Problem 47 - Advanced Engineering Mathematics

Question

Assume that \(f\) and \(g\) are differentiable functions of two variables. Prove the given identity. $$ \nabla(f g)=f \nabla g+g \nabla f $$

Step-by-Step Solution

Answer

The identity \(\nabla(fg) = f \nabla g + g \nabla f\) is proven using the product rule for differentiation.

Step 1: Recall the Gradient Definition

The gradient of a function \(u(x, y)\) is a vector of its partial derivatives: \(abla u = \left( \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y} \right)\).

Step 2: Express the Gradient of the Product

For the function \(fg\), the gradient is \(abla(fg) = \left( \frac{\partial (fg)}{\partial x}, \frac{\partial (fg)}{\partial y} \right)\).

Step 3: Apply the Product Rule for Partial Derivatives

Using the product rule, the partial derivative of \(fg\) with respect to \(x\) is \(f \frac{\partial g}{\partial x} + g \frac{\partial f}{\partial x}\) and with respect to \(y\) is \(f \frac{\partial g}{\partial y} + g \frac{\partial f}{\partial y}\).

Step 4: Write the Gradient Vector for the Product

From the previous step, \(abla(fg) = \left( f \frac{\partial g}{\partial x} + g \frac{\partial f}{\partial x}, f \frac{\partial g}{\partial y} + g \frac{\partial f}{\partial y} \right)\).

Step 5: Compute the Expression for f∇g + g∇f

Calculate \(f abla g\) which is \(\left( f \frac{\partial g}{\partial x}, f \frac{\partial g}{\partial y} \right)\) and \(g abla f\) which is \(\left( g \frac{\partial f}{\partial x}, g \frac{\partial f}{\partial y} \right)\).

Step 6: Add the Vectors from Step 5

Add the vectors from \(f abla g\) and \(g abla f\): \(\left( f \frac{\partial g}{\partial x} + g \frac{\partial f}{\partial x}, f \frac{\partial g}{\partial y} + g \frac{\partial f}{\partial y} \right)\).

Step 7: Compare Both Expressions

The expression for \(abla(fg)\) from Step 4 is identical to the expression from Step 6. Therefore, \(abla(fg) = f abla g + g abla f\).

Key Concepts

Differentiable Functions

Differentiable functions are a fundamental concept in calculus. These functions have derivatives at every point in their domain. Here, derivatives capture the rate of change of the function. Understanding when a function is differentiable is important in many areas of math and its applications. Differentiability requires the function to be smooth, with no sharp corners or breaks. This means:

• The function can be continuously traced without lifting a pen.
• Both the first and higher-order derivatives exist.

In practice, checking the differentiability over multiple variables often involves looking at partial derivatives. Differentiable functions allow us to apply tools like gradients and use calculus rules effectively, such as the product and chain rules.

Partial Derivatives

When dealing with functions of multiple variables, like in our original problem, partial derivatives are your go-to tool. They represent the rate of change of a function with respect to one variable, holding the other variables constant. For example, if you have a function of two variables, say, \(u(x, y)\), the partial derivatives are represented as \(\frac{\partial u}{\partial x}\) and \(\frac{\partial u}{\partial y}\). These tell us how \(u\) changes as an individual variable changes. Partial derivatives are quite similar to regular derivatives but they focus on just one dimension, or variable, at a time.In vector calculus, partial derivatives allow us to construct concepts like the gradient, which is crucial for analyzing multi-variable functions across their entire domain.

• Use them to understand the local behavior of functions in each direction.
• Essential for optimization and understanding function topography.

Product Rule

The product rule is a differentiation rule essential for finding the derivative of a product of two functions. It states that the derivative of two multiplied functions \(f(x)\cdot g(x)\) can be found with the formula:\[ (fg)' = f'g + fg' \]In the context of partial derivatives, this becomes essential when both functions \(f\) and \(g\) are of two variables. Whether you're differentiating with respect to \(x\) or \(y\), the rule applies similarly:

• With respect to \(x\): \(f \frac{\partial g}{\partial x} + g \frac{\partial f}{\partial x}\).
• With respect to \(y\): \(f \frac{\partial g}{\partial y} + g \frac{\partial f}{\partial y}\).

The product rule is a cornerstone of vector calculus, enabling manipulation and calculation of gradients for product functions, ultimately proving vital identities and theorems.

Vector Calculus

Vector calculus is a vital extension of calculus that deals with functions of multiple variables and is grounded in vectors. It covers operations like gradient, divergence, and curl, which are integral in physics and engineering.The gradient, in particular, is key, representing the direction and magnitude of the steepest increase of a scalar function. Written as \(abla f\), it consists of the partial derivatives and maps changes in multidimensional space:- \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \)Vector calculus allows us not only to describe physical phenomena that vary in space (like temperature or fluid flow) but also to solve complex mathematical problems involving multiple dimensions.

• It bridges scalar and vector fields, enhancing our tools for mathematical descriptions.
• Utilizes gradients to find maximum and minimum values of functions efficiently through methods like gradient descent.