Problem 9 - Calculus

Question

Find \(\Delta y / \Delta x\) for \(y(x)=x+x^{2}\). Then find \(d y / d x\).

Step-by-Step Solution

Answer

\(\Delta y / \Delta x = 1 + 2x + \Delta x\); \(dy/dx = 1 + 2x\).

Step 1: Identify the Function Components

The function given is \(y(x) = x + x^2\). Identify the components: \(x\) is a linear term, and \(x^2\) is a quadratic term.

Step 2: Calculate \(\Delta y\) and \(\Delta x\)

To find \(\Delta y / \Delta x\), consider small changes. Assume \(x\) changes to \(x + \Delta x\), then \(\Delta x = \Delta x\) and \(\Delta y = (x + \Delta x + (x + \Delta x)^2) - (x + x^2)\).

Step 3: Simplify \(\Delta y\) Expression

Expand \((x + \Delta x)^2\) to get \(x^2 + 2x\Delta x + (\Delta x)^2\). Now, \(\Delta y = (x + \Delta x + x^2 + 2x\Delta x + (\Delta x)^2) - (x + x^2)\), simplify to get \(\Delta y = \Delta x + 2x\Delta x + (\Delta x)^2\).

Step 4: Compute \(\Delta y / \Delta x\)

Divide \(\Delta y = \Delta x + 2x\Delta x + (\Delta x)^2\) by \(\Delta x\) to get \(\Delta y / \Delta x = 1 + 2x + \Delta x\).

Step 5: Identify the Limit and Find \(\frac{dy}{dx}\)

To find the derivative \(\frac{dy}{dx}\), take the limit of \(\Delta y / \Delta x\) as \(\Delta x\) approaches 0. This removes the \(\Delta x\) term, resulting in \(\frac{dy}{dx} = 1 + 2x\).

Key Concepts

Change in Y Over Change in X

Before diving deeper, let's understand the basic concept of "change in y over change in x," often written as \(\Delta y / \Delta x\). This represents the average rate of change of the function \(y(x)\) over a small interval around a point on the function's graph. Imagine you are climbing a hill and you want to know how steep the incline is at every step.
This "change" tells us the steepness or slope between two points:

• \(\Delta y\) is the change in the output of the function (vertical change).
• \(\Delta x\) is the change in the input value (horizontal change).

For the function \(y(x) = x + x^2\), calculating \(\Delta y / \Delta x\) helps us understand how the function behaves as \(x\) changes. Essentially, this measure gives us a sneak peek into how the slope varies over different intervals of \(x\).

Limit Process

The limit process is a fundamental concept in calculus and is crucial for finding the derivative. It involves taking the limit of \(\Delta y / \Delta x\) as \(\Delta x\) becomes very small, approaching 0. Think of it as zooming in on the graph of \(y(x)\) to just one point to find the instantaneous rate of change. This is what differentiates the derivative from the average rate of change.
By using limits, we transition from an average rate of change over a segment to an instantaneous rate at a single point:

• Taking the limit allows us to focus on an exact point on the graph.
• It effectively cancels out any extra terms that disappear when \(\Delta x\) is so small it is nearly zero.

For the function \(y(x) = x + x^2\), applying the limit to \(1 + 2x + \Delta x\) as \(\Delta x \rightarrow 0\), we simplify to the derivative \(\frac{dy}{dx} = 1 + 2x\). This derivative represents the exact slope at any point \(x\).

Linear Term

In the context of the given function \(y(x) = x + x^2\), the term \(x\) is identified as the linear term of the function. Linear terms are those with the highest exponent of 1, and they typically describe a straight line, like the equation of a line \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.
Linear terms provide crucial insights:

• They dictate the initial slope of the function when all other terms are minimized or zero.
• The coefficient (here, implicit 1) tells us how much \(y\) will increase per unit increase in \(x\).

In this specific function, the linear term \(x\) contributes to the slope, making the derivative start with a base of 1. Understanding linear terms is key to recognizing their role in the overall behavior of more complex functions.

Quadratic Term

The term \(x^2\) in the function \(y(x) = x + x^2\) is what we call the quadratic term. Quadratic terms are identified by their squared factor, where the exponent is 2. They introduce a parabolic curvature to the function's graph and affect the rate of change more significantly than linear terms.
Quadratic terms possess distinct characteristics:

• They add a curved shape to the graph, differentiating the function from being a straight line.
• The coefficient of \(x^2\) influences how "wide" or "narrow" the parabola opens.

In this function, the quadratic term \(x^2\) modifies the slope dramatically as \(x\) increases or decreases. This means that as \(x\) grows, the rate of change accelerates due to \(2x\) from the derivative \(\frac{dy}{dx} = 1 + 2x\). This showcases how quadratic terms significantly impact the behavior and dynamics of polynomial functions.