Problem 15 - Calculus Volume 3

Question

Find the gradient vector field of each function \(f\). $$ f(x, y)=x \sin y+\cos y $$

Step-by-Step Solution

Answer

The gradient vector field is \( (\sin y, x \cos y - \sin y) \).

Step 1: Understanding the Gradient

The gradient of a scalar function is a vector field. For a function of two variables, \( f(x, y) \), the gradient is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). This means we will take the partial derivative of the function with respect to both \( x \) and \( y \).

Step 2: Partial Derivative with respect to x

To find \( \frac{\partial f}{\partial x} \), differentiate \( f(x, y) = x \sin y + \cos y \) with respect to \( x \). Here, \( x \) only appears in the term \( x \sin y \), and \( \sin y \) is treated as a constant:\[ \frac{\partial}{\partial x}(x \sin y) = \sin y \]. The term \( \cos y \) does not involve \( x \), so its derivative is \( 0 \). Thus, \( \frac{\partial f}{\partial x} = \sin y \).

Step 3: Partial Derivative with respect to y

To find \( \frac{\partial f}{\partial y} \), differentiate \( f(x, y) = x \sin y + \cos y \) with respect to \( y \). For \( x \sin y \), treat \( x \) as a constant and differentiate \( \sin y \):\[ \frac{\partial}{\partial y}(x \sin y) = x \cos y \].For \( \cos y \), the derivative is:\[ \frac{\partial}{\partial y}(\cos y) = -\sin y \].Thus, \( \frac{\partial f}{\partial y} = x \cos y - \sin y \).

Step 4: Writing the Gradient Vector Field

Combine the results from Steps 2 and 3 to create the gradient vector field. The gradient of \( f(x, y) \) is \( abla f(x, y) = (\sin y, x \cos y - \sin y) \).

Key Concepts

Partial Derivatives

Partial derivatives involve taking the derivative of a function with more than one variable, one variable at a time. It helps us understand how the function changes as just one of the variables changes while keeping others constant.
Here’s how it works:

• We identify the variable with respect to which we want to differentiate. For instance, if given a function \( f(x, y) \), this could be \( x \) or \( y \).
• Other variables are treated as constants during differentiation.

For the function \( f(x, y) = x \sin y + \cos y \), the partial derivative with respect to \( x \) gives us \( \sin y \) because \( \sin y \) is like a constant with respect to \( x \). Partial derivatives are useful for constructing gradient vector fields and understanding the direction of rate changes in multi-variable functions.

Scalar Function

A scalar function is simply a function that associates a single value (a scalar) with every point in some space. This concept is important as it serves as the basis for calculating gradients and performing other vector calculus operations.
In our example, the scalar function is \( f(x, y) = x \sin y + \cos y \). It takes two inputs (\( x \) and \( y \)) and outputs one scalar value.
Key characteristics of scalar functions include:

• Scalar outputs, meaning the result is a singular numerical value rather than a vector.
• Use in computing gradients, which transform the scalar function into a vector field.

Scalar functions are foundational in calculus and physics, helping us analyze diverse phenomena through variable relationships and transformations.

Vector Field

Vector fields represent mathematical constructs which assign a vector to every point in the space. These vectors might represent various physical and mathematical phenomena, like the gradient of temperature in a room or the velocity of a fluid.
In the case of our scalar function \( f(x, y) = x \sin y + \cos y \), its gradient is a vector field given by \( abla f = (\sin y, x \cos y - \sin y) \).
Important features of vector fields:

• They allow us to visualize the direction and magnitude of quantities which change over space.
• Derived from scalar functions through gradients for detailed spatial analysis.

Understanding vector fields is crucial in physics and engineering, where they are used to describe a wide array of dynamic systems.

Differentiation with respect to variables

Differentiation with respect to variables involves calculating the rate of change of a function concerning one of its variables while keeping the other variables constant. This process helps us understand how the function’s output changes in various directions across its domain.
In our given function \( f(x, y) = x \sin y + \cos y \), differentiating with respect to \( x \) gives \( \sin y \), and with respect to \( y \) gives \( x \cos y - \sin y \).
By following these steps:

• Choose the variable of interest for differentiation.
• Treat other variables as constants to simplify the differentiation process.
• Compute partial derivatives sequentially to form components of the gradient.

These concepts are pivotal in multivariable calculus, enabling solutions to more complex problems involving multiple changing variables.