Problem 41

Question

Find formulas for $d y$ and $\Delta y$. $$y=x^{2}-2 x+1$$

Step-by-Step Solution

Verified

Answer

$dy = (2x - 2)dx$; $\Delta y = 2x\Delta x + (\Delta x)^2 - 2\Delta x$.

1Step 1: Differentiate the function

We start by differentiating the function with respect to $x$ to find $dy$. The derivative of $y = x^2 - 2x + 1$ is obtained by applying basic differentiation rules. The derivative of $x^2$ is $2x$, the derivative of $-2x$ is $-2$, and the derivative of a constant term $1$ is $0$. Thus, $\frac{dy}{dx} = 2x - 2$.

2Step 2: Solve for dy

To express $dy$, multiply both sides of the derivative equation by $dx$. This gives: \[dy = (2x - 2)dx.\] This formula helps in finding the infinitesimal change in $y$ with respect to an infinitesimal change in $x$.

3Step 3: Express the incremental change in y, $\Delta y$

The incremental change in $y$, known as $\Delta y$, is expressed as $\Delta y = f(x + \Delta x) - f(x)$. Substitute the given function $y = x^2 - 2x + 1$ into this expression: $ f(x + \Delta x) = (x + \Delta x)^2 - 2(x + \Delta x) + 1$.

4Step 4: Expand and simplify the expression

Expand $(x + \Delta x)^2$ to $x^2 + 2x\Delta x + (\Delta x)^2$, and $-2(x + \Delta x)$ to $-2x - 2\Delta x$. Combine these in the $f(x + \Delta x)$ expression: \[((x + \Delta x)^2 - 2(x + \Delta x) + 1) = x^2 + 2x\Delta x + (\Delta x)^2 - 2x - 2\Delta x + 1.\]

5Step 5: Compute $\Delta y$

Substitute the original function value $f(x) = x^2 - 2x + 1$ and the expanded expression for $f(x + \Delta x)$ into the $\Delta y$ formula. This results in: \[\Delta y = (x^2 + 2x\Delta x + (\Delta x)^2 - 2x - 2\Delta x + 1) - (x^2 - 2x + 1).\] Simplify to get: \[\Delta y = 2x\Delta x + (\Delta x)^2 - 2\Delta x.\]

Key Concepts

Understanding the DerivativeInfinitesimal Change and dyIncremental Change and Delta y

Understanding the Derivative

The derivative of a function is a key concept in differential calculus. It represents the rate at which a function is changing at any given point. Think of it as the slope of the tangent line to the curve at a point. For the function given, $ y = x^2 - 2x + 1 $, we used differentiation rules to find its derivative: $ \frac{dy}{dx} = 2x - 2 $. This derivative tells us how the function $ y $ changes with respect to small changes in $ x $.

Rate of Change: The derivative $ 2x - 2 $ signifies how $ y $ increases or decreases as $ x $ changes.
Tangent Slope: It provides the slope of the tangent line to the graph of $ y $ at any point $ x $.

Understanding derivatives is crucial for interpreting the behavior of functions and applying calculus effectively to real-world problems.

Infinitesimal Change and dy

In calculus, infinitesimal change refers to an extremely small change in a variable. When we talk about $ dy $, we are considering the infinitesimal change in $ y $ corresponding to an infinitesimal change in $ x $, denoted as $ dx $.To find $ dy $ for the function $ y = x^2 - 2x + 1 $, we multiplied the derivative by $ dx $, leading to $ dy = (2x - 2)dx $. This formula expresses how $ y $ changes momentarily around a specific $ x $ value.

Infinitesimal Change: $ dy $ gives us an insight into how the output changes for a minuscule input variation.
Local Linear Approximation: It offers a linear approximation of the function at very small scales, almost like "zooming in" to see the curve as a straight line.

This concept is instrumental in understanding how functions behave in the infinitesimally small neighborhoods of points.

Incremental Change and Delta y

Incremental change, represented by $ \Delta y $, can be visualized as the difference in the function's outcome over a finite change in the input. It is a broader measure than $ dy $ because $ \Delta x $ is not infinitesimally small but still possibly small enough to examine the function's behavior.Steps were taken to compute $ \Delta y $ for the function $ y = x^2 - 2x + 1 $ over an interval $ \Delta x $. We found that $ \Delta y = 2x\Delta x + (\Delta x)^2 - 2\Delta x $. This represents the actual change in $ y $ when $ x $ is increased by $ \Delta x $.

Finite Change: Unlike $ dy $, $ \Delta y $ is about noticeable changes over specific intervals.
Approximation Improvements: When $ \Delta x $ is small, $ \Delta y \approx dy $ providing a practical approximation of the actual change.

Grasping the idea of incremental changes in functions is beneficial for understanding how real-world scenarios evolve over measurable differences.

Problem 41

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