Problem 20 - Calculus Volume 3

Question

Find the gradient vector field of each function \(f\). $$ f(x, y, z)=x \cos \left(\frac{y}{z}\right) $$

Step-by-Step Solution

Answer

The gradient vector field is \( \nabla f(x, y, z) = \left( \cos \left( \frac{y}{z} \right), -\frac{x}{z} \sin \left( \frac{y}{z} \right), \frac{xy}{z^2} \sin \left( \frac{y}{z} \right) \right) \).

Step 1: Identify the Function

The function given is \( f(x, y, z) = x \cos \left( \frac{y}{z} \right) \). Our task is to find its gradient vector field.

Step 2: Understand the Gradient

The gradient (\( abla f \)) of a function \( f(x, y, z) \) in three dimensions is a vector field given by the partial derivatives with respect to \( x, y, \text{ and } z \). Specifically, \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \).

Step 3: Find Partial Derivative with respect to x

Calculate \( \frac{\partial f}{\partial x} \). Since \( x \) is outside the cosine function, its derivative is simply \(abla f = \cos \left( \frac{y}{z} \right) \).

Step 4: Find Partial Derivative with respect to y

To find \( \frac{\partial f}{\partial y} \), we apply the chain rule. Differentiating \( \cos \left( \frac{y}{z} \right) \) with respect to \( y \) gives \(-\frac{x}{z} \sin \left( \frac{y}{z} \right) \).

Step 5: Find Partial Derivative with respect to z

Differentiate with respect to \( z \), again using the chain rule and quotient rule where needed. The result is \( \frac{xy}{z^2} \sin \left( \frac{y}{z} \right) \), taking account of the negative sign from the derivative of \( \sin \).

Step 6: Combine the Components

Combine all the partial derivatives into the gradient vector field. \[ abla f(x, y, z) = \left( \cos \left( \frac{y}{z} \right), -\frac{x}{z} \sin \left( \frac{y}{z} \right), \frac{xy}{z^2} \sin \left( \frac{y}{z} \right) \right) \].

Key Concepts

Partial Derivatives

In multivariable calculus, partial derivatives are used to find how a function changes, as one of several variables changes, while other variables are held constant. For instance, in the function \( f(x, y, z) = x \cos\left(\frac{y}{z}\right) \), the partial derivative with respect to \(x\) examines how \(f\) changes if only \(x\) varies, leaving \(y\) and \(z\) constant.

To solve for the partial derivative \( \frac{\partial f}{\partial x} \), you treat \(x\) as the variable and the rest as constants. This results in \( \cos \left( \frac{y}{z} \right) \). Partial derivatives are pivotal in constructing a gradient vector, as they offer the rate of change along each axis:

• \( \frac{\partial f}{\partial x} \) accounts for changes along the \(x\)-axis.
• \( \frac{\partial f}{\partial y} \) accounts for changes along the \(y\)-axis.
• \( \frac{\partial f}{\partial z} \) accounts for changes along the \(z\)-axis.

Chain Rule

The chain rule is a fundamental tool in differentiation, especially when dealing with compositions of functions. It reveals how to differentiate a function that is nested within another function. For example, in the function \( f(x, y, z) = x \cos\left(\frac{y}{z}\right) \), we use the chain rule to differentiate with respect to \(y\) and \(z\).

While finding \( \frac{\partial f}{\partial y} \), the internal function \( \frac{y}{z} \) influences \( \cos \). By applying the chain rule, we derive the expression for \( \frac{\partial f}{\partial y} = -\frac{x}{z} \sin \left( \frac{y}{z} \right) \). Similarly, applying the chain rule for \( \frac{\partial f}{\partial z} \) tackles the quotient \( \frac{y}{z} \). Combining the rules, this derivative becomes \( \frac{xy}{z^2} \sin \left( \frac{y}{z} \right) \).

• Helps compute derivatives of nested functions.
• Essential for handling functions depending on other variables indirectly.

Multivariable Calculus

Multivariable calculus extends calculus to functions with more than one variable, like \( f(x, y, z) \). Unlike single-variable calculus, here you explore how changes in variables influence the function.

Key concepts include:

• Partial derivatives, which provide the rate of change for each variable.
• The gradient vector, indicating the direction and rate of fastest ascent.

Such functions are common in real-world scenarios, optimizing outputs depending on interdependent inputs.

Being proficient in multivariable calculus requires understanding how variables interact in complex environments.

Gradient in Three Dimensions

The concept of the gradient vector in three dimensions is essential in understanding how to identify the slope of a function in space. For a function \( f(x, y, z) \), the gradient \( abla f \) is a vector composed of all the partial derivatives.

The gradient represents the direction of steepest ascent and its magnitude tells us how steep that ascent is. In our problem, the gradient vector field of \( f(x, y, z) = x \cos\left(\frac{y}{z}\right) \) is:
\[abla f(x, y, z) = \left( \cos \left( \frac{y}{z} \right), -\frac{x}{z} \sin \left( \frac{y}{z} \right), \frac{xy}{z^2} \sin \left( \frac{y}{z} \right) \right)\]

• The x-component \( \cos \left( \frac{y}{z} \right) \) describes change in the x-axis.
• The y-component \( -\frac{x}{z} \sin \left( \frac{y}{z} \right) \) explains change in the y-axis.
• The z-component \( \frac{xy}{z^2} \sin \left( \frac{y}{z} \right) \) addresses change in the z-axis.

Understanding this gradient helps in fields requiring optimization, like physics and engineering.