Problem 41 - Advanced Engineering Mathematics

Question

Use the Chain Rule to find the indicated partial derivatives. $$ z=4 x-5 y^{2} ; \quad x=u^{4}-8 v^{3}, y=(2 u-v)^{2} ; \quad \frac{\partial z}{\partial u}, \frac{\partial z}{\partial v} $$

Step-by-Step Solution

Answer

\( \frac{\partial z}{\partial u} = 16u^3 - 40(2u-v)^3 \), \( \frac{\partial z}{\partial v} = -96v^2 + 20(2u-v)^3 \).

Step 1: Understand the Function Relationships

We have the function \( z = 4x - 5y^2 \) with intermediate functions \( x = u^4 - 8v^3 \) and \( y = (2u - v)^2 \). To find the partial derivatives \( \frac{\partial z}{\partial u} \) and \( \frac{\partial z}{\partial v} \), we will use the chain rule, identifying \( z \) as a function of \( x \) and \( y \) which themselves depend on \( u \) and \( v \).

Step 2: Apply the Chain Rule for \(\frac{\partial z}{\partial u}\)

Using the chain rule for partial differentiation, the derivative \( \frac{\partial z}{\partial u} \) can be written as \( \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u} \). Compute these partial derivatives:- \( \frac{\partial z}{\partial x} = 4 \)- \( \frac{\partial z}{\partial y} = -10y \)Now, \( \frac{\partial x}{\partial u} = 4u^3 \) and \( \frac{\partial y}{\partial u} = 4(2u - v) \).Substitute these into the chain rule expression:\[\frac{\partial z}{\partial u} = 4 \cdot 4u^3 + (-10y) \cdot 4(2u - v) \].

Step 3: Simplify \(\frac{\partial z}{\partial u}\)

First, simplify \( y = (2u - v)^2 \), so \( -10y = -10(2u - v)^2 \). Substitute back:\[\frac{\partial z}{\partial u} = 16u^3 - 40(2u - v)^2(2u - v) \]Simplify further:\[\frac{\partial z}{\partial u} = 16u^3 - 40(2u - v)^3 \].

Step 4: Apply the Chain Rule for \(\frac{\partial z}{\partial v}\)

Use the chain rule again for \( \frac{\partial z}{\partial v} \) as:\( \frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial v} \).Calculate the derivatives:- \( \frac{\partial x}{\partial v} = -24v^2 \)- \( \frac{\partial y}{\partial v} = -2(2u - v) \)Plug these into the expression:\[\frac{\partial z}{\partial v} = 4(-24v^2) + (-10y)(-2(2u - v)) \].

Step 5: Simplify \(\frac{\partial z}{\partial v}\)

Simplify the expression:\[\frac{\partial z}{\partial v} = -96v^2 + 20(2u - v)^2(2u - v) \].This becomes:\[\frac{\partial z}{\partial v} = -96v^2 + 20(2u - v)^3 \].

Key Concepts

Chain Rule

The Chain Rule is a fundamental tool in calculus, especially when dealing with complex functions of several variables. It allows us to differentiate a composite function by breaking it down into simpler parts. In the context of partial derivatives, it is applied when we have functions that are themselves functions of other variables. For our specific exercise, we have the function \( z = 4x - 5y^2 \) where \( x \) and \( y \) are each functions of \( u \) and \( v \).

To find the partial derivatives \( \frac{\partial z}{\partial u} \) and \( \frac{\partial z}{\partial v} \), the chain rule helps by providing a systematic approach:

• Identify the outer function, \( z \), and the intermediate functions, \( x \) and \( y \).
• Calculate the derivative of \( z \) with respect to \( x \) and \( y \).
• Determine how \( x \) and \( y \) change with \( u \) and \( v \), and use these to find the overall change of \( z \) with \( u \) and \( v \).

This method breaks down the differentiation process into smaller, manageable steps, making it easier to compute even complex derivatives.

Multivariable Calculus

Multivariable Calculus extends the concepts of single-variable calculus to functions of more than one variable. It involves studying derivatives and integrals in higher dimensions. This branch of calculus is crucial because it provides the mathematical framework for analyzing real-world systems with multiple inputs and outputs. In our exercise, we're dealing with a function \( z = 4x - 5y^2 \) where both \( x \) and \( y \) are functions of variables \( u \) and \( v \).

To address problems in multivariable calculus, you often work with:

• Partial Derivatives, which measure the rate of change of a function with respect to one of its variables, holding others constant.
• Concepts such as gradient vectors, which give the direction of greatest increase of a function.

In our specific case, calculating \( \frac{\partial z}{\partial u} \) and \( \frac{\partial z}{\partial v} \), involves partial derivatives and requires understanding how each variable influences the overall function. Multivariable calculus offers the tools necessary to systematically manage such complexity.

Intermediate Functions

Intermediate functions serve as the building blocks in a chain of functions, allowing for complex relationships between variables to be methodically analyzed. In problems requiring the Chain Rule, functions like \( x = u^4 - 8v^3 \) and \( y = (2u - v)^2 \) are considered intermediate.

Such functions provide the crucial linkage in the relationship between the dependent variable \( z \) and the independent variables \( u \) and \( v \). Understanding the role of intermediates is essential in

• Breaking down the problem into simpler derivatives that can be solved step by step.
• Systematically applying the chain rule, as each layer of function affects the differentiation of the more complex outer function.

By isolating these intermediate functions in our exercise, we can clearly outline how changes in \( u \) and \( v \) affect \( x \) and \( y \), and consequently, \( z \). Recognizing the interdependencies between these functions is key to successfully applying the chain rule and solving multivariable calculus problems.