Question

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

Short answer

Question: Prove that the gradient of the function \(f = \frac{1}{|\mathbf{r}|^p}\) is given by \(\nabla f = \frac{-p \mathbf{r}}{|\mathbf{r}|^{p+2}}\), where \(\mathbf{r} = \langle x, y, z \rangle\). Answer: We have calculated the gradient of the function \(f\) as follows: $$ \nabla f = \langle -\frac{p x}{(x^2+y^2+z^2)^{p/2+1}}, -\frac{p y}{(x^2+y^2+z^2)^{p/2+1}}, -\frac{p z}{(x^2+y^2+z^2)^{p/2+1}} \rangle $$ Since this matches the given expression, we have proved that \(\nabla\left(\frac{1}{|\mathbf{r}|^{p}}\right)=\frac{-p \mathbf{r}}{|\mathbf{r}|^{p+2}}\).

Step-by-step solution

Find the partial derivative of f with respect to x

To find the partial derivative of \(f\) with respect to \(x\), we use the chain rule: $$ \frac{\partial f}{\partial x} = (\frac{1}{(x^2+y^2+z^2)^{p/2}})'(x^2+y^2+z^2)'= -\frac{p(x^2+y^2+z^2)^{-p/2-1}}{2} \cdot 2x = -\frac{p x}{(x^2+y^2+z^2)^{p/2+1}} $$

Find the partial derivative of f with respect to y

Similarly, for the partial derivative of \(f\) with respect to \(y\): $$ \frac{\partial f}{\partial y} = -\frac{p y}{(x^2+y^2+z^2)^{p/2+1}} $$

Find the partial derivative of f with respect to z

Lastly, the partial derivative of \(f\) with respect to \(z\): $$ \frac{\partial f}{\partial z} = -\frac{p z}{(x^2+y^2+z^2)^{p/2+1}} $$

Write the gradient

Now we can write the gradient of \(f\) by combining the partial derivatives from steps 1, 2, and 3: $$ \nabla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \rangle = \langle -\frac{p x}{(x^2+y^2+z^2)^{p/2+1}}, -\frac{p y}{(x^2+y^2+z^2)^{p/2+1}}, -\frac{p z}{(x^2+y^2+z^2)^{p/2+1}} \rangle $$

Verify the result

Finally, we can check if the calculated gradient matches the given expression: $$ \frac{-p \mathbf{r}}{|\mathbf{r}|^{p+2}} = \frac{-p \langle x, y, z \rangle}{(x^2+y^2+z^2)^{\frac{p}{2}+1}} = \langle -\frac{p x}{(x^2+y^2+z^2)^{p/2+1}}, -\frac{p y}{(x^2+y^2+z^2)^{p/2+1}}, -\frac{p z}{(x^2+y^2+z^2)^{p/2+1}} \rangle $$ Since the calculated gradient matches the given expression, we have proved that $$ \nabla\left(\frac{1}{|\mathbf{r}|^{p}}\right)=\frac{-p \mathbf{r}}{|\mathbf{r}|^{p+2}} $$

Key concepts

Partial Derivatives

Understanding partial derivatives is a key step in vector calculus. Partial derivatives help analyze how a multivariable function changes as each variable is varied independently. For example, with a function like \( f(x, y, z) \), we can look at how \( f \) changes with just \( x \), while keeping \( y \) and \( z \) constant.

• The partial derivative of \( f \) with respect to \( x \), denoted \( \frac{\partial f}{\partial x} \), tells us how much \( f \) changes per unit change in \( x \).
• In the original exercise, finding partial derivatives of a function related to a vector \( \mathbf{r} = \langle x, y, z \rangle \) involves differentiating an expression that includes \( x, y, \) and \( z \).

This detailed breakdown of the roles played by each variable helps in constructing a full gradient vector. Understanding this concept is the first step toward analyzing how functions behave in three-dimensional space.

Chain Rule

The Chain Rule is a fundamental tool in calculus, used when differentiating composite functions. It helps us find the derivative of a function that is nested inside another function, by working through the layers of composition.

• For functions in vector calculus, where you often encounter composite functions, the chain rule is essential. For example, differentiating \( (x^2 + y^2 + z^2)^{p/2} \) involves both the power rule and the chain rule.
• In the exercise, the chain rule is applied to find the derivative of \( \frac{1}{(x^2+y^2+z^2)^{p/2}} \) with respect to \( x \).

This specific application helps deconstruct complex, nested functions into simpler parts that are easier to manage. Thus, the chain rule is a powerful technique that simplifies the evaluation of derivatives for multivariable functions in calculus.

Vector Calculus

Vector calculus is a branch of mathematics focused on vector fields and operations involving vectors. A vector is essentially a quantity that has both a direction and a magnitude.

• In the context of the exercise, we deal with the gradient vector \( abla f \), which points in the direction of the greatest rate of increase of the function f and whose magnitude is that rate of increase.
• Vector calculus includes operations such as gradient, divergence, and curl, which help analyze and visualize how functions behave in multi-dimensional spaces.

Using vector calculus, we determine properties like field lines or potential maps, critical in physics and engineering. The gradient calculation in the exercise is a practical example of applying vector calculus to demonstrate how vector operations work seamlessly with derivatives.