Question

If \(y=x^{2}-3\), find the values of \(\Delta y\) and \(d y\) in each case. (a) \(x=2\) and \(d x=\Delta x=0.5\) (b) \(x=3\) and \(d x=\Delta x=-0.12\)

Short answer

(a) Δy = 2.25, dy = 2.0; (b) Δy = -0.7056, dy = -0.72.

Step-by-step solution

Understanding the Problem

We are given a function, \(y = x^2 - 3\). We need to find the change in \(y\) denoted by \(\Delta y\), and the differential of \(y\) denoted by \(dy\), for two different values of \(x\) and \(dx = \Delta x\).

Calculation for Case (a) Δy

For \(x = 2\) and \(dx = \Delta x = 0.5\), compute \(\Delta y\). First, find the new value of \(y\):1. Initial \(x = 2\), so \(y = 2^2 - 3 = 1\).2. New \(x = 2 + 0.5 = 2.5\), so the new \(y = 2.5^2 - 3 = 3.25\).3. Therefore, \(\Delta y = 3.25 - 1 = 2.25\).

Calculation for Case (a) dy

Use the formula \(dy = \frac{dy}{dx} \cdot dx\), where \(\frac{dy}{dx}\) is the derivative. For \(y = x^2 - 3\), the derivative \(\frac{dy}{dx} = 2x\).1. At \(x = 2\), \(\frac{dy}{dx} = 2 \cdot 2 = 4\).2. Calculate \(dy = 4 \cdot 0.5 = 2.0\).

Calculation for Case (b) Δy

For \(x = 3\) and \(dx = \Delta x = -0.12\), compute \(\Delta y\).1. Initial \(x = 3\), so \(y = 3^2 - 3 = 6\).2. New \(x = 3 - 0.12 = 2.88\), so the new \(y = 2.88^2 - 3 = 5.2944\).3. Therefore, \(\Delta y = 5.2944 - 6 = -0.7056\).

Calculation for Case (b) dy

Using \(dy = \frac{dy}{dx} \cdot dx\), where \(\frac{dy}{dx} = 2x\).1. At \(x = 3\), \(\frac{dy}{dx} = 2 \cdot 3 = 6\).2. Calculate \(dy = 6 \cdot (-0.12) = -0.72\).

Key concepts

Function Derivatives

Derivatives are a fundamental concept in calculus. They represent the rate at which a function's output changes as its input changes. Essentially, a derivative tells us how much the function value is increasing or decreasing at a specific point on the graph.
For any given function, such as the one in our problem, \[y = x^2 - 3\]calculating the derivative \(\frac{dy}{dx}\) provides a mathematical expression that describes this rate of change.
To find the derivative of \(y = x^2 - 3\), simply apply the power rule of differentiation: if\( y = x^n \), then \( \frac{dy}{dx} = nx^{n-1} \). This means for our function,\[ \frac{dy}{dx} = 2x. \]This derivative tells us that the rate of change of \(y\) depends linearly on \(x\), multiplying by 2.

Differential Calculus

Differential calculus focuses on the concept of the derivative and the process of differentiation. This branch of mathematics allows us to understand how variables change in relation to each other. It answers questions like 'how fast is this value changing?' or 'what would happen if this input changed slightly?'
In our exercise, differential calculus helps us calculate \(dy\), known as the differential of \(y\). The formula \(dy = \frac{dy}{dx} \cdot dx\) is used to estimate small changes in the function's output. The variable \(dx\) represents a tiny change in \(x\). This emphasizes how closely related differentiated values are to the original function.

• For our function, when \(x = 2\)and \(dx = 0.5\), the derivative at \(x\) is 4, and \(dy = 4 \cdot 0.5 = 2.0\).
• Similarly, when \(x = 3\)and \(dx = -0.12\), the derivative is 6, resulting in \(dy = 6 \cdot (-0.12) = -0.72\).

This gives us an approximate answer to the question of how \(y\) changes with a tiny movement in \(x\).

Change in Variable

Change in variable notation comprises \(\Delta y\) and is a primary concern in problems discussing calculus differentials. \(\Delta y\) or 'change in \(y\)' refers to the actual change in the function's output when the input varies from one point to another.
In practical terms, when dealing with functions such as \(y = x^2 - 3\), finding \(\Delta y\) involves computing the difference between the new \(y\) and its original value:

• For instance, with \(x = 2\)and \(\Delta x = 0.5\):
  Initially, \(y = 1\), and after change, \(y = 3.25\), therefore\(\Delta y = 2.25\).
• Similarly for \(x = 3\)and \(\Delta x = -0.12\):
  The initial \(y = 6\) changes to \(y = 5.2944\). Thus \(\Delta y = -0.7056\).

This demonstrates the actual change in the output, making it different from the approximate result given by the differential, \(dy\). Such calculations are crucial for understanding precise changes in real-world contexts.