Problem 53 - Calculus Early Transcendentals

Question

Consider the following surfaces specified in the form \(z=f(x, y)\) and the curve \(C\) in the \(x y\) -plane given parametrically in the form \(x=g(t), y=h(t)\). a. In each case, find \(z^{\prime}(t)\). b. Imagine that you are walking on the surface directly above the curve \(C\) in the direction of increasing t. Find the values of \(t\) for which you are walking uphill (that is, \(z\) is increasing). $$z=x^{2}+4 y^{2}+1, C: x=\cos t, y=\sin t ; 0 \leq t \leq 2 \pi$$

Step-by-Step Solution

Answer

Question: Determine the time intervals when the curve \(z = x^2 + 4y^2 + 1\) is increasing, where \(x = \cos t\) and \(y = \sin t\). Answer: The curve is increasing in the intervals \((0, \frac{\pi}{2})\) and \((\pi, \frac{3\pi}{2})\).

Step 1: Compute derivative z'(t)

Given that \(z = x^2 + 4y^2 + 1\) where \(x = \cos t\) and \(y = \sin t\), differentiate \(z\) with respect to \(t\) by taking partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) and then using chain rule: \begin{align*} \frac{\partial z}{\partial t} &= \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt} \\ &= 2x \cdot (-\sin t) + 8y \cdot \cos t \\ &= -2\cos t \sin t + 8\sin t\cos t \\ &= 6\cos t \sin t. \end{align*} So, we find that \(z^{\prime}(t) = 6\cos t \sin t\).

Step 2: Find values of t for which z'(t) > 0

We are looking for the increasing slope of z, then we need \(z^{\prime}(t)>0\) (i.e., positive slope). Set \(z^{\prime}(t) > 0\) and solve for \(t\): \begin{align*} z^{\prime}(t) > 0 & \Leftrightarrow 6\cos t \sin t > 0 \\ & \Leftrightarrow 3\sin (2t) > 0. \end{align*} The equation \(\sin(2t) > 0\) is true for \(0 < 2t < \pi\) or \(2\pi < 2t < 3\pi\). Hence, the interval for \(t\) when \(z^(t)>0\) (the curve is going uphill) is $$0 < t <\frac{\pi}{2}\ \text{or} \ \pi < t <|frac{3\pi}{2}.$$ Final answer: \(z^{\prime}(t) = 6\cos t \sin t\), and the values of \(t\) for which \(z\) is increasing are in the intervals \((0, \frac{\pi}{2})\) and \((\pi, \frac{3\pi}{2})\).

Key Concepts

Parametric Curves

Parametric curves are fascinating because they allow us to describe a curve in the plane using a parameter. A common parameter is time, denoted as \( t \). In our exercise, the curve \( C \) is given by the parametric equations \( x = \cos t \) and \( y = \sin t \).
These equations describe a path on the \( xy \)-plane, creating a circle with a radius of 1, since both sine and cosine represent circular motion. This curve becomes a backbone for exploring more complex scenarios, such as moving along surfaces in three-dimensional space. Understanding parametric curves is crucial because they provide a versatile way to represent complex functions and paths in more than one dimension.
A parametric curve isn't just a list of points; it offers information on direction and position simultaneously, which helps in comprehending how functions behave in motion.

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus, allowing us to understand how a function changes as one of its variables changes. Imagine you have a surface defined by \( z = f(x, y) \), like the one in our example where \( z = x^2 + 4y^2 + 1 \). A partial derivative, like \( \frac{\partial z}{\partial x} \), indicates how \( z \) changes when \( x \) changes while keeping \( y \) constant.
This can be thought of as a way to "slice" through the multi-dimensional space to examine more straightforward changes along a single axis. In our exercise, the partial derivatives \( \frac{\partial z}{\partial x} = 2x \) and \( \frac{\partial z}{\partial y} = 8y \) provide rates of change along the \( x \) and \( y \) directions, respectively.
These derivatives are integral in calculating how such surfaces slope, bend, or twist, providing an understanding of the geometry and behavior of multi-variable functions.

Chain Rule

The chain rule is a vital tool when dealing with composite functions, particularly in multivariable calculus. Let's say we have a situation where a variable depends on some others, such as \( z = f(x, y) \) and each of \( x \) and \( y \) depends on yet another parameter \( t \). This dependency creates a need to calculate the derivative of \( z \) with respect to \( t \) through the chain rule.
In our exercise, you use partial derivatives and the chain rule to find \( z'(t) \), where:

• \( \frac{dx}{dt} = -\sin t \)
• \( \frac{dy}{dt} = \cos t \)

The chain rule combines these to relate changes in \( t \) to changes in \( z \) by using:
\[ \frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt}. \]
This gives us \( z'(t) = 6\cos t \sin t \), helping to track how quickly \( z \) changes as you move along your path in terms of \( t \). Mastery of the chain rule is critical for navigating complex equations where variables interact in nested forms.

Analyzing Functions of Two Variables

Analyzing functions of two variables, like \( z = f(x, y) \), requires understanding how these functions can alter with changes in both \( x \) and \( y \). In multidimensional space, such functions create surfaces that can slope, plateau, or escalate in various directions.
Our task involves tracking the behavior of \( z \) and determines where it increases—specifically, where \( z'(t) > 0 \). This highlights areas on the surface where walking along the path causes you to go uphill, dependent on whether \( t \) is within specific intervals like \((0, \frac{\pi}{2})\) and \((\pi, \frac{3\pi}{2})\).
By plotting and evaluating these intervals, you visually and mathematically interpret the surface's shape. This process exemplifies the connection between mathematical calculations and real-world applications, where one must analyze conditions for increases or decreases within complex systems.