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Prove the following identities. Use Theorem 17.13 (Product Rule) whenever possible. $$\nabla\left(\frac{1}{|\mathbf{r}|^{2}}\right)=-\frac{2 \mathbf{r}}{|\mathbf{r}|^{4}}$$

Short Answer

Expert verified
Answer: The gradient of the function \(\frac{1}{|\mathbf{r}|^2}\) is given by the expression \(-\frac{2\mathbf{r}}{|\mathbf{r}|^4}\).

Step by step solution

01

Review the definition of the gradient and Product Rule

According to the definition, the gradient of a scalar function \(f\) is a vector whose components are the partial derivatives of the function with respect to each coordinate: $$\nabla f = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z})$$. Now, let's recall the Product Rule from Theorem 17.13: $$\nabla(g \cdot h) = h \cdot \nabla g + g \cdot \nabla h$$, where \(g\) and \(h\) are scalar functions.
02

Rewrite the given function using the Product Rule

We want to find the gradient of \(\frac{1}{|\mathbf{r}|^2}\). We can rewrite this function as: $$\frac{1}{|\mathbf{r}|^2} = \frac{1}{(x^2+y^2+z^2)}$$ Now, we can make use of the Product Rule. Let's take \(g(x,y,z)=1\), and \(h(x,y,z)=x^2+y^2+z^2\). Hence, the given function is the product of \(g\) and \(h^{-1}\). Using the Product Rule, we get: $$\nabla \left(\frac{1}{x^2+y^2+z^2}\right) = -(x^2+y^2+z^2)^{-2} \cdot \nabla (x^2+y^2+z^2)$$
03

Calculate the gradient of the sum of squares

Now we need to calculate the gradient of \(x^2+y^2+z^2\). Recall the definition of the gradient: $$\nabla (x^2+y^2+z^2) = (\frac{\partial}{\partial x}(x^2+y^2+z^2), \frac{\partial}{\partial y}(x^2+y^2+z^2), \frac{\partial}{\partial z}(x^2+y^2+z^2))$$ Differentiate component wise: $$\nabla (x^2+y^2+z^2) = (2x, 2y, 2z) = 2\mathbf{r}$$
04

Substitute the gradient of the sum of squares into the expression

Now, substitute the gradient of \((x^2+y^2+z^2)\) back into our expression from Step 2: $$\nabla \left(\frac{1}{x^2+y^2+z^2}\right) = -(x^2+y^2+z^2)^{-2} \cdot 2\mathbf{r}$$
05

Simplify the expression

Finally, rewrite \((x^2+y^2+z^2)^{-2}\) as \(\frac{1}{(x^2+y^2+z^2)^2}\) and simplify the expression: $$\nabla \left(\frac{1}{x^2+y^2+z^2}\right) = -\frac{2\mathbf{r}}{(x^2+y^2+z^2)^2} = -\frac{2\mathbf{r}}{|\mathbf{r}|^4}$$ Thus, we have proven the identity: $$\nabla\left(\frac{1}{|\mathbf{r}|^{2}}\right)=-\frac{2\mathbf{r}}{|\mathbf{r}|^{4}}$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product Rule
A crucial tool in calculus, especially when dealing with differentiating products of functions, is the Product Rule. The Product Rule is a formula used to find the derivative of the product of two functions. It states that the gradient of the product of two scalar functions, say \( g \) and \( h \), is given by the expression:
  • \( abla(g \cdot h) = h \cdot abla g + g \cdot abla h \)
In simpler terms, this rule tells us to differentiate the first function, then multiply it by the second function, differentiate the second function, and then add the two results together.
Imagine we have two functions, \( a(x) \) and \( b(x) \). The derivative of the product \( a(x)b(x) \) is:
  • \( a'(x)b(x) + a(x)b'(x) \)
This principle is especially important in Vector Calculus when dealing with vector fields and scalar functions.
Vector Calculus
Vector calculus extends calculus to vector fields, which are functions that assign a vector to each point in space. If you're studying forces, fluid flow, or electromagnetism, vector calculus is highly applicable.
In vector calculus, we often work with operators that help explore how functions change in vector fields. The most common operators are:
  • Gradient - This transforms a scalar function into a vector field, pointing in the direction of the greatest rate of increase of the function.
  • Divergence - It measures how much a vector field spreads out from a point.
  • Curl - It quantifies how much a vector field rotates around a point.

Let's focus on the gradient. Written as \( abla f \), it represents how a scalar function \( f \) changes with each spatial dimension. It's calculated by taking partial derivatives of \( f \) with respect to each coordinate and produces a vector that points in the direction of the highest rate of increase of \( f \).
If \( f(x, y, z) \) is a scalar function, its gradient is noted as:
  • \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \)
Partial Derivatives
Partial derivatives are a key tool in multivariable calculus. They show us how a function changes as each variable changes independently. When you have a function \( f \) that depends on multiple variables, you can examine how \( f \) changes with respect to just one of those variables while keeping the others constant.
For instance, if we have a function \( f(x, y) \), the partial derivative of \( f \) with respect to \( x \) is expressed as \( \frac{\partial f}{\partial x} \). It's the derivative of \( f \) treating \( y \) as a constant.
Similarly, the partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} \). This shows how \( f \) changes as only \( y \) varies.
  • These derivatives are useful for calculating gradients, which are vectors composed of all the partial derivatives of a scalar function.
  • In physics and engineering, understanding these changes helps in studying systems that depend on multiple factors.
Taking partial derivatives is a step-by-step process that is central to many more advanced topics in calculus.

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Most popular questions from this chapter

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