Problem 3 - Calculus Early Transcendentals

Question

Suppose \(w\) is a function of \(x, y,\) and \(z,\) which are each functions of t. Explain how to find \(\frac{d w}{d t}\).

Step-by-Step Solution

Answer

Question: Given the functions \(w = w(x, y, z)\), \(x = x(t)\), \(y = y(t)\), and \(z = z(t)\), find the derivative of \(w\) with respect to \(t\). Answer: Using the chain rule, the derivative of \(w\) with respect to \(t\) is given as \(\frac{d w}{d t} = \frac{\partial w}{\partial x}\frac{d x}{d t} + \frac{\partial w}{\partial y}\frac{d y}{d t} + \frac{\partial w}{\partial z}\frac{d z}{d t}\), where \(\frac{\partial w}{\partial x}\), \(\frac{\partial w}{\partial y}\), and \(\frac{\partial w}{\partial z}\) are partial derivatives of \(w\) with respect to \(x, y, z\), respectively, and \(\frac{d x}{d t}\), \(\frac{d y}{d t}\), and \(\frac{d z}{d t}\) are derivatives of \(x, y, z\) with respect to \(t\), respectively. To find the specific expression for \(\frac{d w}{d t}\), we need the formulas for \(w(x, y, z)\), \(x(t)\), \(y(t)\), and \(z(t)\).

Step 1: Write down the given functions.

The given functions are: 1. \(w = w(x, y, z)\) 2. \(x = x(t)\) 3. \(y = y(t)\) 4. \(z = z(t)\)

Step 2: Apply the Chain Rule.

The chain rule states that if a variable \(z\) depends on \(y\), which depends on \(x\), then the derivative of \(z\) with respect to \(x\) is the product of the derivatives of \(z\) with respect to \(y\) and \(y\) with respect to \(x\). In our case, we want to find \(\frac{d w}{d t}\). This can be computed as: \(\frac{d w}{d t} = \frac{\partial w}{\partial x}\frac{d x}{d t} + \frac{\partial w}{\partial y}\frac{d y}{d t} + \frac{\partial w}{\partial z}\frac{d z}{d t}\) Where \(\frac{\partial w}{\partial x}\), \(\frac{\partial w}{\partial y}\), and \(\frac{\partial w}{\partial z}\) are partial derivatives of \(w\) with respect to \(x, y, z\), respectively, and \(\frac{d x}{d t}\), \(\frac{d y}{d t}\), and \(\frac{d z}{d t}\) are derivatives of \(x, y, z\) with respect to \(t\), respectively.

Step 3: Compute the necessary derivatives.

To compute all the necessary derivatives, we need the specific formulas for \(w(x, y, z)\), \(x(t)\), \(y(t)\), and \(z(t)\). However, they are not given. But, in general, to find the derivative \(\frac{d w}{d t}\), we need to: 1. Compute the partial derivatives of the function \(w(x, y, z)\) with respect to \(x, y, z\). 2. Compute the derivatives of the functions \(x(t)\), \(y(t)\), and \(z(t)\) with respect to \(t\).

Step 4: Substitute and simplify.

Substitute the computed derivatives in the chain rule formula for \(\frac{d w}{d t}\) and simplify if necessary. Once we have all the necessary derivatives, the final step is to substitute them in the chain rule formula and simplify the result.

Key Concepts

Partial Derivatives

Partial derivatives are a key concept when working with functions of multiple variables. They allow us to understand how a function changes as we vary one variable while keeping others constant. For example, if you have a function like \(w = w(x, y, z)\), you can find how \(w\) changes with respect to \(x\), \(y\), or \(z\) individually.
To find a partial derivative of \(w\) with respect to \(x\), denoted as \(\frac{\partial w}{\partial x}\), treat \(y\) and \(z\) as constants and differentiate \(w\) as you normally would with respect to \(x\).

• \(\frac{\partial w}{\partial x}\): The rate of change of \(w\) as \(x\) changes, keeping \(y\) and \(z\) constant.
• \(\frac{\partial w}{\partial y}\) and \(\frac{\partial w}{\partial z}\) follow the same logic, treating other variables as constants when differentiating.

Understanding partial derivatives is crucial in fields like physics and engineering, where functions often depend on many variables.

Multivariable Calculus

Multivariable calculus extends the ideas of single-variable calculus to functions of multiple variables. By doing so, it helps us analyze complex systems, such as those found in physics, engineering, and other sciences.
In multivariable calculus, functions can have more than one input and correspondingly more output dimensions. For instance, \(w = w(x, y, z)\) has three inputs \(x, y, z\), which can be considered in their own separate axes in three-dimensional space.

• Visualizing graphs of multivariable functions often involves surfaces or contours rather than simple curves.
• Differentiation involves partial derivatives and concepts like gradients, which describe the direction of fastest increase of a function.

These components allow us to better understand and navigate higher-dimensional space, making multivariable calculus an essential tool for modeling real-world phenomena.

Composite Functions

Composite functions combine two or more functions to create a new function. When dealing with derivatives of composite functions in multivariable calculus, the chain rule becomes a powerful tool.
In our original problem, the function \(w = w(x, y, z)\) is dependent on variables \(x, y, z\) that are themselves functions of \(t\). This creates a nested, or composite, structure where changes in \(t\) affect \(x, y, z\), and consequently \(w\).

• The chain rule formula \(\frac{d w}{d t} = \frac{\partial w}{\partial x}\frac{d x}{d t} + \frac{\partial w}{\partial y}\frac{d y}{d t} + \frac{\partial w}{\partial z}\frac{d z}{d t}\) elegantly breaks down this complex situation.
• Each term represents the product of a partial derivative and the derivative of the intermediate function, capturing how changes rip through the layers of functions.

Composite functions and the chain rule help us unravel these layered dependencies, making them indispensable in advanced calculus and practical applications.