Problem 6 - APEX Calculus

Question

In Exercises \(5-8,\) find the total differential \(d z\). $$ z=\left(2 x^{2}+3 y\right)^{2} $$

Step-by-Step Solution

Answer

The total differential is \( dz = 8x(2x^2 + 3y) \, dx + 6(2x^2 + 3y) \, dy \).

Step 1: Identify the Function

The function given is \( z = (2x^2 + 3y)^2 \). We need to find the total differential \( dz \). For this, we must first identify the dependent and independent variables. Here, \( z \) depends on \( x \) and \( y \).

Step 2: Differentiate with respect to x

To find the partial derivative of \( z \) with respect to \( x \), apply the chain rule to \( z = (2x^2 + 3y)^2 \):\[ \frac{\partial z}{\partial x} = 2(2x^2 + 3y) \cdot (\frac{d}{dx}(2x^2 + 3y)) = 2(2x^2 + 3y) \cdot 4x = 8x(2x^2 + 3y) \]

Step 3: Differentiate with respect to y

Now, differentiate \( z \) with respect to \( y \) using the chain rule:\[ \frac{\partial z}{\partial y} = 2(2x^2 + 3y) \cdot (\frac{d}{dy}(2x^2 + 3y)) = 2(2x^2 + 3y) \cdot 3 = 6(2x^2 + 3y) \]

Step 4: Write the Total Differential

The total differential \( dz \) is given by:\[ dz = \frac{\partial z}{\partial x} \, dx + \frac{\partial z}{\partial y} \, dy \]Substitute the expressions we found:\[ dz = 8x(2x^2 + 3y) \, dx + 6(2x^2 + 3y) \, dy \]

Key Concepts

Partial Derivatives

When you're dealing with functions of multiple variables, like our function for the variable \(z\), you use partial derivatives to understand how \(z\) changes with respect to one variable while holding the others constant. For our exercise, \(z\) is a function of both \(x\) and \(y\). Partial derivatives allow us to capture the influence of each individual variable on \(z\).

• To find \(\frac{\partial z}{\partial x}\), treat \(y\) as constant. Here, you apply the chain rule to decompose the job into manageable parts.
• Similarly, \(\frac{\partial z}{\partial y}\) involves treating \(x\) as constant and applying the chain rule to identify how \(z\) changes when only \(y\) varies.

Using these partial derivatives, you can see what happens to \(z\) when either \(x\) or \(y\) shifts a little.

Chain Rule

The chain rule is an essential tool in calculus, especially when you're dealing with composite functions. In our function \(z = (2x^2 + 3y)^2\), it's the squaring aspect that makes it composite.
When employing the chain rule:

• You first differentiate the outside function—here, squaring.
• Then, you multiply by the derivative of the inside function—\(2x^2 + 3y\).

This process helps unravel complicated expressions by breaking them down into simpler parts, allowing for straightforward differentiation.

Multivariable Calculus

Multivariable calculus extends the concepts of basic calculus into more than one dimension. When functions depend on several variables, it becomes essential to understand how changes in each of them affect the function.

• The total differential \(dz\) gives the combined effect of small changes in both \(x\) and \(y\) on \(z\).
• It's represented as: \[ dz = \frac{\partial z}{\partial x} \, dx + \frac{\partial z}{\partial y} \, dy \]

This offers crucial insights, particularly in fields where multiple conditions or parameters dynamically interact, like physics or economics.

Calculus

Calculus, in its essence, is the mathematics of change. It provides techniques for finding derivatives, which help analyze rates of change in various applications. With two key branches, differential calculus focuses on the concept of differentiation, which is the core of our exercise.

• Using differentiation, you can handle curves instead of just straight lines, as normal algebra would.
• Specifically, you'll find rates at which any function increases or decreases.

In our context, calculus helps calculate how a complex surface, defined by \(z\), changes as each input variable increases or decreases.