Problem 15 - 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- $$ z=5 x+2 y, \quad x=2 \cos t+1, \quad y=\sin t-3 $$

Step-by-Step Solution

Answer

The values of \(t\) are \(t = \arctan\left(\frac{1}{5}\right) + n\pi\), where \(n\) is an integer.

Step 1: Understand the problem

We are given functions in terms of two parameters: \(z=f(x,y)\), \(x=g(t)\), and \(y=h(t)\). The task is to find the values of \(t\) for which \(\frac{dz}{dt}=0\).

Step 2: Express z in terms of t

Substitute \(x=2\cos t+1\) and \(y=\sin t-3\) into the function \(z=5x+2y\). This gives \(z=5(2\cos t+1) + 2(\sin t-3)\). Simplify the expression to get \(z=10\cos t + 5 + 2\sin t - 6\), which further simplifies to \(z=10\cos t + 2\sin t - 1\).

Step 3: Find the derivative of z with respect to t

Compute \(\frac{dz}{dt}\) using the derivative of the function: \(z=10\cos t + 2\sin t - 1\). Using basic derivatives, we get \(\frac{dz}{dt} = -10\sin t + 2\cos t\).

Step 4: Set the derivative equal to zero

Set the equation \(-10\sin t + 2\cos t = 0\) to solve for \(t\). This simplifies to \(\tan t = \frac{1}{5}\).

Step 5: Solve for t

Using the equation \(\tan t = \frac{1}{5}\), solve for \(t\). The solutions will be \(t = \arctan(\frac{1}{5}) + n\pi\) where \(n\) is an integer.

Key Concepts

Partial Derivatives

Partial derivatives are an essential part of calculus, particularly when dealing with functions of multiple variables. They represent how a function changes as one of its input variables changes, while the others are held constant. In the context of our exercise, we deal with the function \( z = f(x, y) \). By partially differentiating \( z \) with respect to \( x \) and \( y \), we can observe how changes in these variables independently affect \( z \).
When you see the notation \( \frac{\partial z}{\partial x} \), it means we're looking at the rate of change of \( z \) in the \( x \) direction. Similarly, \( \frac{\partial z}{\partial y} \) finds the rate of change in the \( y \) direction. The total change in \( z \), when both \( x \) and \( y \) change, combines these partial derivatives with the derivatives of \( x \) and \( y \) with respect to \( t \), using the chain rule.

Chain Rule

The chain rule is a fundamental concept in calculus that allows us to differentiate composite functions effectively. It's particularly useful when dealing with functions that are not directly defined in terms of a single variable but through other functions. In our exercise, to find \( \frac{d z}{d t} \), we apply the chain rule:

• First, differentiate \( z \) with respect to \( x \) and \( y \). This gives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \).
• Next, find \( \frac{d x}{d t} \) and \( \frac{d y}{d t} \), the derivatives of \( x \) and \( y \) with respect to \( t \).
• Finally, combine these results using the chain rule: \( \frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt} \).

Through this process, you can see how the interrelationship between \( x \), \( y \), and \( t \) affects \( z \) as \( t \) changes.

Trigonometric Functions

Trigonometric functions such as sine and cosine often appear in calculus problems, especially when dealing with periodic or oscillating phenomena. In this exercise, both \( x \) and \( y \) are expressed in terms of trigonometric functions of \( t \): \( x = 2 \cos(t) + 1 \) and \( y = \sin(t) - 3 \).
These functions are crucial when evaluating the derivative of \( z \) with respect to \( t \). When differentiating \( 10 \cos t + 2 \sin t \), we recall the derivatives \( \frac{d}{dt}(\cos t) = -\sin t \) and \( \frac{d}{dt}(\sin t) = \cos t \).
These trigonometric derivatives are fundamental pieces in evaluating \( \frac{dz}{dt} \) and ensure we understand the behavior of \( z \) as \( t \) varies. If the exercise involves finding critical points, knowing how these derivatives influence \( z \) is critical in determining where \( \frac{dz}{dt} = 0 \).

Critical Points

Critical points in a function are where its derivative is zero or undefined. These points are significant as they often correspond to maximum, minimum, or saddle points of the function. In our problem, finding critical points involves identifying where \( \frac{dz}{dt} = 0 \).
The equation \(-10 \sin t + 2 \cos t = 0\) provides us with a condition for critical points. By rearranging terms, we get \( \tan t = \frac{1}{5} \). This indicates where the slope of \( z \) with respect to \( t \) is zero.
The solutions, \( t = \arctan \left( \frac{1}{5} \right) + n\pi \), for integer \( n \), represent the possible values of \( t \) where the changes in \( z \) momentarily pause. This can reflect turning points or simply flat regions, depending on the broader context of the function.