Problem 4 - Single Variable Calculus

Question

Let \(y=1 / x\). a. Find \(\Delta x\) and \(\Delta y\) if \(x\) changes from 1 to \(1.02\). b. Find the differential \(d y\), and use it to approximate \(\Delta y\) if \(x\) changes from 1 to \(1.02\). c. Compute \(\Delta y-d y\), the error in approximating \(\Delta y\) by \(d y\).

Step-by-Step Solution

Answer

a) ∆x = 0.02 and ∆y ≈ -0.0196 b) dy = -0.02 and dy ≈ -0.0196 when x changes from 1 to 1.02 c) Error ≈ 0.0004

Step 1: Find ∆x

The change in x (∆x) can be calculated by subtracting the initial value of x from the final value. So, given that x changes from 1 to 1.02, that gives us: ∆x = 1.02 - 1 ∆x = 0.02

Step 2: Find ∆y

To find the change in y when x changes from 1 to 1.02, we can first find the values of y when x = 1 and x = 1.02, using the given function: y = 1/x y(x = 1) = 1/1 = 1 y(x = 1.02) = 1/1.02 ≈ 0.9804 Now, we can calculate ∆y: ∆y = y(x = 1.02) - y(x = 1) ∆y = 0.9804 - 1 ∆y ≈ -0.0196

Step 3: Compute dy

To find the differential dy, we need to calculate the derivative of the function y = 1/x: y = 1/x y = x^(-1) Using the power rule, we can find the derivative: y' = -x^(-2) Now, we'll evaluate y' at x = 1: y'(x=1) = -(1)^(-2) y'(1) = -1 Now that we have the derivative y'(x)=-1, we can find dy: dy = y'(x) ∆x dy = -1 * 0.02 dy = -0.02

Step 4: Approximate ∆y using dy

We already have the values for dy and ∆y. Now, let's compare them: Approximated ∆y = dy = -0.02 Actual ∆y ≈ -0.0196

Step 5: Compute the error between ∆y and dy

Finally, we will compute the error between the actual change in y and the approximated value using the differential: Error = ∆y - dy Error ≈ -0.0196 - (-0.02) Error ≈ 0.0004 So, the error in approximating ∆y by dy is approximately 0.0004.

Key Concepts

Derivatives

In differential calculus, derivatives are a vital tool for understanding how functions change. The derivative of a function gives us the rate at which one variable changes in relation to another. In simple terms, it tells us the slope of the tangent line to the function at any given point.
For the function \(y = \frac{1}{x}\), we want to find how \(y\) changes as \(x\) changes. To find the derivative, we use the power rule. Rewriting \(y = \frac{1}{x}\) as \(y = x^{-1}\), the derivative, \(y'\), is found by bringing down the exponent and decreasing it by one.
Hence, \(y' = -x^{-2}\). This derivative tells us that as \(x\) increases, \(y\) decreases, which makes sense since division causes a number to drop when its denominator rises.

Approximations

Approximations are used in calculus to estimate values that might be difficult to calculate precisely. The differential provides an approximation of how much a function changes. It can be particularly useful when dealing with small changes.
Instead of recalculating function values for each little change, we can use the differential as an estimate, which simplifies the process.
In our problem, to approximate \(\Delta y\) as \(x\) changes from 1 to 1.02, we use the differential \(dy\), calculated by multiplying the derivative \(y'(x)\) by the change in \(x\) or \(\Delta x\). Here, \(dy\) gives us \(-0.02\), a convenient and quick way to estimate the actual change \(\Delta y \approx -0.0196\).

Differentials

Differentials are an essential concept in calculus. They represent a linear approximation of how a function's output changes as its input changes slightly.
In our exercise, the differential \(dy\) is calculated using the formula \(dy = y'(x) \Delta x\). By substituting \(y'(x) = -x^{-2}\) and \(\Delta x = 0.02\), we found \(dy = -0.02\).
This linear approximation gives us a handy way to predict the behavior of the function as \(x\) changes, without needing to compute the function value repeatedly. Differentials show the primary impact of changes in \(x\) on \(y\) over small intervals.

Error Analysis

Error analysis is crucial when using differentials to approximate real changes. It helps to quantify how close the approximation is to the actual change.
We calculated the error by subtracting \(dy\) from \(\Delta y\). Since \(\Delta y \approx -0.0196\) and \(dy = -0.02\), the error was calculated as:

• Error = \(\Delta y - dy \)
• Error \approx -0.0196 + 0.02
• Error \approx 0.0004

This shows that the differential gave a very close approximation. Analyzing such errors tells us how good our approximation is, and it's essential for ensuring the reliable application of differentials in real-world scenarios.