Problem 96 - Advanced Calculus

Question

Find the Jacobian matrix and the differential of the vector Function \(\mathrm{f}: \mathrm{R}^{2} \rightarrow \mathrm{R}^{3}\) defined by \(|\mathrm{x}(\mathrm{u}, \mathrm{v})| \quad|\cos \mathrm{u} \cos \mathrm{v}|\) \(\mathrm{f}(\mathrm{u}, \mathrm{v})=|\mathrm{y}(\mathrm{u}, \mathrm{v})|=|\cos \mathrm{u} \sin \mathrm{v}|\) \(|\mathbf{z}(\mathrm{u}, \mathrm{v})| \quad \sin \mathrm{u}\)

Step-by-Step Solution

Answer

The Jacobian matrix \(J(u, v)\) of the given vector function \(f(u, v)=\begin{bmatrix} \cos(u)\cos(v) \\ \cos(u)\sin(v) \\ \sin(u) \end{bmatrix}\) is: \[ J(u, v) = \begin{bmatrix} -\sin(u)\cos(v) & -\cos(u)\sin(v) \\ -\sin(u)\sin(v) & \cos(u)\cos(v) \\ \cos(u) & 0 \end{bmatrix}. \] The differential of the vector function, \(df(u, v)\), is: \[ df(u, v) = \begin{bmatrix} -\sin(u)\cos(v) & -\cos(u)\sin(v) \\ -\sin(u)\sin(v) & \cos(u)\cos(v) \\ \cos(u) & 0 \end{bmatrix} \begin{bmatrix} d(u) \\ d(v) \end{bmatrix}. \]

Step 1: Write down the given function with three components

The given function can be expressed as: \[ f(u, v) = \begin{bmatrix} x(u, v) \\ y(u, v) \\ z(u, v) \end{bmatrix}= \begin{bmatrix} \cos (u)\cos (v) \\ \cos (u)\sin (v) \\ \sin (u) \end{bmatrix}. \]

Step 2: Find the partial derivatives with respect to u and v

Now, we have to calculate the partial derivatives of each function component with respect to both u and v. In total, we need 6 partial derivatives: Components with respect to u: \[ \frac{\partial x(u, v)}{\partial u} = -\sin(u)\cos(v), \quad \frac{\partial y(u, v)}{\partial u} = -\sin(u)\sin(v), \quad \frac{\partial z(u, v)}{\partial u} = \cos(u). \] Components with respect to v: \[ \frac{\partial x(u, v)}{\partial v} = -\cos(u)\sin(v), \quad \frac{\partial y(u, v)}{\partial v} = \cos(u)\cos(v), \quad \frac{\partial z(u, v)}{\partial v} = 0. \]

Step 3: Arrange the partial derivatives to form the Jacobian matrix

The Jacobian matrix for the given function is a 3x2 matrix in which the first column contains the partial derivatives with respect to u, and the second column has the partial derivatives with respect to v: \[ J(u, v) = \begin{bmatrix} \frac{\partial x(u, v)}{\partial u} & \frac{\partial x(u, v)}{\partial v} \\ \frac{\partial y(u, v)}{\partial u} & \frac{\partial y(u, v)}{\partial v} \\ \frac{\partial z(u, v)}{\partial u} & \frac{\partial z(u, v)}{\partial v} \end{bmatrix} = \begin{bmatrix} -\sin(u)\cos(v) & -\cos(u)\sin(v) \\ -\sin(u)\sin(v) & \cos(u)\cos(v) \\ \cos(u) & 0 \end{bmatrix}. \]

Step 4: Write down the differential of the vector function

Finally, we can write the differential of the vector function, f, as: \[ df(u, v) = J(u, v)\,d\mathbf{r} \] where d\mathbf{r} is a 2x1 column vector, representing the differentials of u and v: \[ d\mathbf{r} = \begin{bmatrix} d(u) \\ d(v) \end{bmatrix}. \] So, \[ df(u, v) = \begin{bmatrix} -\sin(u)\cos(v) & -\cos(u)\sin(v) \\ -\sin(u)\sin(v) & \cos(u)\cos(v) \\ \cos(u) & 0 \end{bmatrix} \begin{bmatrix} d(u) \\ d(v) \end{bmatrix}. \]

Key Concepts

Partial Derivatives

When we delve into functions with multiple variables, understanding partial derivatives becomes essential. A partial derivative measures how a function changes as only one variable is varied, holding the others constant. For the vector function given, \[f(u, v) = \begin{bmatrix} \cos (u)\cos (v) \ \cos (u)\sin (v) \ \sin (u) \end{bmatrix}\], we find partial derivatives with respect to \(u\) and \(v\).

This involves calculating how each component of the vector function changes:

• With respect to \(u\): these derivatives account for changes along the \(u\)-axis while holding \(v\) constant. For example, \(\frac{\partial x(u, v)}{\partial u}\) captures how \(x\) changes when \(u\) changes.
• With respect to \(v\): these derivatives examine variations along the \(v\)-axis with \(u\) being constant. It explains, for example, the sensitivity of \(y\) with changes in \(v\).

Computing these derivatives leads us to construct the Jacobian matrix, a key tool in multivariable calculus.

Vector Function

A vector function assigns a vector to each point in its domain. In the exercise, the function \(f: \mathbb{R}^{2} \rightarrow \mathbb{R}^{3}\) maps every pair of variables \((u, v)\) to a vector in three-dimensional space.

Such functions are vital in modeling systems with multiple outputs. Key components to understand:

• Dimensions: The domain \(\mathbb{R}^{2}\) indicates inputs \((u, v)\) are from a 2D space, while \(\mathbb{R}^{3}\) shows the output is a vector in 3D.
• Applications: Vector functions appear in physics, engineering, and computer graphics, modeling anything from mechanical forces to motion paths.

Understanding how these functions operate sets the foundation for exploring more complex systems and their behaviors.

Differential Calculus

Differential calculus focuses on the rate at which quantities change. When applied to multivariable functions, we deal with concepts like gradients, partial derivatives, and differentials.

In this context, finding the differential of \(f(u, v)\) involves calculating how small changes in \(u\) and \(v\) affect the function's output. We use the Jacobian matrix to express this in a linear form:\[df(u, v) = J(u, v)\,d\mathbf{r}\]
Where:

• Jacobian Matrix: Contains all partial derivatives, providing a linear approximation of how outputs change with respect to inputs.
• \(d\mathbf{r}\): Represents infinitesimal changes in \(u\) and \(v\).

This approach offers insights into how even tiny variations contribute to overall function changes.

Multivariable Calculus

Multivariable calculus extends regular calculus to functions with more than one variable. This involves studying concepts like partial derivatives, multiple integrals, and vector functions.

Key elements of multivariable calculus include:

• Multiple Variables: Functions can have several independent variables, like \(u\) and \(v\), influencing the output vector.
• Jacobian Matrix: Plays a crucial role in transformations, optimizing outputs relative to inputs, and helps in finding differentials.
• Real-World Applications: Used in fields like economics for modeling markets or in engineering for analyzing structures.

By understanding these principles, students can tackle complex problems involving multi-dimensional systems.