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Chapter 8: Problem 44

Are length Find the length of the curve \(y=\ln (\sec x)\) \(0 \leq x \leq \pi / 4\)

Short Answer

Expert verified
The length of the curve is 1 unit.

Step by step solution

01

Understand the Formula for Arc Length

The formula to find the length of a curve given by \( y=f(x) \) from \( x=a \) to \( x=b \) is:\[ L = \int_a^b \sqrt{1 + \left(f'(x)\right)^2} \, dx \]In this exercise, the curve is \( y = \ln(\sec x) \) and we need to find its length from \( x=0 \) to \( x=\frac{\pi}{4} \).
02

Compute the Derivative

Find the derivative of the function \( y=\ln(\sec x) \). Using the chain rule, the derivative is:\[ y' = \frac{d}{dx}(\ln(\sec x)) = \sec(x)\tan(x) \]
03

Set Up the Integrand

Substitute \( y' = \sec(x)\tan(x) \) into the integrand of the arc length formula:\[ \sqrt{1 + \left(\sec(x)\tan(x)\right)^2} = \sqrt{1 + \sec^2(x)\tan^2(x)} = \sqrt{\sec^2(x) + \sec^2(x)\tan^2(x)} = \sqrt{\sec^2(x)(1 + \tan^2(x))} = \sqrt{\sec^2(x)\sec^2(x)} = \sec^2(x)\]This is because \( 1 + \tan^2(x) = \sec^2(x) \).
04

Integrate to Find the Arc Length

Integrate \( \sec^2(x) \) over the interval \( [0, \frac{\pi}{4}] \):\[ L = \int_0^{\pi/4} \sec^2(x) \, dx = \left[ \tan(x) \right]_0^{\pi/4} = \tan\left(\frac{\pi}{4}\right) - \tan(0) = 1 - 0 = 1 \]
05

Confirm and Validate the Solution

The integral evaluation simplifies to 1, confirming that our application of the derivative and integral rules was correct. Therefore, the length of the curve \( y = \ln(\sec x) \) from \( x=0 \) to \( x=\frac{\pi}{4} \) is 1 unit.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration
Integration is a fundamental concept in calculus, often likened to the process of adding up an infinite number of small parts to find a whole. It is used in various applications like calculating areas under curves and finding arc lengths, as seen in our example with the curve \( y = \ln(\sec x) \).
The core idea of integration is to solve the integral:
  • Definite Integrals: These are used to calculate the area under a curve from one point to another, integrating over a specific interval. In our example, the integral of \( \sec^2(x) \) was evaluated from \( 0 \) to \( \pi/4 \).

  • Indefinite Integrals: These general forms include a constant and are used when no specific intervals are set, representing a family of functions.

When finding the arc length of a curve, you integrate the square root of \( 1 + (f'(x))^2 \), which in this exercise simplifies to integrating \( \sec^2(x) \).
Derivative
The derivative represents the rate at which a function is changing at any given point, essentially the slope of the function. It's a pivotal tool in calculus as it tells us how the function behaves. In our exercise, the derivative of \( y = \ln(\sec x) \) was found using the chain rule.
The derivative is calculated as follows:
  • Determine the outer function, in this case, the natural logarithm, \( \ln() \).

  • Differentiate the inside function, \( \sec(x) \), then apply the derivative of the outer function to it.

  • The result is \( y' = \sec(x)\tan(x) \), which indicates how steeply \( y = \ln(\sec(x)) \) is inclined at any point \( x \).

In computing arc length, the derivative is essential in transforming the equation before integrating.
Chain Rule
The chain rule is a critical method used in taking derivatives and is especially important in functions composed of multiple layers of functions, such as \( y = \ln(\sec x) \). It helps us find the derivative of a composite function by differentiating the outer function and then multiplying it by the derivative of the inner function.
Here's how it works:
  • Recognize the composition: Here, we have the composition of \( \ln() \) and \( \sec(x) \).

  • Differentiate the outer part: The derivative of \( \ln(u) \) is \( \frac{1}{u} \).

  • Multiply it by the derivative of the inner part: The derivative of \( \sec(x) \) is \( \sec(x)\tan(x) \).

  • Combined, they yield \( y' = \sec(x)\tan(x) \).

This rule enables us to efficiently break down and solve the derivative of composed functions, crucial for further computations like integrating in arc length.
Definite Integral
A definite integral is one of the main tools in calculus used to compute exact areas, volumes, and, as in our example, the arc length of a curve from one point to another. It involves integrating a function within specified limits.
For our exercise, we calculated the definite integral to find the arc length between specific limits:\[ L = \int_0^{\pi/4} \sec^2(x) \, dx \]
This process involves:
  • Evaluating the integral of the function \( \sec^2(x) \).

  • Applying the application "evaluate the integral from the lower limit to the upper limit," which converts the antiderivative back into a context-dependent value.

  • Concluding with \( \tan(\frac{\pi}{4}) - \tan(0) = 1 - 0 \), which yields 1 unit as the arc length.

This result shows how definite integrals can be used to find precise measurements, such as the length of a curve between two points.

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Most popular questions from this chapter

Normal probability distribution The function $$f(x)=\frac{1}{\sigma \sqrt{2 \pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}}$$ is called the normal probability density function with mean \(\mu\) and standard deviation \(\sigma .\) The number \(\mu\) tells where the distribution is centered, and \(\sigma\) measures the "scatter" around the mean. (See Section 8.9. $$\begin{array}{c}{\text { From the theory of probability, it is known that }} \\\ {\int_{-\infty}^{\infty} f(x) d x=1}\end{array}$$ In what follows, let \(\mu=0\) and \(\sigma=1.\) a. Draw the graph of \(f .\) Find the intervals on which \(f\) is increasing, the intervals on which \(f\) is decreasing, and any local extreme values and where they occur. b. Evaluate $$\int_{-n}^{n} f(x) d x$$ for \(n=1,2,\) and 3. c. Give a convincing argument that $$\int_{-\infty}^{\infty} f(x) d x=1.$$ (Hint: Show that \(0 < f(x) < e^{-x / 2}\) for \(x > 1,\) and for \(b >1,\) $$\int_{b}^{\infty} e^{-x / 2} d x \rightarrow 0 \quad \text { as } \quad b \rightarrow \infty_{ .} )$$

Your engineering firm is bidding for the contract to construct the tunnel shown here. The tunnel is 300 \(\mathrm{ft}\) long and 50 \(\mathrm{ft}\) wide at the base. The cross-section is shaped like one arch of the curve \(y=25 \cos (\pi x / 50) .\) Upon completion, the tunnel's inside surface (excluding the roadway) will be treated with a waterproof sealer that costs \(\$ 2.35\) per square foot to apply. How much will it cost to apply the sealer? (Hint: Use numerical integration to find the length of the cosine curve.)

Use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$\int_{0}^{\pi / 2} \cot \theta d \theta$$

Evaluate $$\int_{3}^{\infty} \frac{d x}{x \sqrt{x^{2}-9}}.$$

Use integration, the Direct Comparison Test, or the Limit Comparison Test to test the integrals for convergence. If more than one method applies, use whatever method you prefer. $$\int_{1}^{\infty} \frac{1}{\sqrt{e^{x}-x}} d x$$
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