Problem 3 - Thomas Calculus

Question

Assume that \(y=5 x\) and \(d x / d t=2 .\) Find \(d y / d t\)

Step-by-Step Solution

Answer

\( \frac{dy}{dt} = 10 \).

Step 1: Identify the Given Information

We are given that the equation is \( y = 5x \) and that \( \frac{dx}{dt} = 2 \). We need to find \( \frac{dy}{dt} \).

Step 2: Apply the Chain Rule

We will differentiate both sides of the equation \( y = 5x \) with respect to \( t \). The chain rule tells us that \( \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} \).

Step 3: Differentiate the Equation

The derivative of \( y = 5x \) with respect to \( x \) is \( \frac{dy}{dx} = 5 \).

Step 4: Substitute Known Values into the Chain Rule

Substitute \( \frac{dy}{dx} = 5 \) and \( \frac{dx}{dt} = 2 \) into the equation from Step 2: \( \frac{dy}{dt} = 5 \cdot 2 \).

Step 5: Calculate \( \frac{dy}{dt} \)

Multiply the values to obtain \( \frac{dy}{dt} = 10 \).

Key Concepts

Understanding Differentiation

Differentiation is a core concept in calculus used to determine the rate at which a quantity changes. When you differentiate a function, you are essentially finding its derivative, which represents the slope or rate of change of a function's curve.
Differentiation finds practical applications in various real-world problems, from physics to economics. For instance, in the exercise above, we are calculating how fast a quantity, y, changes relative to time using another variable, x.

• To differentiate a simple linear equation like \( y = 5x \), we find the rate at which \( y \) changes with respect to \( x \).
• The process of differentiation involves applying specific rules or formulas, with the chain rule being one crucial tool in our toolkit.

Exploring Derivatives

A derivative provides a way of representing how a function changes as its input changes. In the simplest terms, a derivative identifies the slope of a tangent line to the curve of a function at any point.

• In our exercise, the derivative of \( y = 5x \) with respect to \( x \) is \( \frac{dy}{dx} = 5 \), showing that for every unit increase in \( x \), \( y \) increases by 5 units monthly.
• The derivative is calculated using differentiation rules, which help us handle different types of equations efficiently.

By finding \( \frac{dy}{dt} \) as 10, we have effectively determined the rate in which \( y \) changes as \( x \) changes over time (here, time is modeled by \( t \)).

Calculus and its Application

Calculus is a branch of mathematics that studies continuous change. It focuses on two main concepts: differentiation and integration.
In solving the problem, we primarily deal with differentiation, an essential tool in calculus, which involves determining how variables change in relation to each other over time. The origin of this study lies in the need to understand patterns of motion and change—core to fields like physics, engineering, and beyond.

• The chain rule, a crucial idea in calculus, helps find the derivative of composite functions, like determining \( \frac{dy}{dt} \).
• It's particularly useful when dealing with problems where change depends on multiple interconnected rates.

This exercise offers a simple demonstration of how calculus helps analyze dynamic systems and rate-related issues by breaking down problems into manageable parts using rules like the chain rule.