Problem 2 - Calculus Early Transcendentals

Question

Let \(z\) be a function of \(x\) and \(y,\) while \(x\) and \(y\) are functions of \(t\). Explain how to find \(\frac{d z}{d t}\).

Step-by-Step Solution

Answer

Answer: The general formula for finding the derivative of a function \(z(x,y)\) with respect to \(t\) is: \(\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}\), which is derived using the chain rule.

Step 1: Understand the Chain Rule

The chain rule is a method for finding the derivative of a composite function. In general, if we have a function \(z(x,y)\) where \(x\) and \(y\) are both functions of a variable \(t\), that is \(x(t)\) and \(y(t)\), then the derivative of \(z\) with respect to \(t\) can be found using the chain rule as follows: \(\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}\) The chain rule effectively tells us to differentiate \(z\) with respect to \(x\) and \(y\), and then multiply these derivatives with the derivatives of \(x\) and \(y\) with respect to \(t\), and finally add the results.

Step 2: Apply the Chain Rule

To find \(\frac{dz}{dt}\), follow these steps: 1. Find the partial derivatives of \(z\) with respect to \(x\) and \(y\), denoted by \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\). Remember that when taking the partial derivative with respect to a certain variable, we treat all other variables as constants. 2. Find the derivatives of \(x\) and \(y\) with respect to \(t\), denoted by \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). 3. Multiply \(\frac{\partial z}{\partial x}\) and \(\frac{dx}{dt}\), and then multiply \(\frac{\partial z}{\partial y}\) and \(\frac{dy}{dt}\). 4. Add the results from step 3 to find the final answer, \(\frac{dz}{dt}\). It's important to note that without knowing the specific functions \(z(x,y)\), \(x(t)\), and \(y(t)\), we cannot explicitly calculate \(\frac{dz}{dt}\). However, understanding and applying the chain rule as described above will enable you to find \(\frac{dz}{dt}\) for any given functions.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. When dealing with functions of several variables, like in our case with the function \( z(x,y) \), partial derivatives help us understand how the function changes as one of the variables changes while keeping other variables constant.

To find the partial derivative, you focus on one variable at a time:

• \( \frac{\partial z}{\partial x} \): This measures how \( z \) changes with respect to \( x \), treating \( y \) as a constant.
• \( \frac{\partial z}{\partial y} \): This measures how \( z \) changes with respect to \( y \), treating \( x \) as a constant.

In our chain rule application, the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) are vital as they are the pieces that link the direct changes in the function \( z \) due to changes in \( x \) and \( y \).
Remember, partial derivatives are just like regular derivatives, but they look at one piece of the multi-variable puzzle at a time.

Composite Functions

Composite functions come into play when you have one function nested inside another. In our scenario, \( z \) is a function of \( x \) and \( y \), and both \( x \) and \( y \) can be expressed in terms of another variable \( t \). This results in a composite function because the output of one function becomes the input of another.

Here's how you can identify composite functions in this context:

• Function \( z \) takes inputs \( x(t) \) and \( y(t) \).
• The outputs of \( x(t) \) and \( y(t) \) feed into the function \( z \), making \( z \) a composite function of \( t \).

Understanding composite functions helps in applying the chain rule, as the composition of functions requires you to consider how each variable contributes to changes in the overarching function. Essentially, each layer of function depends on the previous one, and this layering structure is what the chain rule helps you unravel to find derivatives.

Implicit Differentiation

While the concept in your problem does not directly use implicit differentiation, it's often confused with chain rule applications as both involve derivatives of intertwined functions. Implicit differentiation is used when you're dealing with equations not solved for a particular variable. In cases when a function is defined implicitly rather than explicitly, implicit differentiation allows us to find derivatives without isolating the dependent variable explicitly.

Here's a simple reminder about how implicit differentiation works:

• Consider all variables as functions of another variable (like \( t \)).
• Differentiate every term on both sides of an equation with respect to that variable.
• Solve the resulting equation for the desired derivative.

In chain rule contexts like this one, you work with explicit expressions for the dependent variable in terms of independent ones. However, understanding implicit differentiation can assist in solving more complex problems where the relationships aren’t neatly laid out.