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). $$z=4 x^{2}-y^{2}+1, C: x=\cos t, y=\sin t ; 0 \leq t \leq 2 \pi$$

Short answer

Question: On a certain planet in a different solar system, you are walking on the surface directly above a curve C lying in the xy-plane given by the parametric equations \(x=g(t)=\cos t, y=h(t)=\sin t, 0 \leq t \leq 2\pi\). At each point (x, y), the temperature z at that point on the surface is given by the scalar function \(z=4x^2-y^2+1\). Find the intervals of t for which you are walking uphill on the surface. Answer: You are walking uphill on the surface when t is in the intervals \((0, \pi/2)\) and \((\pi, 3\pi/2)\).

Step-by-step solution

Express z in terms of t

Using given, \(x=\cos t\) and \(y=\sin t\), substitute into the function \(z=4x^2-y^2+1\): $$z(t) = 4(\cos t)^{2} - (\sin t)^{2} + 1$$

Differentiate z with respect to t using chain rule

In order to find \(z'(t)\), we need to differentiate z(t) with respect to t: $$z'(t) = \frac{d}{dt} [4(\cos t)^{2} - (\sin t)^{2} + 1]$$ Using the chain rule, we get: $$z'(t) = 8\cos t(-\sin t) - 2 \sin t(\cos t)$$

Simplify the derivative

Now simplify the expression for \(z'(t)\): $$z'(t) = -10\sin t \cos t$$

Find the intervals where z'(t) > 0

In order to find out the ranges of t for which z'(t) > 0 (uphill), solve the inequality \(z'(t) > 0\): $$-10\sin t \cos t > 0$$ We know that, 0 is a sin or cos value at certain points between \(0 \leq t \leq 2\pi\). So we should concentrate on: $$\sin t \cos t < 0$$ This inequality is true when two conditions happen: sin(t) and cos(t) are less than zero or both are greater than zero: $$(\sin t < 0 \; \text{and} \; \cos t < 0) \;\; \text{or} \;\; (\sin t > 0 \; \text{and} \; \cos t > 0)$$ This gives us two intervals for t according to the unit circle: $$t ∈ (\pi, 3\pi/2) \; \text{or} \; t ∈ (0, \pi/2)$$ In conclusion, you are walking uphill on the surface directly above the curve C when t is in the intervals \((0, \pi/2)\) and \((\pi, 3\pi/2)\).

Key concepts

Parametric Equations

Parametric equations are a way of defining a relationship between variables using one or more parameters. In our exercise, we used parametric equations to specify the curve \( C \) in the \( xy \)-plane. Specifically, we have \( x = \cos t \) and \( y = \sin t \) as our parametric equations for the curve. Here, \( t \) is the parameter, and it represents a point on the curve as its value changes.
Parametric equations offer flexibility because they can define more complex curves that might be challenging to describe with a single equation in \( x \) and \( y \). They are especially useful for describing paths, such as an object's motion along a path over time. The parameter, like \( t \) here, often represents time or another changing quantity.
Overall, parametric equations provide a unique perspective on understanding relationships in coordinates by expressing them in terms of an additional variable. This enhances our ability to analyze such curves, both graphically and algebraically.

Derivatives

Derivatives are a fundamental concept in calculus used to determine the rate of change or the slope of a function. In this exercise, we need to find the derivative of \( z(t) \), which is expressed in terms of another variable \( t \), rather than \( x \) or \( y \) alone.
To find \( z'(t) \), we differentiate \( z \) with respect to \( t \). Differentiating allows us to see how \( z \) changes as \( t \) changes. This is crucial for understanding the behavior of the function and for finding where the function is increasing or decreasing.
Derivatives have many applications, including determining maxima and minima, optimizing functions, and understanding motion. In our problem, calculating the derivative helps us determine when we are moving uphill on the curve relative to the surface.

Chain Rule

The chain rule is a rule in calculus used for differentiating compositions of functions. When we have a function \( z \) that depends on \( x \) and \( y \), which are, in turn, functions of \( t \), the chain rule allows us to take the derivative of \( z \) with respect to \( t \).
In our exercise, the chain rule was used to take the derivative of \( z(t) = 4(\cos t)^{2} - (\sin t)^{2} + 1 \). The chain rule tells us to multiply the derivative of the outer function by the derivative of the inner function. So, using the chain rule, we simplified the expression to \( z'(t) = -10\sin t \cos t \).
By applying the chain rule, we can navigate complex functions and ensure we correctly account for all dependencies when finding derivatives. This is essential for problems involving parametric equations, where variables are interconnected.

Inequalities

Inequalities express the relative size or order of values and are a valuable tool in calculus for analyzing the behavior of functions. In this exercise, we focused on solving the inequality \( z'(t) > 0 \) to find when the surface \( z \) is increasing as we walk over the curve \( C \).
The inequality \( -10\sin t \cos t > 0 \) simplifies to \( \sin t \cos t < 0 \), showing when \( z \) increases as \( t \) changes. Understanding this inequality helped us find intervals where \( t \) meets certain conditions – when either both sin(t) and cos(t) are positive or both are negative.
In calculus, solving inequalities is crucial for understanding when functions increase or decrease and for identifying intervals over which certain conditions hold true. This allows us to make informed predictions about the behavior of mathematical models in real-world scenarios.