Problem 8 - Calculus for Biology and Medicine

Question

In Problems \(1-8\), find the gradient of each function. $$ f(x, y)=\cos \left(3 x^{2}-2 y^{2}\right) $$

Step-by-Step Solution

Answer

The gradient is \( \nabla f(x, y) = \left(-6x \sin(3x^2 - 2y^2), 4y \sin(3x^2 - 2y^2)\right) \).

Step 1: Understand the Problem

We need to find the gradient of the function given by \( f(x, y) = \cos(3x^2 - 2y^2) \). The gradient is a vector consisting of partial derivatives with respect to \( x \) and \( y \).

Step 2: Compute Partial Derivative with respect to x

To find \( \frac{\partial f}{\partial x} \), use the chain rule. The derivative of \( \cos(u) \) with respect to \( u \) is \( -\sin(u) \). Then multiply by the derivative of \( u = 3x^2 - 2y^2 \) with respect to \( x \), which is \( 6x \):\[\frac{\partial f}{\partial x} = -\sin(3x^2 - 2y^2) \cdot 6x = -6x \sin(3x^2 - 2y^2)\]

Step 3: Compute Partial Derivative with respect to y

To find \( \frac{\partial f}{\partial y} \), use the chain rule similarly. The derivative of \( u = 3x^2 - 2y^2 \) with respect to \( y \) is \( -4y \):\[\frac{\partial f}{\partial y} = -\sin(3x^2 - 2y^2) \cdot (-4y) = 4y \sin(3x^2 - 2y^2)\]

Step 4: Write the Gradient Vector

Combine these partial derivatives to get the gradient vector:\[abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = \left( -6x \sin(3x^2 - 2y^2), 4y \sin(3x^2 - 2y^2) \right)\]

Key Concepts

Partial Derivatives

Partial derivatives are crucial when dealing with multivariable functions, such as our given function \( f(x, y) = \cos(3x^2 - 2y^2) \). Instead of taking the derivative with respect to one single variable, partial derivatives help us explore how the function changes with respect to each variable individually.

For example, the partial derivative \( \frac{\partial f}{\partial x} \) examines the rate of change of the function with respect to \( x \) while keeping \( y \) constant. Similarly, \( \frac{\partial f}{\partial y} \) does the same with respect to \( y \).

• Performing Partial Derivatives: In taking partial derivatives, treat other variables as constants and follow conventional differentiation rules.
• Importance: Understanding the behavior in multi-variable systems, especially when optimizing and problem solving in various scientific and engineering fields.

Chain Rule

The chain rule is a fundamental calculus concept used when differentiating composite functions. It allows us to "chain" together the derivatives of nested functions. In our exercise, it's crucial because the function \( f(x, y) = \cos(3x^2 - 2y^2) \) is composed of an outer function (cosine) and an inner function \( u = 3x^2 - 2y^2 \).

Here's how the chain rule works in this context:

• Outer Function Derivative: Differentiate the outer function \( \cos(u) \), which gives \( -\sin(u) \).
• Inner Function Derivative: Differentiate the inner function \( u \) with respect to the variable of interest \( x \) or \( y \), yielding derivatives like \( 6x \) or \(-4y\).
• Combine Them: Multiply these derivatives to get the final result, exemplified by the calculation for \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).

Understanding and applying the chain rule correctly is key to handling complex functions effectively.

Gradient Vector

The gradient vector is a powerful tool in calculus, especially for functions with two or more variables. It provides a way to consolidate information about the rate of change of a function with respect to each of its variables. In simpler terms, a gradient vector generalizes the concept of a derivative into a multidimensional context.

For a function like \( f(x, y) = \cos(3x^2 - 2y^2) \):

• Forming the Gradient: The gradient is formed by taking the partial derivatives with respect to each variable. In our example, the gradient vector is \( abla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\).
• Applications: The gradient vector points in the direction of the greatest rate of increase of the function. This is particularly useful in optimization tasks and understanding the behavior of surfaces or fields in physics.
• Visual Understanding: Visualizing the gradient vector as an arrow on a graph helps in picturing how the function behaves at and around any given point \((x, y)\).

By leveraging the concept of the gradient vector, we gain insights into how to efficiently navigate and solve complex problems in real-world situations.