Problem 6 - Calculus: Early Transcendentals

Question

Let \(f(x)=\sqrt{x}\). If \(a=1\) and \(d x=\Delta x=1 / 10,\) what are \(\Delta y\) and \(d y ?\)

Step-by-Step Solution

Answer

\( \Delta y \approx 0.0488 \) and \( dy = 0.05 \).

Step 1: Identify Given Information

We are given a function \( f(x) = \sqrt{x} \), with \( a = 1 \) and \( dx = \Delta x = \frac{1}{10} = 0.1 \). We need to find \( \Delta y \) and \( dy \).

Step 2: Define Change in Y, \(\Delta y\)

\( \Delta y \) is the actual change in the function value, calculated as \( f(a + \Delta x) - f(a) \). Substitute the values to find \( \Delta y = f(1.1) - f(1) \).

Step 3: Calculate Function Values

Calculate \( f(1) = \sqrt{1} = 1 \) and \( f(1.1) = \sqrt{1.1} \).

Step 4: Calculate \(\Delta y\)

Substitute the function values into \( \Delta y = \sqrt{1.1} - 1 \). Upon evaluating \( \sqrt{1.1} \approx 1.0488 \), thus \( \Delta y \approx 1.0488 - 1 = 0.0488 \).

Step 5: Right Differential Formula

The differential \( dy \) is approximated by \( f'(a)\cdot dx \), where \( f'(x) \) is the derivative of \( f(x) \).

Step 6: Find Derivative of \(f(x)\)

Find \( f'(x) = \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} \).

Step 7: Evaluate Derivative at \(a=1\)

Evaluate the derivative at \( a = 1 \) to find \( f'(1) = \frac{1}{2\sqrt{1}} = \frac{1}{2} \).

Step 8: Calculate \(dy\)

Substitute into the formula \( dy = f'(1) \cdot dx = \frac{1}{2} \cdot 0.1 = 0.05 \).

Key Concepts

Function Derivatives

Function derivatives are essential tools in calculus that help measure how a function's output value changes as its input changes. In our exercise, the function is given as \( f(x) = \sqrt{x} \). To find its derivative, we perform differentiation. Differentiation gives us another function that describes the rate of change of \( f(x) \) with respect to \( x \).
This process involves rules like the power rule, chain rule, among others. For \( f(x) = \sqrt{x} \), it can be rewritten as \( x^{1/2} \). Applying the power rule, where we multiply by the exponent and subtract one, we get:

• The derivative, \( f'(x) = \frac{1}{2}x^{-1/2} \) or equivalently, \( \frac{1}{2\sqrt{x}} \).

This derivative tells us for small changes in \( x \), how much \( f(x) \) will change. For our specific case, we evaluate this derivative at \( x = 1 \), providing us with \( f'(1) = \frac{1}{2} \). This means at the point \( x = 1 \), for every unit change in \( x \), the function changes by half a unit.

Approximation

Approximation is a valuable concept in calculus, allowing us to estimate the behavior of functions. In our exercise, we find that the differential \( dy \) serves as an approximation of the change in \( y \) when \( x \) changes. The differential is derived from the derivative we discussed, using the relation:

• \( dy = f'(a) \cdot dx \)

We use this formula to estimate how the function \( f(x) = \sqrt{x} \) behaves around the point \( x = 1 \) when there's a small change \( dx = 0.1 \). We find that \( dy = \frac{1}{2} \cdot 0.1 = 0.05 \).
This approximation process can be quite useful when computing exact values is challenging. In practice, approximations like this help in predicting changes in functions without actually having to recalculate the function for every small incremental change.

Change in Functions

In calculus, understanding the change in functions is fundamental. It occurs when inputs change, which results in outputs changing too. The exercise asks us to calculate both \( \Delta y \) and \( dy \) to grasp actual versus approximated changes in values.
\( \Delta y \) measures the actual change in the function's value when \( x \) changes from \( a \) to \( a + \Delta x \). It is calculated as follows:

• \( \Delta y = f(a + \Delta x) - f(a) \)

Substituting our values, we calculate \( \Delta y = \sqrt{1.1} - 1 \approx 0.0488 \). Meanwhile, \( dy \) provides an estimated change using the derivative, which in our case is \( dy = 0.05 \).
While both values are close, \( \Delta y \) is the more accurate computation of change, while \( dy \) serves as a quick approximation. Understanding these concepts allows students to make subtle predictions about how functions behave based on small changes, a skill that's essential in many fields of science and engineering.