Problem 2 - 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. $$ \begin{array}{l}{\text { Equation: } x+4 y=3 .} \\ {\text { (a) Given that } d x / d t=1, \text { find } d y / d t \text { when } x=2 .} \\ {\text { (b) Given that } d y / d t=4, \text { find } d x / d t \text { when } x=3 \text { . }}\end{array} $$

Step-by-Step Solution

Answer

(a) \( \frac{dy}{dt} = -\frac{1}{4} \); (b) \( \frac{dx}{dt} = -16 \).

Step 1: Differentiate the equation implicitly with respect to t

The equation given is \( x + 4y = 3 \). We differentiate each term with respect to \( t \):1. The derivative of \( x \) with respect to \( t \) is \( \frac{dx}{dt} \).2. The derivative of \( 4y \) with respect to \( t \) is \( 4 \frac{dy}{dt} \).3. The derivative of the constant \( 3 \) is \( 0 \).Putting it all together, the differentiated equation is:\[ \frac{dx}{dt} + 4 \frac{dy}{dt} = 0 \]

Step 2: Solve for \( \frac{dy}{dt} \) using the given \( \frac{dx}{dt} \)

For part (a), we are given \( \frac{dx}{dt} = 1 \) and need to find \( \frac{dy}{dt} \) when \( x = 2 \):- Substitute \( \frac{dx}{dt} = 1 \) into the differentiated equation: \[ 1 + 4 \frac{dy}{dt} = 0 \]- Solve for \( \frac{dy}{dt} \): \[ 4 \frac{dy}{dt} = -1 \] \[ \frac{dy}{dt} = -\frac{1}{4} \]

Step 3: Solve for \( \frac{dx}{dt} \) using the given \( \frac{dy}{dt} \)

For part (b), we are given \( \frac{dy}{dt} = 4 \) and need to find \( \frac{dx}{dt} \) when \( x = 3 \):- Substitute \( \frac{dy}{dt} = 4 \) into the differentiated equation: \[ \frac{dx}{dt} + 4(4) = 0 \]- This simplifies to: \[ \frac{dx}{dt} + 16 = 0 \]- Solve for \( \frac{dx}{dt} \): \[ \frac{dx}{dt} = -16 \]

Key Concepts

Related Rates

Related rates is an essential concept in calculus that involves finding the rate at which one quantity changes concerning another quantity. In the given exercise, we are interested in how the rate of change of one variable affects the other. For instance, if you are given how rapidly the value of \(x\) changes over time \(\frac{dx}{dt}\), you might be asked to find how \(y\) changes over time \(\frac{dy}{dt}\).

These problems often involve physical scenarios, such as the radius of a balloon increasing and you needing to find how its surface area changes. While using related rates, you typically follow these steps:

• Identify what rates are given and what rates you need to find.
• Differentiate the given equation or formula concerning time (often involving implicit differentiation).
• Solve for the unknown rate.

In our exercise, we apply related rates to the equation \(x + 4y = 3\). By looking at how \(x\) and \(y\) change with respect to \(t\), we can determine the needed rates of change.

Differentiation

Differentiation allows us to find the rate of change of variables and is a fundamental tool in calculus. In implicit differentiation, unlike regular differentiation, you differentiate both sides of an equation containing two or more variables, often with respect to time \(t\).

In the exercise at hand, we differentiate \(x + 4y = 3\) implicitly. The term \(x\) becomes \(\frac{dx}{dt}\) and \(4y\) becomes \(4 \frac{dy}{dt}\). The constant \(3\) differentiates to \(0\). This step is crucial for setting up the equation of derivatives \(\frac{dx}{dt} + 4 \frac{dy}{dt} = 0\), which contains the rates related by the initial equation.

Implicit differentiation helps us when dealing with equations where variables are dependent on one another. Here you find derivatives without solving explicitly for one variable.

Function of a Parameter

The concept of a function of a parameter deals with how variables depend on parameters, often showing time-dependent behaviors. Variables such as \(x\) and \(y\) describe this dependency on the parameter \(t\), like in the equation we have, where each change in \(t\) directly influences other variables.

When dealing with functions of parameters in related rates problems, it’s important to:

• Understand both \(x\) and \(y\) as functions of \(t\).
• Interpret how changes in \(t\) affect these functions, reflected through their derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

In the scope of this exercise, by knowing either \(\frac{dx}{dt}\) or \(\frac{dy}{dt}\), we can determine the rate of change of the other variable using the differentiated equation. This understanding of the dependency on the parameter \(t\) is crucial for solving related rates problems effectively.