Problem 51 - Calculus - AP Edition

Question

Use the differential \(d y\) to approximate \(\Delta y\) when \(x\) changes as indicated. $$y=\sqrt{3 x-2} ; \text { from } x=2 \text { to } x=2.03$$

Step-by-Step Solution

Answer

\( \Delta y \approx 0.0225 \)

Step 1: Find the derivative of the function

The function given is \( y = \sqrt{3x - 2} \). To find the differential \( dy \) , we first need to find the derivative \( \frac{dy}{dx} \). Using the chain rule, the derivative of \( y \) with respect to \( x \) is \( \frac{d}{dx}(\sqrt{3x - 2}) = \frac{1}{2\sqrt{3x - 2}} \cdot 3 = \frac{3}{2\sqrt{3x - 2}} \).

Step 2: Calculate dy using the differential formula

The formula for the differential \( dy \) is \( dy = \frac{dy}{dx} \cdot dx \). Here, \( dx \) is the change in \( x \), which is \( 2.03 - 2 = 0.03 \). Plugging in the values and \( x = 2 \) into the derivative, we get \( dy = \frac{3}{2\sqrt{3(2) - 2}} \cdot 0.03 \).

Step 3: Evaluate dy at x = 2

Substitute \( x = 2 \) into the derivative to calculate the differential: \( dy = \frac{3}{2\sqrt{3(2) - 2}} \cdot 0.03 = \frac{3}{2}\cdot \frac{1}{\sqrt{4}} \cdot 0.03 = \frac{3}{4}\cdot 0.03 = 0.0225 \).

Step 4: Approximate \( \Delta y \) using dy

The differential \( dy \) gives an approximation of the change in \( y \), denoted by \( \Delta y \), which is \( 0.0225 \). This means that as \( x \) changes from \( 2 \) to \( 2.03 \), \( y \) is approximately increased by 0.0225.

Key Concepts

Derivative Calculations

Derivative calculations form the cornerstone of calculus, enabling us to find the rate of change of a function as its input changes. In the given exercise, we start by finding the derivative of the function \( y = \sqrt{3x - 2} \). The derivative, symbolically represented as \( \frac{dy}{dx} \), tells us how the function \( y \) changes as \( x \) varies.

To compute this derivative, we apply differentiation rules. The chain rule, specifically, helps to differentiate composite functions like \( \sqrt{3x - 2} \), where an inner function \( 3x - 2 \) is plugged into an outer square root function. By using the chain rule, we note that the derivative of \( \sqrt{u} \), where \( u = 3x - 2 \), is \( \frac{1}{2\sqrt{u}} \). Then, we multiply this by the derivative of the inner function \( u = 3x - 2 \), which is \( 3 \).

This combination delivers us the derivative: \( \frac{d}{dx}(\sqrt{3x - 2}) = \frac{3}{2\sqrt{3x - 2}} \). This step is crucial because it provides the necessary expression to determine how fast \( y \) changes with \( x \). Understanding this process makes dealing with different functional forms easier.

Chain Rule

The chain rule is a fundamental technique in calculus used for differentiating compositions of functions—where one function is nested inside another. It allows us to systematically break down and tackle problems where functions are intertwined.

In this exercise, \( y = \sqrt{3x - 2} \), we identify two layers: the outer layer \( f(u) = \sqrt{u} \) and the inner layer \( u(x) = 3x - 2 \). To apply the chain rule, we first take the derivative of the outer function \( \frac{df}{du} = \frac{1}{2\sqrt{u}} \) while keeping the inner function intact. Next, we multiply by the derivative of the inner function \( \frac{du}{dx} = 3 \).

This gives:

• \( \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{3x - 2}} \cdot 3 = \frac{3}{2\sqrt{3x - 2}} \)

The chain rule thus creates a clear method for handling complicated derivatives, effectively managing layers in composite functions. This simplification is a powerful tool in calculus, allowing accurate rate of change evaluations for nested functions, like those presented in our problem.

Differential Approximation

Differential approximation is a technique that uses derivatives to estimate changes in a function's value as its input changes. It is specifically helpful when the change in the input (\( dx \)) is small, providing a quick and efficient way to approximate change.

In our problem, after finding the derivative, we use it to calculate the differential \( dy \), which serves as an approximation for the actual change in \( y \), denoted \( \Delta y \), over the interval from \( x = 2 \) to \( x = 2.03 \).

We use the formula:

• \( dy = \frac{dy}{dx} \cdot dx \)

where \( \frac{dy}{dx} = \frac{3}{2\sqrt{3\cdot2 - 2}} \) was computed earlier. The small increment, \( dx = 0.03 \), plugs into this, giving us \( dy = 0.0225 \).

This value tells us that as \( x \) moves slightly from 2 to 2.03, \( y \) rises approximately by 0.0225. While not exact, differential approximation is practical for swiftly gauging changes in values, leveraging the derivative's rate of change information.