Problem 38 - Calculus

Question

Estimate the change in \(y\) for the given change in \(x\).$$y=f(x), f^{\prime}(100)=0.4, x \text { increases from } 100 \text { to } 101$$.

Step-by-Step Solution

Answer

The estimated change in \(y\) is 0.4.

Step 1: Understand the Problem

The problem asks us to estimate the change in the value of \(y\) given the change in \(x\). It provides \(f^{\prime}(x)\), the derivative of \(f(x)\), evaluated at \(x=100\), which is \(0.4\). This represents the rate of change of \(y\) with respect to \(x\). The task is to estimate the change in \(y\) as \(x\) changes from \(100\) to \(101\).

Step 2: Use Linear Approximation

Using the concept of linear approximation or differentials, the change in \(y\), denoted as \(\Delta y\), can be estimated by the derivative at the point of interest multiplied by the change in \(x\), \(\Delta x\). The formula used here is: \[ \Delta y \approx f^{\prime}(x_0) \cdot \Delta x \] where \(x_0\) is the point where the derivative is known, which in this case is \(100\).

Step 3: Calculate the Change in \(x\)

The change in \(x\), denoted as \(\Delta x\), is the difference between the new \(x\) value and the initial \(x\) value. Given that \(x\) increases from \(100\) to \(101\), we calculate:\[ \Delta x = 101 - 100 = 1 \].

Step 4: Estimate the Change in \(y\)

Now substitute \(f^{\prime}(100) = 0.4\) and \(\Delta x = 1\) into the linear approximation formula to find \(\Delta y\):\[ \Delta y \approx 0.4 \cdot 1 = 0.4 \]

Step 5: Interpret the Result

The estimated change in \(y\) when \(x\) changes from \(100\) to \(101\) is \(0.4\). This means that as \(x\) increases by \(1\) unit, \(y\) is expected to increase by approximately \(0.4\) units.

Key Concepts

Differentials

Differentials are a key concept in calculus, helping us approximate changes in function values. When we talk about differentials, we refer to small changes or increments represented mathematically. In the context of a function, if we have a small change in the input value, known as \( \Delta x \), this will produce a corresponding change in the output value, \( \Delta y \).

Differentials make it simple to estimate how much a function's output will change in response to a minor change in the input. We often use the derivative of a function to find the differential. If we know the derivative of a function \( f \) at a particular point, we denote this small change by using the following approximation:

• \( \Delta y \approx f'(x) \cdot \Delta x \)

For example, in the exercise, the derivative at \( x=100 \) is \( f'(100)=0.4 \). This tells us that around this point, a small change in \( x \) will result in a proportional change in \( y \), approximately \(0.4 \) times the change in \( x \).

Differentials are handy in real-life applications where precise calculations are tough, but quick estimates are needed. They allow us to think about small parts of the graph as straight lines, which are easier to work with.

Derivative

The derivative is a fundamental concept in calculus that represents the rate at which a function is changing at any given point. In simpler terms, it gives us a "snapshot" of how the function behaves at an exact point. When someone says "derivative," it usually concerns how sensitive the function is to changes in its input.

In mathematical notation, if \( y = f(x) \), then the derivative of \( y \) with respect to \( x \) is denoted as \( f'(x) \) or sometimes \( \frac{dy}{dx} \). The purpose of finding the derivative is to know how fast \( y \) is changing as \( x \) changes. This is incredibly useful in problems like the one in our exercise, where we're interested in estimating changes in \( y \) given changes in \( x \).

The value of the derivative tells us the "steepness" or slope of the tangent line to the function at a particular point. In the exercise, \( f'(100) = 0.4 \) means that at \( x=100 \), the slope of the function is \( 0.4 \). This indicates that when \( x \) increases by 1 unit, and if the function behaves linearly for small changes, \( y \) is expected to increase by approximately 0.4 units.

Understanding derivatives is a stepping-stone to more advanced calculus topics and is essential for solving real-world problems where change is constant.

Rate of Change

The rate of change of a function provides insights into how one quantity varies in relation to another. In practical terms, it's the speed at which one variable changes relative to another. Often, the rate of change is explored using derivatives, as they precisely tell us this rate at specific points.

In our exercise, the rate of change of the function \( y=f(x) \) with respect to \( x \) at the point \( x=100 \) is given as \( 0.4 \). This means that for every single unit increase in \( x \) near this point, \( y \) is expected to increase by about 0.4 units. Therefore, the derivative at a point is a measure of the rate of change of the function at that exact point.

When we discuss "rate of change," it can relate to different contexts. It's not just restricted to mathematics. For instance:

• Economics: How profit changes as costs increase.
• Physics: How distance changes as time progresses.
• Biology: How a population grows in relation to time.

The concept is vital because it helps explain natural phenomena and makes predictions over small intervals based on available data. Understanding the rate of change empowers us to predict behavior and make informed decisions in numerous fields.