Problem 12 - Applied Calculus

Question

Find the partial derivatives in problems. The variables are restricted to a domain on which the function is defined. \(\frac{\partial A}{\partial h}\) if \(A=\frac{1}{2}(a+b) h\)

Step-by-Step Solution

Answer

The partial derivative \( \frac{\partial A}{\partial h} = \frac{1}{2}(a+b) \).

Step 1: Understand the Problem

To find \( \frac{\partial A}{\partial h} \), we need to compute the partial derivative of the function \( A = \frac{1}{2}(a+b) h \) with respect to the variable \( h \). This means we'll treat \( a \) and \( b \) as constants.

Step 2: Apply the Partial Derivative

When taking a partial derivative with respect to \( h \), treat all other variables as constants. Thus, we differentiate \( A = \frac{1}{2}(a+b)h \) with respect to \( h \).

Step 3: Find the Derivative Expression

The expression \( A = \frac{1}{2}(a+b)h \) can be seen as a simple product of the constant \( \frac{1}{2}(a+b) \) and the variable \( h \). The partial derivative with respect to \( h \) is simply the coefficient of \( h \), which is \( \frac{1}{2}(a+b) \).

Step 4: Conclusion

Therefore, the partial derivative \( \frac{\partial A}{\partial h} = \frac{1}{2}(a+b) \). This represents the rate of change of the area \( A \) with respect to the height \( h \) in this formula.

Key Concepts

Multivariable Calculus

Multivariable calculus is an extension of calculus involving functions of multiple variables. In contrast to single-variable calculus, where functions depend on a single input, multivariable calculus tackles situations where numerous inputs are involved. This is crucial for understanding and modeling real-world phenomena that depend on several different factors simultaneously.

You'll often deal with functions like \(f(x, y, z)\) that depend on multiple variables. A key idea in multivariable calculus is analyzing how these functions change when just one of the inputs is altered while holding others constant. This is where **partial derivatives** come into play. They help in understanding the behavior of these functions in the multidimensional space.

• The function in the exercise is a straightforward example of a multivariable function, it has the form \(A = \frac{1}{2}(a+b)h\), where \(A\) is dependent on multiple variables \(a, b, h\).
• The goal is to understand how \(A\) changes specifically with respect to \(h\), showing practical use of partial derivatives in multivariable scenarios.
• In this context, \(a\) and \(b\) are treated as constants while analyzing the effect of \(h\).

Rate of Change

Rate of change is a fundamental concept in calculus that describes how a quantity changes over time or in relation to another variable. When dealing with multiple variables, partial derivatives offer a tool to evaluate this rate in relation to each variable independently.

In the given problem, the task is to find the partial derivative of the area \(A\) with respect to height \(h\). This can be understood as finding the rate at which the area changes as the height increases or decreases, while assuming the other parameters \(a\) and \(b\) remain constant.

• The partial derivative \(\frac{\partial A}{\partial h}\) gives us a single value which tells how much the area \(A\) increases for every one-unit increase in height \(h\).
• This metric is invaluable, especially when one needs to predict or optimize such changes in areas involving parameters like height, which might be commonplace in engineering or architecture.
• Understanding the rate of change via partial derivatives allows engineers and scientists to fine-tune designs and models much more effectively.

Differentiation Rules

Differentiation rules provide the backbone for finding derivatives and partial derivatives. These rules guide how we differentiate various types of functions, ensuring accurate and consistent results across different problems.

In this exercise, since the function is linear in \(h\), we apply basic rules of differentiation. The primary rule used here is that the derivative of \(ch\) with respect to \(h\) is \(c\), where \(c\) is a constant.

• By treating \(a\) and \(b\) as constants, the expression \(\frac{1}{2}(a+b)h\) simplifies, and its height-related part can be differentiated directly.
• The differentiation rule used here allows a straightforward computation of \(\frac{\partial A}{\partial h} = \frac{1}{2}(a+b)\).
• These rules are straightforward but powerful, serving as building blocks to tackle more complex functions seen in multivariable calculus.

Gaining mastery over differentiation rules enhances one's ability to handle more intricate mathematical exploration, thus broadening analytical skill sets.