Question

Prove the following identities. Use Theorem 14.11 (Product Rule) whenever possible. $$\nabla \cdot \nabla\left(\frac{1}{|\mathbf{r}|^{2}}\right)=\frac{2}{|\mathbf{r}|^{4}} \quad \text { (use Exercise } 36)$$

Short answer

Question: Prove the following identity using the Product Rule (Theorem 14.11) if possible: $$\nabla \cdot \nabla \left(\frac{1}{|\mathbf{r}|^{2}}\right) = \frac{2}{|\mathbf{r}|^{4}}$$ Answer: We have proved that $$\nabla \cdot \nabla\left(\frac{1}{|\mathbf{r}|^{2}}\right) = \frac{2}{|\mathbf{r}|^{4}}$$

Step-by-step solution

Find the gradient of the function

First, we need to find the gradient of the function \(\frac{1}{|\mathbf{r}|^{2}} = \frac{1}{x^2+y^2+z^2}\) with respect to each coordinate: $$\nabla\left(\frac{1}{|\mathbf{r}|^{2}}\right) = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right) \left(\frac{1}{x^{2}+y^{2}+z^{2}}\right)$$ Now, we find the x-component of the gradient: $$\frac{\partial}{\partial x}\left(\frac{1}{x^{2}+y^{2}+z^{2}}\right) = \frac{-2x}{(x^{2}+y^{2}+z^{2})^{2}}$$ Similarly, to find the y- and z-components of the gradient, we get: $$\frac{\partial}{\partial y}\left(\frac{1}{x^{2}+y^{2}+z^{2}}\right) = \frac{-2y}{(x^{2}+y^{2}+z^{2})^{2}}$$ $$\frac{\partial}{\partial z}\left(\frac{1}{x^{2}+y^{2}+z^{2}}\right) = \frac{-2z}{(x^{2}+y^{2}+z^{2})^{2}}$$ Thus, the gradient can be written as: $$\nabla\left(\frac{1}{|\mathbf{r}|^{2}}\right) = \left(\frac{-2x}{(x^{2}+y^{2}+z^{2})^{2}},\frac{-2y}{(x^{2}+y^{2}+z^{2})^{2}},\frac{-2z}{(x^{2}+y^{2}+z^{2})^{2}}\right)$$

Find the divergence of the gradient

Now, we need to find the divergence of the gradient: $$\nabla \cdot \nabla \left(\frac{1}{|\mathbf{r}|^{2}}\right) = \frac{\partial}{\partial x}\left(\frac{-2x}{(x^{2}+y^{2}+z^{2})^{2}}\right) + \frac{\partial}{\partial y}\left(\frac{-2y}{(x^{2}+y^{2}+z^{2})^{2}}\right) + \frac{\partial}{\partial z}\left(\frac{-2z}{(x^{2}+y^{2}+z^{2})^{2}}\right)$$ We'll find the partial derivatives one by one: $$\frac{\partial}{\partial x}\left(\frac{-2x}{(x^{2}+y^{2}+z^{2})^{2}}\right) = \frac{-2(3x^{2}+y^{2}+z^{2})}{(x^{2}+y^{2}+z^{2})^{3}}$$ $$\frac{\partial}{\partial y}\left(\frac{-2y}{(x^{2}+y^{2}+z^{2})^{2}}\right) = \frac{-2(x^{2}+3y^{2}+z^{2})}{(x^{2}+y^{2}+z^{2})^{3}}$$ $$\frac{\partial}{\partial z}\left(\frac{-2z}{(x^{2}+y^{2}+z^{2})^{2}}\right) = \frac{-2(x^{2}+y^{2}+3z^{2})}{(x^{2}+y^{2}+z^{2})^{3}}$$ Adding the partial derivatives, we get: $$\nabla \cdot \nabla\left(\frac{1}{|\mathbf{r}|^{2}}\right) = \frac{2(x^{2} + y^{2} + z^{2})}{(x^{2} + y^{2} + z^{2})^{3}} = \frac{2}{(x^{2} + y^{2} + z^{2})^{2}} = \frac{2}{|\mathbf{r}|^{4}}$$ Thus we have proved the required identity: $$\nabla \cdot \nabla\left(\frac{1}{|\mathbf{r}|^{2}}\right) = \frac{2}{|\mathbf{r}|^{4}}$$

Key concepts

Gradient of a Function

Understanding the gradient of a function is fundamental in vector calculus. The gradient transforms a scalar function into a vector field, pointing in the direction of the steepest ascent of the function. For a function \(f(x,y,z)\), the gradient is given by \(abla f = \(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\)\). Each component of this vector represents the rate of change of \(f\) with respect to one of the variables, while keeping the others constant. In terms of physical intuition, it is like feeling the slope of a hill under your feet; the gradient vector shows how steep the hill is and in which direction it is steepest.

In the context of the given exercise, the gradient of \(\frac{1}{|\mathbf{r}|^{2}}\) reveals how this function changes in three-dimensional space. By calculating each component, as shown in the solution, students can visualize the 'flow' of the function in terms of its rate of change along the x, y, and z directions.

Divergence of a Vector Field

The divergence of a vector field is another critical concept in vector calculus, representing the net rate of ‘flow’ emanating from a given point. If you imagine the vector field as representing the velocity of a fluid, the divergence at a point gives you an idea of whether the fluid is compressing (negative divergence) or expanding (positive divergence) at that point. The divergence of a vector field \(\mathbf{F} = (P, Q, R)\) is a scalar function given by \(abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}\).

In the exercise, finding the divergence of the gradient of \(\frac{1}{|\mathbf{r}|^{2}}\) requires the application of the divergence operator to the already computed gradient. The solution provided demonstrates this process, which involves taking the partial derivatives of each component of the gradient and summing them up to get the scalar function that represents the divergence.

Partial Derivatives

The concept of partial derivatives plays a crucial role in multivariable calculus. When we deal with functions that have more than one variable, we often want to understand how the function changes as we vary one of these variables while keeping the others fixed. This is where partial derivatives come in. For a function \(f(x, y, z)\), the partial derivative with respect to \(x\) is denoted \(\frac{\partial f}{\partial x}\), and it measures the rate at which \(f\) changes with \(x\), all else being constant.

The computation of partial derivatives is evident in the solution steps for both the gradient and the divergence of the function. By carefully evaluating the partial derivatives of the given function, students unravel how the function's change is directed in relation to each independent variable.

Theorem 14.11 Product Rule

The Product Rule, a vital theorem in both single-variable and multivariate calculus, describes how to differentiate a product of two functions. In single-variable calculus, the Product Rule states that the derivative of a product of two functions is given by \(\frac{d}{dx}(u \cdot v) = u'v + uv'\). The multivariate generalization, Theorem 14.11 in the context of the exercise, expresses how to differentiate the product of scalar and vector functions. It is an indispensable tool when dealing with the differentiation of products involving vectors.

The Product Rule is especially useful when dealing with products involving gradients, divergences, and other vector operators in calculus. Although it isn’t used explicitly in the solution of the given exercise, understanding this rule is essential when dealing with more complex vector calculus problems that involve products of multiple functions.