Problem 22 - Calculus and its applications

Question

Find \(\frac{\partial f}{\partial b}\) and \(\frac{\partial f}{\partial m}\). $$\begin{aligned}f(b, m)=& 5 m^{2}-m b^{2}-3 b+(2 m+b-8)^{2} \\\&+(3 m+b-9)^{2}\end{aligned}$$

Step-by-Step Solution

Answer

\( \frac{\partial f}{\partial b} = -2mb + 10m + 4b - 37 \), \( \frac{\partial f}{\partial m} = 36m - b^2 + 10b - 86 \).

Step 1: Write Down the Function

We need to find the partial derivatives of the function \( f(b, m) = 5m^2 - mb^2 - 3b + (2m + b - 8)^2 + (3m + b - 9)^2 \) with respect to \( b \) and \( m \).

Step 2: Differentiate with Respect to \( b \)

To find \( \frac{\partial f}{\partial b} \), apply the differentiation rules treating \( m \) as a constant. Differentiate each term individually:- The term \( -mb^2 \) becomes \( -2mb \).- The term \( -3b \) becomes \( -3 \).- Differentiate \( (2m + b - 8)^2 \) using the chain rule: \( 2(2m + b - 8) \cdot 1 \) is \( 2(2m + b - 8) \).- Differentiate \( (3m + b - 9)^2 \) using the chain rule: \( 2(3m + b - 9) \cdot 1 \) is \( 2(3m + b - 9) \).The derivative \( \frac{\partial f}{\partial b} = -2mb - 3 + 2(2m + b - 8) + 2(3m + b - 9) \).

Step 3: Simplify \( \frac{\partial f}{\partial b} \)

Simplify the expression:- Combine like terms: \( -2mb - 3 + 4m + 2b - 16 + 6m + 2b - 18 \).- This simplifies to \( (4m + 6m - 2mb) + (2b + 2b) - 3 - 16 - 18 \).- The result is \( -2mb + 10m + 4b - 37 \).

Step 4: Differentiate with Respect to \( m \)

To find \( \frac{\partial f}{\partial m} \), differentiate treating \( b \) as a constant:- The term \( 5m^2 \) becomes \( 10m \).- The term \( -mb^2 \) becomes \( -b^2 \).- Differentiate \( (2m + b - 8)^2 \) using the chain rule: \( 2(2m + b - 8) \cdot 2 \) becomes \( 4(2m + b - 8) \).- Differentiate \( (3m + b - 9)^2 \) using the chain rule: \( 2(3m + b - 9) \cdot 3 \) becomes \( 6(3m + b - 9) \).The derivative \( \frac{\partial f}{\partial m} = 10m - b^2 + 4(2m + b - 8) + 6(3m + b - 9) \).

Step 5: Simplify \( \frac{\partial f}{\partial m} \)

Simplify the expression:- Expand and combine terms: \( 10m - b^2 + 8m + 4b - 32 + 18m + 6b - 54 \).- This simplifies to \( (10m + 8m + 18m) + (-b^2) + (4b + 6b) - 32 - 54 \).- The result is \( 36m - b^2 + 10b - 86 \).

Key Concepts

Multivariable Calculus

Understanding multivariable calculus involves dealing with functions that have more than one variable. In our exercise, the function is defined as \( f(b,m) = 5m^2 - mb^2 - 3b + (2m + b - 8)^2 + (3m + b - 9)^2 \). This means our function, \( f \), depends on variables \( b \) and \( m \).

In single-variable calculus, you find a derivative to determine how a function changes as its variable changes. In multivariable calculus, partial derivatives help us understand how the function changes with respect to one variable, keeping the other constant. This concept is crucial in fields where multiple factors influence outcomes, like physics, economics, and engineering.

• The first variable, \( b \), can be held constant while examining changes with \( m \) and vice versa.
• Each variable's partial derivative shows its own direct influence on the function's value, independent of the other variables at any moment.

This foundational understanding sets the stage for using techniques like the chain rule and differentiation rules in more complex derivative problems.

Chain Rule

The chain rule in calculus is essential when differentiating composite functions. In our example, we have terms like \((2m + b - 8)^2\) and \((3m + b - 9)^2\) in the function \( f(b, m) \). These are composite functions because they contain expressions in \( m \) and \( b \) raised to a power.

When finding a partial derivative of these terms, we apply the chain rule. This involves taking the derivative of the outer function and multiplying it by the derivative of the inner function:

• For \((2m + b - 8)^2\), the derivative is found by first differentiating \((u^2)\) to get \(2u\) where \(u = 2m + b - 8\), then multiplying by the derivative of \(u\) with respect to the chosen variable.
• The expression \((3m + b - 9)^2\) is handled similarly, emphasizing that regardless of the complexity of a composite function, the chain rule methodically breaks down its differentiation into manageable steps.

This approach simplifies the problem, reducing errors and making computations clearer, especially in multivariable calculus.

Differentiation Rules

In the context of our exercise, understanding differentiation rules is key to finding partial derivatives of the given function. We apply basic rules of differentiation to each term of the function \( f(b, m) \).

For example:

• The term \( 5m^2 \) becomes \( 10m \) when differentiating with respect to \( m \), using the power rule which states that the derivative of \( ax^n \) is \( n \, ax^{n-1} \).
• The term \( -mb^2 \) becomes \( -2mb \) when differentiating with respect to \( b \), employing both the power rule and treating \( m \) as a constant.
• Constant terms like \(-3b\) become \(-3\) when differentiated with respect to \( b \). If we differentiate it with respect to \( m \), it will be zero since it does not contain \( m \).

These rules highlight the systematic approach to tackling terms independently, an efficient strategy for solving more comprehensive derivative problems. Comprehending and applying such differentiation rules are vital, not just for theoretical understanding but also for practical problem-solving in mathematics and its applications.