Chapter 7: Problem 55
Find the lengths of the curves. $$y=\left(x^{2} / 8\right)-\ln x, \quad 4 \leq x \leq 8$$
Short Answer
Expert verified
The curve length is approximately 11.040 units.
Step by step solution
01
Understand Curve Length Formula
The length of a curve from \(x = a\) to \(x = b\) is given by the formula \(L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\). We need to find the derivative of the function first.
02
Compute the Derivative
Given \(y=\frac{x^{2}}{8}-\ln x\), find the derivative with respect to \(x\). The derivative is \(\frac{dy}{dx} = \frac{1}{4}x - \frac{1}{x}\).
03
Set Up the Integral for Arc Length
Substitute \(\frac{dy}{dx}\) into the arc length formula: \[L = \int_{4}^{8} \sqrt{1 + \left(\frac{1}{4}x - \frac{1}{x}\right)^2} \, dx\]
04
Simplify and Evaluate the Expression
Simplify the expression under the square root: \[1 + \left(\frac{1}{4}x - \frac{1}{x}\right)^2 = \frac{x^4}{16} - \frac{x^2}{2} + \frac{1}{x^2} + 1\]. Thus, \[L = \int_{4}^{8} \sqrt{\frac{x^4}{16} - \frac{x^2}{2} + \frac{1}{x^2} + 1} \, dx\]
05
Calculate the Integral
Evaluate the definite integral \(\int_{4}^{8} \sqrt{\frac{x^4}{16} - \frac{x^2}{2} + \frac{1}{x^2} + 1} \, dx\) using numerical methods or a calculator, since this integral does not have a simple antiderivative.
06
Present the Length
After evaluating the integral, the length of the curve is obtained. The approximate length of the curve from \(x = 4\) to \(x = 8\) is 11.040 units.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Arc Length Formula
To calculate the length of a curve, we use the arc length formula. This formula is incredibly useful because it allows us to find the actual distance along a curved line between two points. The formula is given by:\[ L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]This formula calculates the length from \(x = a\) to \(x = b\). It combines the idea of a derivative, which describes the slope or steepness of the curve at any point, into an integral, which sums up these small slopes into a total length over an interval.
- \( \frac{dy}{dx} \) represents the derivative of the curve function, indicating how steep the curve is at each point.
- The square root component \( \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \) adjusts for changes in both horizontal movement and vertical elevation, offering a complete arc length measure.
Definite Integral
A definite integral is essential in computing the length or area over a particular range. In our context, it's used to find the length of a curve, specifically from one \(x\) value to another. In terms of arc length, the integral represents the total of all infinitesimally small line segments that form the curve.Consider the set-up:\[ L = \int_{4}^{8} \sqrt{1 + \left(\frac{1}{4}x - \frac{1}{x}\right)^2} \, dx \]
- The limits of integration, here 4 and 8, define where our calculation begins and ends on the curve.
- The integral with a square root captures each small segment of the curve and sums them across the specified limits.
- This results in what is called a 'numerical approximation." Many definite integrals do not have straightforward formulas for solutions; hence calculators or numerical methods are commonly used.
Derivative Computation
Calculating the derivative is a fundamental step in deriving the arc length. The derivative tells us how the curve behaves, allowing us to adjust our arc length formula accordingly.For the function provided:\[ y = \frac{x^2}{8} - \ln x \]The derivative, \( \frac{dy}{dx} \), is computed as:\[ \frac{dy}{dx} = \frac{1}{4}x - \frac{1}{x} \]
- The term \( \frac{1}{4}x \) arises from differentiating \( \frac{x^2}{8} \), conforming to the power rule.
- \( -\frac{1}{x} \) comes from differentiating \(-\ln x\), utilizing a standard derivative rule for logarithms.