Problem 40 - Advanced Engineering Mathematics

Question

Use the Chain Rule to find the indicated partial derivatives. $$ z=u^{2} \cos 4 v ; u=x^{2} y^{3}, v=x^{3}+y^{3} ; \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y} $$

Step-by-Step Solution

Answer

\( \frac{\partial z}{\partial x} = 4x^3 y^6 \cos(4(x^3 + y^3)) - 12x^4 y^6 \sin(4(x^3 + y^3)) \), \( \frac{\partial z}{\partial y} = 6x^4 y^5 \cos(4(x^3 + y^3)) - 12x^4 y^8 \sin(4(x^3 + y^3)) \)

Step 1: Rewrite the Function

Given the function \[z = u^2 \cos(4v)\]where \[u = x^2 y^3, \quad v = x^3 + y^3\]. To find \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \), we first need to express \( u \) and \( v \) in terms of \( x \) and \( y \).

Step 2: Determine Partial Derivatives of Intermediate Variables

Calculate the partial derivatives of \( u \) and \( v \) with respect to \( x \) and \( y \):\[\frac{\partial u}{\partial x} = 2xy^3, \quad \frac{\partial u}{\partial y} = 3x^2y^2\]\[\frac{\partial v}{\partial x} = 3x^2, \quad \frac{\partial v}{\partial y} = 3y^2\]

Step 3: Apply Chain Rule for \( \frac{\partial z}{\partial x} \)

Using the chain rule for multiple variables, we have: \[\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial x} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial x}\]First, calculate the partial derivatives of \( z \):\[\frac{\partial z}{\partial u} = 2u \cos(4v), \quad \frac{\partial z}{\partial v} = -4u^2 \sin(4v)\]Substitute in the derivatives found earlier:\[\frac{\partial z}{\partial x} = (2u \cos(4v))(2xy^3) + (-4u^2 \sin(4v))(3x^2)\]Simplify to get:\[\frac{\partial z}{\partial x} = 4u x y^3 \cos(4v) - 12u^2 x^2 \sin(4v)\]

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

Using the chain rule for \( \frac{\partial z}{\partial y} \):\[\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u}\frac{\partial u}{\partial y} + \frac{\partial z}{\partial v}\frac{\partial v}{\partial y}\]Substitute the partial derivatives:\[\frac{\partial z}{\partial y} = (2u \cos(4v))(3x^2y^2) + (-4u^2 \sin(4v))(3y^2)\]Simplify:\[\frac{\partial z}{\partial y} = 6u x^2 y^2 \cos(4v) - 12u^2 y^2 \sin(4v)\]

Step 5: Substitute Expressions for u and v

Now substitute \( u = x^2 y^3 \) and \( v = x^3 + y^3 \) into our partial derivatives:\[\frac{\partial z}{\partial x} = 4(x^2 y^3) x y^3 \cos(4(x^3 + y^3)) - 12(x^2 y^3)^2 x^2 \sin(4(x^3 + y^3))\]\[\frac{\partial z}{\partial y} = 6(x^2 y^3) x^2 y^2 \cos(4(x^3 + y^3)) - 12(x^2 y^3)^2 y^2 \sin(4(x^3 + y^3))\]

Step 6: Final Expressions

The final expressions for the partial derivatives are:\[\frac{\partial z}{\partial x} = 4x^3 y^6 \cos(4(x^3 + y^3)) - 12x^4 y^6 \sin(4(x^3 + y^3))\]\[\frac{\partial z}{\partial y} = 6x^4 y^5 \cos(4(x^3 + y^3)) - 12x^4 y^8 \sin(4(x^3 + y^3))\]

Key Concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus used to differentiate composite functions. In the context of partial derivatives, it allows us to find how a change in one variable affects another variable through several intermediate variables. Imagine you have a function defined in terms of other functions, like our function where \( z = u^2 \cos(4v) \) with \( u \) and \( v \) expressed in terms of \( x \) and \( y \). The Chain Rule helps find the derivative of \( z \) with respect to these variables when they are not directly in the form of \( x \) or \( y \).
Here’s a simple way to think of the Chain Rule when applied to multiple variables:

• You first take the partial derivative of the outer function with respect to the intermediate variables.
• Then multiply it by the partial derivative of these intermediate variables with respect to the original variables.
• Finally, you sum up those products if there are more than one intermediate variable affecting the outcome.

This method allows for efficient calculations of derivatives in complex systems that involve multiple layers of variables.

multivariable calculus

Multivariable calculus is an extension of single-variable calculus to functions with more than one variable. When dealing with functions like \( z = u^2 \cos(4v) \), it’s important to consider how each variable contributes to the behavior of the function. Here, \( z \), \( u \), and \( v \) are all dependent on \( x \) and \( y \).
In this realm of calculus, you will deal with concepts like gradients, partial derivatives, and multiple integrals. These concepts aim to extend calculus results from one-dimensional space to multidimensional ones. It means understanding how a tiny change in one dimension affects the overall system.
Here's why multivariable calculus is essential:

• It helps describe systems where multiple factors influence outcomes, like physical phenomena in physics or chemistry.
• It allows us to develop methods for optimization, like finding the highest point on a surface defined by two independent variables.

In essence, knowing multivariable calculus enables you to tackle more complex, real-world problems where many variables are at play.

intermediate variables

In multivariable calculus, intermediate variables like \( u = x^2 y^3 \) and \( v = x^3 + y^3 \) help break down complicated functions into easier parts. They act as stepping stones to relate dependent and independent variables, making the problem more manageable.
Here's how they function in a calculus problem:

• Break down the function into layers, where intermediate variables exist between the input (\( x \) and \( y \)) and the output (\( z \)).
• Calculate how a change in each independent variable affects these intermediates.
• Use them within the Chain Rule to find the derivative efficiently.

Intermediate variables are invaluable in simplification because they reduce the complexity of differentiating composite functions directly. This approach not only simplifies calculation but also facilitates better understanding and analysis of how changes in one part of the system affect the whole.