Problem 27 - Calculus Early Transcendentals

Question

Use a tree diagram to write the required Chain Rule formula. \(w\) is a function of \(z,\) where \(z\) is a function of \(x\) and \(y,\) each of which is a function of \(t .\) Find \(d w / d t\).

Step-by-Step Solution

Answer

Using the Chain Rule and the relationships between the functions, we derived the following formula to calculate \(\frac{dw}{dt}\): \(\frac{dw}{dt} = \frac{dw}{dz} (\frac{dz}{dx} \frac{dx}{dt} + \frac{dz}{dy} \frac{dy}{dt})\) The general form of the solution is provided, but more information about the specific functions and their derivatives must be given to find the actual value of \(\frac{dw}{dt}\).

Step 1: Draw a tree diagram

A tree diagram is used to visually represent the relationships between the functions. In this case, we have the following relationships: 1. \(w = w(z)\) 2. \(z = z(x, y)\) 3. \(x = x(t)\) 4. \(y = y(t)\) The tree diagram would look like this: w | z / \ x y \ / t

Step 2: Write the Chain Rule formula for \(\frac{dw}{dt}\)

The Chain Rule states that the derivative of a composite function with respect to an outer variable can be found by multiplying the derivatives of all the inner functions and their relationships: \(\frac{dw}{dt} = \frac{dw}{dz} \frac{dz}{dt}\) Now, since \(z\) is a function of both \(x\) and \(y\), we have to expand the Chain Rule formula for \(\frac{dz}{dt}\) as well: \(\frac{dz}{dt} = \frac{dz}{dx} \frac{dx}{dt} + \frac{dz}{dy} \frac{dy}{dt}\) Combining these two components, we get the required Chain Rule formula for \(\frac{dw}{dt}\): \(\frac{dw}{dt} = \frac{dw}{dz} (\frac{dz}{dx} \frac{dx}{dt} + \frac{dz}{dy} \frac{dy}{dt})\)

Step 3: Calculate \(\frac{dw}{dt}\)

To calculate \(\frac{dw}{dt}\) using the Chain Rule formula derived in Step 2, we need to know the derivatives \(\frac{dw}{dz}\), \(\frac{dz}{dx}\), \(\frac{dx}{dt}\), \(\frac{dz}{dy}\), and \(\frac{dy}{dt}\). Since we are not given this information, the solution is left in its general form: \(\frac{dw}{dt} = \frac{dw}{dz} (\frac{dz}{dx} \frac{dx}{dt} + \frac{dz}{dy} \frac{dy}{dt})\) With this formula, if more information is given about the functions \(w(z)\), \(z(x, y)\), \(x(t)\), and \(y(t)\), you would be able to find the desired derivative \(\frac{dw}{dt}\).

Key Concepts

Composite Functions

Understanding composite functions is crucial when dealing with complex derivatives. A composite function is essentially a function that is built from two or more functions. In the exercise provided, we see this concept at work with multiple layers of functions. For example:

• The outer function is defined as \(w = w(z)\).
• The intermediate function is \(z = z(x, y)\) which takes its inputs from the underlying functions in \(x\) and \(y\).
• Lastly, both \(x\) and \(y\) are composite functions of \(t\).

This layer-by-layer structure allows us to apply the Chain Rule to differentiate these functions with respect to the deepest variable \(t\). By strategically deciphering these relationships, you can effectively navigate through each layer and find the derivative required for composite functions. Composite functions illustrate how altering one small input at the beginning impacts results further down the flow of the function.

Tree Diagram

Tree diagrams are excellent tools for visualizing the processes in calculus, particularly for composite functions. They provide a clear view of how each function connects with others. In our exercise, imagine a tree with branches representing the hierarchical structure of the functions:

• At the top of the tree, we find \(w\), which depends on the function \(z\).
• Moving down the tree, \(z\) branches into \(x\) and \(y\).
• Finally, each branch \(x\) and \(y\) ends with leaves representing function \(t\).

This branching structure helps break down the complex relationship into manageable steps, detailing how changes in one function affect another. By following the paths from the root to the leaves, we can see how each component function contributes to the final derivative calculation. This visualization simplifies understanding and applying the Chain Rule effectively.

Derivatives

Derivatives are fundamental in calculus as they represent the rate of change of a function concerning one of its variables. In the context of this exercise, we apply derivatives through the Chain Rule, which is invaluable when dealing with composite functions. Here's how it works:Once you identify the chain of dependencies among the functions, derivatives help in calculating how a small change in \(t\) alters \(w\). The Chain Rule formula breaks this process into steps, first considering how \(w\) changes with \(z\) using \(\frac{dw}{dz}\). Next, it examines \(z\)'s dependency on \(x\) and \(y\) via \(\frac{dz}{dx}\), \(\frac{dz}{dy}\) and their subsequent changes with \(t\) through \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).Calculating the derivative \(\frac{dw}{dt}\) entails using these smaller derivatives in conjunction and adding them up as prescribed by the Chain Rule: \[\frac{dw}{dt} = \frac{dw}{dz} \left( \frac{dz}{dx} \frac{dx}{dt} + \frac{dz}{dy} \frac{dy}{dt} \right)\]This captures how modifying the innermost variable \(t\) influences the outcome \(w\), illustrating the important role derivatives play in evaluating real-world dynamic systems.