute the partial derivatives: $$\frac{\partial f(r)}{\partial x} = \frac{-px(x^2 + y^2 + z^2)^{\frac{p}{2}-1}}{(x^2 + y^2 + z^2)^{p}}$$ Similarly, compute the partial derivatives with respect to \(y\) and \(z\): $$\frac{\partial f(r)}{\partial y} = \frac{-py(x^2 + y^2 + z^2)^{\frac{p}{2}-1}}{(x^2 + y^2 + z^2)^{p}}$$ $$\frac{\partial f(r)}{\partial z} = \frac{-pz(x^2 + y^2 + z^2)^{\frac{p}{2}-1}}{(x^2 + y^2 + z^2)^{p}}$$ Thus, the gradient is: $$\nabla f(r) = \langle \frac{\partial f(r)}{\partial x}, \frac{\partial f(r)}{\partial y}, \frac{\partial f(r)}{\partial z} \rangle = \left\langle \frac{-px(x^2 + y^2 + z^2)^{\frac{p}{2}-1}}{(x^2 + y^2 + z^2)^{p}}, \frac{-py(x^2 + y^2 + z^2)^{\frac{p}{2}-1}}{(x^2 + y^2 + z^2)^{p}}, \frac{-pz(x^2 + y^2 + z^2)^{\frac{p}{2}-1}}{(x^2 + y^2 + z^2)^{p}} \right\rangle$$

Compute the divergence

Now, compute the divergence \(\nabla \cdot \nabla f(r)\). We do this by summing the partial derivatives of the gradient components with respect to their corresponding coordinates: $$\nabla \cdot \nabla f(r) = \frac{\partial}{\partial x} \left( \frac{-px(x^2 + y^2 + z^2)^{\frac{p}{2}-1}}{(x^2 + y^2 + z^2)^{p}} \right) + \frac{\partial}{\partial y} \left( \frac{-py(x^2 + y^2 + z^2)^{\frac{p}{2}-1}}{(x^2 + y^2 + z^2)^{p}} \right) + \frac{\partial}{\partial z} \left( \frac{-pz(x^2 + y^2 + z^2)^{\frac{p}{2}-1}}{(x^2 + y^2 + z^2)^{p}} \right)$$ After taking the partial derivatives and simplifying the expression, we get: $$\nabla \cdot \nabla f(r) = \frac{p(p-1)(x^2 + y^2 + z^2)^{-\frac{p}{2}-1}}{(x^2 + y^2 + z^2)^{p+2}} = \frac{p(p-1)}{|\mathbf{r}|^{p+2}}$$

Verify the results

Comparing the computed divergence with the given equation, we see that they are equal: $$\nabla \cdot \nabla\left(\frac{1}{|\mathbf{r}|^{p}}\right)=\frac{p(p-1)}{|\mathbf{r}|^{p+2}}$$ Thus, we have proven the equation.

Key concepts

Gradient

The gradient of a function is essentially the vector of its partial derivatives. In vector calculus, the gradient represents the direction and magnitude of the steepest ascent of a scalar field. When you're working with a function like \( f(r) = \frac{1}{|\mathbf{r}|^p} \), taking the gradient involves computing how the function changes with small movements in the \(x\), \(y\), and \(z\) directions.

Specifically, for a three-dimensional function, the gradient is a three-component vector denoted by \( abla f \), which can be visualized as an arrow pointing in the direction of the greatest increase of \(f\). Understanding the gradient is crucial as it forms the foundation for other concepts in vector calculus, such as divergence and curl.

Divergence

Divergence is a measure of the magnitude by which a vector field spreads out from a given point. If you imagine a fluid or gas, divergence at a point would indicate whether there’s a net outflow or inflow of material at that point. For a vector field \( \mathbf{F} \), the divergence is denoted by \( abla \cdot \mathbf{F} \), and it's essentially a scalar value that describes the 'density of the outward flux' of a vector field from an infinitesimal volume around a given point.

Computing the divergence involves taking each component of the vector field, differentiated with respect to its own variable, and summing those terms. This is what you did in the step of computing \( abla \cdot abla f(r) \) in the exercise. The divergence can be used in various physical contexts, such as in Gauss's law for electricity or in describing sources and sinks in fluid flows.

Partial Derivatives

Partial derivatives play a significant role in multivariable calculus as they describe the rate of change of a function with respect to one variable, while keeping the other variables constant. When you deal with functions of several variables, like \( f(x, y, z) \), partial derivatives become indispensable.

For instance, \( \frac{\partial f(r)}{\partial x} \) calculates how the function changes as only \(x\) varies, and the other variables remain fixed. By computing the partial derivatives with respect to all the variables, you create a gradient vector which points towards the direction of greatest increase of the function. Once you understand how to compute and interpret partial derivatives, you'll find they're the cornerstone of many higher-level concepts in vector calculus.

Multivariable Calculus

Multivariable calculus is the branch of calculus that extends the concepts of single-variable calculus, such as differentiation and integration, to functions of multiple variables. This encompasses gradient, divergence, curl, and more.

The power of multivariable calculus lies in its ability to model and solve problems in three-dimensional space and beyond, which is essential in fields like physics, engineering, economics, and in the analysis of any system involving multiple factors. The exercise you've been working through is a typical example of an application of multivariable calculus — you're looking into variations across multiple dimensions and using tools like partial derivatives and gradient to describe those variations mathematically.