Problem 29 - Calculus Early Transcendentals

Question

Use a tree diagram to write the required Chain Rule formula. \(u=f(v),\) where \(v=g(w, x, y), w=h(z), x=p(t, z),\) and \(y=q(t, z) .\) Find \(\partial u / \partial z\)

Step-by-Step Solution

Answer

Question: Calculate the derivative of u with respect to z, given that u depends on v, v depends on w, x, and y, w depends on z, x depends on t and z, and y depends on t and z. Answer: To calculate the derivative of u with respect to z, we can use the Chain Rule formula: $$ \frac{\partial u}{\partial z} = \frac{\partial f}{\partial v} \left( \frac{\partial g}{\partial w} \frac{\partial h}{\partial z} + \frac{\partial g}{\partial x} \frac{\partial p}{\partial z} + \frac{\partial g}{\partial y} \frac{\partial q}{\partial z} \right) $$

Step 1: Draw the tree diagram corresponding to the given problem

Draw a tree diagram with the variable u at the top, and down the branch to v, w, x, and y. From the exercise, follow their dependencies as follows: - u depends on v (u = f(v)) - v depends on w, x, and y (v = g(w, x, y)) - w depends on z (w = h(z)) - x depends on t and z (x = p(t, z)) - y depends on t and z (y = q(t, z)) This tree diagram allows us to visualize the dependencies between the variables and will guide us in applying the Chain Rule.

Step 2: Writing the Chain Rule formula for the desired derivative

We want to find the derivative of u with respect to z, which means \(\frac{\partial u}{\partial z}\). We can use the Chain Rule to write the formula by following the dependencies down the tree. We start with u, which depends on v, and we make our way to z through w, x, and y: $$ \frac{\partial u}{\partial z} = \frac{\partial f}{\partial v} \left( \frac{\partial g}{\partial w} \frac{\partial h}{\partial z} + \frac{\partial g}{\partial x} \frac{\partial p}{\partial z} + \frac{\partial g}{\partial y} \frac{\partial q}{\partial z} \right) $$ The formula was derived by using the Chain Rule for each variable dependency down the tree. This is the required Chain Rule formula to calculate \(\frac{\partial u}{\partial z}\).

Key Concepts

Tree Diagram

A tree diagram is a fantastic visual tool that eases the understanding of relationships between variables in complex scenarios. Imagine it as a branching network where each node represents a variable and branches show dependencies. In our example, starting from the top, the variable \( u \) sits at the peak. This node connects to \( v \) because \( u = f(v) \). Below \( v \), the branches extend to \( w \), \( x \), and \( y \) since \( v = g(w, x, y) \). Going even further, we see:

• \( w \) depends on \( z \) (\( w = h(z) \))
• \( x \) depends on \( t \) and \( z \) (\( x = p(t, z) \))
• \( y \) relies on \( t \) and \( z \) (\( y = q(t, z) \))

Drawing this tree diagram helps by providing a clear roadmap for applying the Chain Rule, guiding you step-by-step from \( u \) down through \( z \). This path will reveal how changes in \( z \) influence \( u \) via these dependencies.

Partial Derivatives

Understanding partial derivatives is essential when dealing with functions of several variables. A partial derivative represents how a function changes as one specific variable changes, while the others are held constant.
In our problem, \[\left( \frac{\partial u}{\partial z} \right)\] is what we're after. It signifies how the variable \( u \) changes as \( z \) changes, with other variables being fixed. By following the chain of dependencies from \( u \) to \( z \), we apply the Chain Rule at each level.
Here's the magic: you break down the derivative into smaller, manageable pieces, starting at the top of the tree:

• Differentiate \( f \) with respect to \( v \).
• Then move lower, differentiating \( g \) concerning \( w \), \( x \), and \( y \).
• Each subsequent branch needs differentiation: \( h \), \( p \), and \( q \) concerning \( z \).

This careful breakdown provides you the comprehensive derivative needed, where each small "change" stacks up to illustrate the complete \( \frac{\partial u}{\partial z} \).

Multivariable Calculus

Multivariable calculus provides the tools to analyze and understand functions with more than one variable. While single-variable calculus deals with functions like \( f(x) \), multivariable calculus extends this idea to encompass functions such as \( g(w, x, y) \).
Think of these functions as taking input values from a multi-dimensional space and mapping them to real numbers. This makes calculus in multiple dimensions both richer and more challenging due to the varied interactions between variables.

• Key concepts include partial derivatives, which look at how functions change with respect to any single variable out of many.
• The Chain Rule, another vital concept, helps in deriving functions when input variables depend on other variables (similar to our problem).
• This necessitates the use of tools like tree diagrams to visualize complex dependencies and systematic differentiation.

Mastering multivariable calculus opens doors to understanding more complex phenomena. It plays a key role in fields such as physics, engineering, and economics, where multiple variables interplay.