Problem 51 - Calculus: Early Transcendental Functions

Question

If \(f\) is a scalar function, \(\mathbf{r}=\langle x, y\rangle\) and \(r=\|\mathbf{r}\|,\) show that $$\nabla f(r)=f^{\prime}(r) \frac{\mathbf{r}}{r}$$

Step-by-Step Solution

Answer

The gradient of the scalar function \(f(r)\) is proven to be equal to \(f'(r) \frac{\mathbf{r}}{r}\), with \(\mathbf{r}\) being the position vector \(\langle x, y\rangle\), \(r\) being the magnitude of \(\mathbf{r}\) and \(f'(r)\) being the derivative of \(f(r)\).

Step 1: Compute the gradient of \(f(r)\)

The gradient of a scalar function \(f(r)\) in two dimensions is a vector of the first order derivatives. The function \(r = \sqrt{x^2 + y^2}\), hence: \(\frac{df}{dx} = \frac{df}{dr} . \frac{dr}{dx}\) and \(\frac{df}{dy} = \frac{df}{dr} . \frac{dr}{dy}\). Now we can use the following derivatives: \(\frac{dr}{dx} = \frac{x}{\sqrt{x^2 + y^2}} = \frac{x}{r}\) and \(\frac{dr}{dy} = \frac{y}{\sqrt{x^2 + y^2}} = \frac{y}{r}\) substituting these into the equations above, we get: \(\frac{df}{dx} = f'(r)\frac{x}{r}\) and \(\frac{df}{dy} = f'(r)\frac{y}{r}\)

Step 2: Form the gradient vector

The gradient vector \(\nabla f(r)\) is formed by putting the first order derivatives into a vector = \(\langle \frac{df}{dx}, \frac{df}{dy}\rangle = \langle f'(r)\frac{x}{r}, f'(r)\frac{y}{r}\rangle\)

Step 3: Simplify the gradient vector

The gradient vector can be simplified by factoring out \(f'(r)\): \(\nabla f(r) = f'(r) \langle \frac{x}{r}, \frac{y}{r}\rangle = f'(r) \frac{\mathbf{r}}{r}\). Hence, the gradient of \(f(r)\) is \(f'(r) \frac{\mathbf{r}}{r}\), which completes the proof.

Key Concepts

Scalar Function

A scalar function is a mathematical expression that assigns a single scalar value to every point in a space. Unlike vectors, which have direction and magnitude, scalar functions derive their value solely from their position.

• Scalar functions are crucial in a wide range of fields, including physics and engineering.
• They often represent quantities like temperature, pressure, or potential energy.

A basic example of a scalar function in a two-dimensional space is the equation for a circle, such as \[f(x, y) = x^2 + y^2\]where each x-y pair is mapped to a single scalar value (the radius squared). This function does not indicate a direction, only the magnitude of the circle's radius.

Gradient Vector

The gradient vector of a scalar function is an essential concept in vector calculus. It provides the direction of the steepest ascent of the function.

• The gradient vector has a direction and length, unlike the scalar function itself.
• This vector points towards the direction in which the function increases most rapidly.

Mathematically, the gradient vector \( abla f(r) \) is composed of the partial derivatives of a scalar function \( f \) with respect to each variable of the function. In the case of our exercise, these derivatives form the components of the gradient vector \( \langle \frac{df}{dx}, \frac{df}{dy} \rangle \). This vector is ubiquitous in fields like optimization and physics since it helps find maxima and minima by following the direction of greatest change.

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and differential operators applied to scalar and vector functions. It's instrumental in physics and engineering, providing tools to model and analyze multivariable systems.

• Vector calculus is centered around vector fields, which are assignments of vector values to points in space.
• It includes operations such as divergence, curl, and the gradient.

In the realm of vector calculus, operations like the gradient can transform scalar field data into vector field data, offering insight into the directional behavior of scalar fields. For example, the gradient can help map the slope and direction of a hill from knowing just the height at various points. Vector calculus is essential for understanding phenomena such as electromagnetic fields and fluid flow.

Partial Derivatives

Partial derivatives are a cornerstone of calculus, especially for functions of multiple variables. They measure how a function changes as one variable changes while keeping others constant.

• Partial derivatives are the building blocks for creating the gradient vector of a scalar function.
• They help understand the impact of each independent variable on the function's behavior.

In our problem, the partial derivatives \( \frac{df}{dx} \) and \( \frac{df}{dy} \) indicate how the scalar function changes along the x-axis and y-axis, respectively. These derivatives combine to form a vector, elucidating both magnitude and direction.Understanding partial derivatives is crucial for tasks such as optimizing functions and solving problems related to changes in physical systems, exemplifying how they support broader applications in analysis.