Problem 18 - APEX Calculus

Question

In Exercises \(13-18,\) functions \(z=f(x, y), x=g(t)\) and \(y=h(t)\) are given. Find the values of \(t\) where \(\frac{d z}{d t}=0 .\) Note: these are the same surfaces/curves as found in Exercises $7- $$ z=\cos x \sin y, \quad x=\pi t, \quad y=2 \pi t+\pi / 2 $$

Step-by-Step Solution

Answer

Values of \(t\) are \(t = \frac{n-1}{3}\) for integer \(n\).

Step 1: Understand the Problem Statement

The problem asks us to find the values of \(t\) where the derivative of \(z\) with respect to \(t\), \(\frac{dz}{dt}\), is equal to zero. We are given the functions \(z = \cos x \sin y\), \(x = \pi t\), and \(y = 2\pi t + \pi/2\).

Step 2: Apply the Chain Rule

Since \(z\) is a function of both \(x\) and \(y\), we need to use the chain rule to differentiate \(z\) with respect to \(t\). The chain rule states: \( \frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt} \).

Step 3: Calculate Partial Derivatives

Compute the partial derivatives of \(z\):\( \frac{\partial z}{\partial x} = -\sin x \sin y \)\( \frac{\partial z}{\partial y} = \cos x \cos y \).

Step 4: Calculate Derivatives of \(x\) and \(y\)

Differentiate \(x\) and \(y\) with respect to \(t\):\( \frac{dx}{dt} = \pi \)\( \frac{dy}{dt} = 2\pi \).

Step 5: Substitute into the Chain Rule

Substitute the partial derivatives and derivatives with respect to \(t\) into the chain rule expression:\[ \frac{dz}{dt} = (-\sin x \sin y)(\pi) + (\cos x \cos y)(2\pi) \]

Step 6: Simplify the Expression

Simplify the expression for \(\frac{dz}{dt}\):\[ \frac{dz}{dt} = -\pi \sin x \sin y + 2\pi \cos x \cos y \]which can be factored as:\[ \pi(-\sin x \sin y + 2\cos x \cos y) \]

Step 7: Set the Derivative Equal to Zero and Solve

Set \(\frac{dz}{dt} = 0\) and solve:\[ \pi(-\sin x \sin y + 2\cos x \cos y) = 0 \]Simplify to:\[ \sin(2y - x) = 0 \]

Step 8: Solve the Trigonometric Equation

Given \( \sin(2y - x) = 0 \), find the general solutions:\( 2y - x = n\pi \), where \(n\) is an integer.Substitute \(x = \pi t\) and \(y = 2\pi t + \pi/2\):\[ 2(2\pi t + \pi/2) - \pi t = n\pi \]\[ 4\pi t + \pi - \pi t = n\pi \]\[ 3\pi t = (n - 1)\pi \]Solve for \(t\):\[ t = \frac{n-1}{3} \]

Step 9: Determine Values of \(t\)

Solve for specific values of \(t\) by substituting integers for \(n\):\( n = 1, 2, 3, \ldots \)This gives the series of solutions for \(t\), such as \(t = 0, \frac{1}{3}, \frac{2}{3}, \ldots \)

Key Concepts

The Chain Rule in Multivariable Calculus

In multivariable calculus, the chain rule is a fundamental tool used when differentiating functions that are composed of other functions. This method becomes crucial when dealing with a function that depends on multiple variables. For example, if we have a function \( z = f(x, y) \), and each of \( x \) and \( y \) is a function of another variable \( t \), we need the chain rule to find the derivative of \( z \) with respect to \( t \). This is expressed by the formula:\[ \frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt} \]where \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) are the partial derivatives of \( z \) with respect to \( x \) and \( y \), respectively.

• Begin by finding the partial derivatives of \( z \) with respect to each variable it is composed of.
• Next, find the derivatives of those composing variables \( x \) and \( y \) with respect to \( t \).
• Substitute these values into the chain rule equation to find \( \frac{dz}{dt} \).

By following these steps, we effectively link the changes in the independent variable \( t \) to changes in \( z \), through the intermediary variables \( x \) and \( y \). This forms the backbone of many dynamic systems analyses.

Understanding Partial Derivatives

Partial derivatives extend the concept of a derivative to functions of multiple variables. When we have a function \( f(x, y) \), the partial derivative with respect to \( x \) denotes how \( f \) changes as \( x \) changes, holding \( y \) constant. Likewise, the partial derivative with respect to \( y \) shows how \( f \) changes as \( y \) varies, with \( x \) unchanged.To compute partial derivatives, you:

• Treat all other variables as constants.
• Proceed as you would with a single-variable function derivative.

For the function \( z = \cos x \sin y \), the partial derivative with respect to \( x \) is \( \frac{\partial z}{\partial x} = -\sin x \sin y \), emphasizing how \( z \) changes with \( x \). For \( y \), it is \( \frac{\partial z}{\partial y} = \cos x \cos y \), capturing the rate of change with \( y \).Partial derivatives are instrumental in optimizing functions, analyzing surfaces, and solving differential equations in physics and engineering.

Solving Trigonometric Equations

Trigonometric equations often appear in calculus as they describe periodic phenomena. Solving these equations requires understanding trigonometric identities and solutions.Consider the equation obtained from the previous steps: \( \sin(2y - x) = 0 \). We aim to find values that make this expression zero, using the identity that \( \sin \theta = 0 \) when \( \theta = n\pi \), where \( n \) is any integer. This leads us to:\[ 2y - x = n\pi \]After substituting for \( x \) and \( y \) from the equations \( x = \pi t \) and \( y = 2\pi t + \pi/2 \), solving gives:

• \[ 4\pi t + \pi - \pi t = n\pi \]
• Simplifying to \[ 3\pi t = (n-1)\pi \]
• Hence, \[ t = \frac{n-1}{3} \]

By choosing integer values for \( n \), you derive specific solutions for \( t \). The skill lies in recognizing periodicity and using identities to solve these kinds of equations effectively.