Problem 3 - Calculus Early Transcendentals

Question

Both \(x\) and \(y\) denote functions of \(t\) that are related by the given equation. Use this equation and the given derivative information to find the specified derivative. $$ \begin{array}{l}{\text { Equation: } 4 x^{2}+9 y^{2}=1} \\ {\text { (a) Given that } d x / d t=3, \text { find } d y / d t \text { when }} \\ {\qquad(x, y)=\left(\frac{1}{2 \sqrt{2}}, \frac{1}{3 \sqrt{2}}\right)}\end{array} $$ $$ \begin{array}{l}{\text { (b) Given that } d y / d t=8, \text { find } d x / d t \text { when }} \\ {\qquad(x, y)=\left(\frac{1}{3},-\frac{\sqrt{5}}{9}\right)}\end{array} $$

Step-by-Step Solution

Answer

(a) \(\frac{dy}{dt} = -2\); (b) \(\frac{dx}{dt} = 6\sqrt{5}\).

Step 1: Differentiate the Given Equation with Respect to t

To find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\), differentiate the given equation \(4x^2 + 9y^2 = 1\) with respect to \(t\). Apply the chain rule: \[\frac{d}{dt}(4x^2) + \frac{d}{dt}(9y^2) = 0.\] This becomes \[8x \frac{dx}{dt} + 18y \frac{dy}{dt} = 0.\]

Step 2: Solve for dy/dt Given dx/dt

For part (a), substitute \(x = \frac{1}{2\sqrt{2}}\), \(y = \frac{1}{3\sqrt{2}}\), and \(\frac{dx}{dt} = 3\) into the differentiated equation \[8x \frac{dx}{dt} + 18y \frac{dy}{dt} = 0.\] This becomes \[8\left(\frac{1}{2\sqrt{2}}\right) \cdot 3 + 18\left(\frac{1}{3\sqrt{2}}\right) \frac{dy}{dt} = 0.\] Simplifying, \[\frac{12}{\sqrt{2}} + \frac{6}{\sqrt{2}} \frac{dy}{dt} = 0.\] Solving for \(\frac{dy}{dt}\), we find \[\frac{dy}{dt} = -2.\]

Step 3: Solve for dx/dt Given dy/dt

For part (b), substitute \(x = \frac{1}{3}\), \(y = -\frac{\sqrt{5}}{9}\), and \(\frac{dy}{dt} = 8\) into the differentiated equation \[8x \frac{dx}{dt} + 18y \frac{dy}{dt} = 0.\] This becomes \[8\left(\frac{1}{3}\right) \frac{dx}{dt} + 18\left(-\frac{\sqrt{5}}{9}\right) \cdot 8 = 0.\] Simplifying, \[\frac{8}{3} \frac{dx}{dt} - 16\sqrt{5} = 0.\] Solving for \(\frac{dx}{dt}\), we find \[\frac{dx}{dt} = 6\sqrt{5}.\]

Key Concepts

Implicit Differentiation

Implicit differentiation is a powerful technique in calculus used when dealing with equations that define a relationship between two variables, where direct differentiation is not straightforward. In cases where one variable is expressed solely in terms of the other, explicit differentiation can be easily performed, but with implicit functions, both variables may need to be treated as functions of a third variable, usually time, denoted as \(t\).

• To perform implicit differentiation, you differentiate each term of the equation with respect to \(t\), while treating one variable as a function of \(t\) and applying the chain rule.
• For example, in the equation \(4x^2 + 9y^2 = 1\), both \(x\) and \(y\) depend on \(t\).
• Each differentiated part should incorporate a derivative like \(\frac{dx}{dt}\) or \(\frac{dy}{dt}\) for understanding how changes in \(t\) affect these variables.

This method grants the ability to find unknown derivatives for functions described implicitly, essential in many real-life problem-solving scenarios.

Chain Rule

The chain rule is a fundamental concept in calculus that is especially useful when differentiating complex expressions that involve composite functions. It's the rule that allows us to differentiate the outer and inner functions separately and then multiply the results.Consider the expression \(4x^2 + 9y^2\). When differentiating with respect to \(t\),:

• The chain rule informs us to first differentiate \(x^2\) and \(y^2\) with respect to \((x)\) and \((y)\). Yields \(8x \cdot \frac{dx}{dt}\) and \(18y \cdot \frac{dy}{dt}\).
• Next, multiply by the derivative of the inside functions \(x\) and \((y)\) with respect to \(t\).

Through this method, we can effectively handle performing calculus on nested functions and solve for connected rates. This rule is crucial in many fields, especially when dealing with physical systems in motion.

Differential Calculus

Differential calculus is the branch of mathematics that deals with the study of rates at which quantities change. It forms the backbone of concepts such as related rates, which involve multiple variables that are changing with respect to time.In problems like our exercise, we utilize differential calculus to:

• Identify the relationship between different variables.
• Apply methods like implicit differentiation to find the rate of change of these variables.
• Formulate a strategy that includes differentiating an equation and solving for unknown rates like \(\frac{dy}{dt}\) or \(\frac{dx}{dt}\).

In essence, differential calculus allows us to quantify how a small change in one variable affects another, providing insight into dynamic systems' behavior.

Parametric Equations

Parametric equations involve expressing a set of quantities as explicit functions of a parameter, often representing time. They offer a unique approach to calculating relationships between variables when they cannot be explicitly solved in terms of each other.These equations allow us to:

• Write both \(x\) and \(y\) as functions of \(t\) independently.
• Examine the behavior of each variable as \(t\) changes, which is fundamental in analyzing the motion of objects or changes in systems.
• Seamlessly apply calculus operations like differentiation to find rates of change relative to the parameter \(t\).

For instance, using parametric forms inherently fits problems requiring related rates because they provide naturally linked derivatives which explain how quantities evolve concurrently over time.