Problem 1 - 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. $$ \begin{array}{l}{\text { Equation: } y=3 x+5 .} \\ {\text { (a) Given that } d x / d t=2, \text { find } d y / d t \text { when } x=1 .} \\ {\text { (b) Given that } d y / d t=-1, \text { find } d x / d t \text { when } x=0 \text { . }}\end{array} $$

Step-by-Step Solution

Answer

(a) \( \frac{dy}{dt} = 6 \), (b) \( \frac{dx}{dt} = -\frac{1}{3} \).

Step 1: Differentiate the Equation with Respect to t

Start with the equation \( y = 3x + 5 \). Differentiate both sides with respect to \( t \). This gives: \( \frac{dy}{dt} = 3 \frac{dx}{dt} \).

Step 2: Solve Part (a)

Given that \( \frac{dx}{dt} = 2 \), substitute this value into the derivative equation from Step 1:\[\frac{dy}{dt} = 3 \times 2 = 6\]So, \( \frac{dy}{dt} = 6 \) when \( x = 1 \).

Step 3: Solve Part (b)

Given that \( \frac{dy}{dt} = -1 \), substitute this into the derivative equation and solve for \( \frac{dx}{dt} \):\[-1 = 3 \frac{dx}{dt}\]Solving for \( \frac{dx}{dt} \), we get:\[\frac{dx}{dt} = \frac{-1}{3}\]Therefore, \( \frac{dx}{dt} = -\frac{1}{3} \) when \( x = 0 \).

Key Concepts

Derivatives

Let's start with the concept of derivatives, which is a fundamental part of calculus. It's a tool that helps us understand how a function changes when the values of its input change. In simpler terms, a derivative represents the rate of change or the slope of the function at any given point.

Think of it as the speed of a car, where the speedometer shows how fast the car is moving. Analogously, in mathematics, the derivative tells us how fast the function's output is changing with respect to changes in input.

• A derivative is represented by symbols like \( \frac{dy}{dx} \) for the derivative of y concerning x.
• The process of finding a derivative is called differentiation.
• Derivatives help us solve various practical problems involving motion, growth, and even temperature changes.

In the context of our exercise, we have a function where y is expressed in terms of x, specifically, \( y = 3x + 5 \). This linear relationship makes differentiation quite straightforward.

Functions of a Variable

A function of a variable is an expression that describes a relationship between two quantities, where one depends on the other. In simple terms, it's like a machine that takes an input, performs some operations on it, and produces an output.

In mathematics, the input variable is often denoted by x, and the output, which depends on x, is usually denoted by y. The equation like \( y = 3x + 5 \) is a perfect example, where x is the variable and y is the function of x.

• Functions can be linear like in our example, or they can be more complex like quadratic or exponential functions.
• The output of a function changes as its input changes, which can be visualized graphically as a curve or a line on a graph.
• Understanding functions is crucial for working with derivatives, as they often describe how quantities change over time or space.

In the exercise, the function ties together y and x, demonstrating how they are interrelated. This relationship is key when applying derivatives to solve real-world problems.

Differentiation with Respect to Time

Differentiation with respect to time is a special case of finding a derivative where the variable of interest changes over time. This is particularly useful in physics and engineering when analyzing how quantities vary as time progresses.

In our exercise, both x and y are functions of time \( t \). Hence, when we differentiate, we do it with respect to t, which means we are observing how these quantities change as time passes.

• The notation \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) indicates the rate of change of x and y over time, respectively.
• By understanding how to manipulate these rates, we can predict future behavior or reconstruct past events based on current data.
• This approach allows us to model dynamic systems where variables are not static but constantly evolving.

In part (a) and part (b) of the exercise, we are tasked with calculating these rates of change using the given information and differentiating the function y = 3x + 5. This situation manifests how dynamic systems are analyzed and is essential for fields like economics, biology, and beyond.