Problem 9 - Calculus Single Variable

Question

Calculate the arc length \(L\) of the graph of the given function over the given interval. $$ f(x)=\ln (\cos (x)) \quad I=[0, \pi / 3] $$

Step-by-Step Solution

Answer

The arc length is \(\ln(2 + \sqrt{3})\).

Step 1: Understand the Problem

We need to determine the arc length \(L\) of the curve defined by \(f(x) = \ln(\cos(x))\) over the interval \([0, \pi/3]\). The formula to find the arc length \(L\) of a function \(f(x)\) over an interval \([a, b]\) is given by \(L = \int_{a}^{b} \sqrt{1 + (f'(x))^2} \, dx\).

Step 2: Find the Derivative

To use the arc length formula, we first need the derivative of \(f(x)\). The derivative of \(f(x) = \ln(\cos(x))\) is \(f'(x) = -\tan(x)\), obtained using the chain rule for derivatives.

Step 3: Set Up the Arc Length Integral

Substitute \(f'(x)\) into the arc length formula: \(L = \int_{0}^{\pi/3} \sqrt{1 + (-\tan(x))^2} \, dx\). Simplifying inside the square root, we have \(L = \int_{0}^{\pi/3} \sqrt{1 + \tan^2(x)} \, dx\).

Step 4: Utilize Trigonometric Identity

Use the trigonometric identity \(1 + \tan^2(x) = \sec^2(x)\). Hence, the integral becomes \(L = \int_{0}^{\pi/3} \sqrt{\sec^2(x)} \, dx = \int_{0}^{\pi/3} \sec(x) \, dx\).

Step 5: Integrate \(\sec(x)\)

The integral of \(\sec(x)\) is \(\ln |\sec(x) + \tan(x)| + C\). Therefore, \(L = \left[ \ln |\sec(x) + \tan(x)| \right]_{0}^{\pi/3}\).

Step 6: Evaluate the Definite Integral

Evaluate \(\left[ \ln |\sec(x) + \tan(x)| \right]_{0}^{\pi/3}\) by plugging in the limits of integration: \(\ln |\sec(\pi/3) + \tan(\pi/3)| - \ln |\sec(0) + \tan(0)|\). Since \(\sec(\pi/3) = 2\), \(\tan(\pi/3) = \sqrt{3}\), \(\sec(0) = 1\), and \(\tan(0) = 0\), the expression becomes \(\ln(2 + \sqrt{3}) - \ln(1)\).

Step 7: Simplify the Result

The result simplifies to \(L = \ln(2 + \sqrt{3}) - 0 = \ln(2 + \sqrt{3})\).

Key Concepts

Integration

Integration is a fundamental concept in calculus. It involves finding a function that describes the accumulation of a quantity, such as area under a curve or arc length of a curve.

• The process of integration is used when you have to find the overall sum or total changing quantity within a given interval or region.
• In this exercise, we apply integration to calculate the arc length of a curve over a specified interval.
• To find this, we need to use the specific formula for arc length, which combines both derivatives and integration.

To find the arc length using integration, you start by setting up the integral with respect to the given function. This setup involves substituting the derivative into the arc length formula and evaluating it over the required interval.

Derivative

The derivative measures how a function changes as its input changes. In our problem, we need to find the derivative of the function given by the logarithm of the cosine function, i.e., \(f(x) = \ln(\cos(x))\).

• To find this, we need to use the chain rule, a powerful technique for differentiating compositions of functions.
• Applying the chain rule, the derivative is \(f'(x) = -\tan(x)\).
• This derivative tells us the slope of the curve \(f(x)\) at any point\(x\).

Once the derivative is known, it can then be used in other calculations, such as finding the arc length, by plugging into the integral formula. Understanding derivatives is crucial because they are the basis for further calculations such as slope determination and rate of change, both integral to calculus applications.

Trigonometric Identities

Trigonometric identities are essential tools that simplify mathematical expressions involving trigonometric functions. In the example of solving for arc length, we used one crucial identity.

• The identity \(1 + \tan^2(x) = \sec^2(x)\) simplifies the integration process significantly.
• This identity helps convert complicated trigonometric expressions into easier-to-integrate forms.

By substituting the identity into our arc length integral, we simplified the square root expression to \(\sqrt{\sec^2(x)} = \sec(x)\). This simplification is critical as it makes the integration step much more straightforward, ultimately leading to solving the problem effectively.

Definite Integral

A definite integral is used to calculate the total accumulation between two points. In the context of arc length, the definite integral computes the total length along a curve.

• Definite integrals provide a numerical result over the chosen interval, from \(a\) to \(b\).
• Here, we evaluated the definite integral \(\int_{0}^{\pi/3} \sec(x) \, dx\).
• After integrating \(\sec(x)\), the function becomes \(\ln|\sec(x) + \tan(x)|\), and we evaluate this from 0 to \(\pi/3\).

By substituting the upper and lower limits of the integral into the antiderivative, we find the difference: \(\ln(2 + \sqrt{3}) - \ln(1)\). In this way, definite integrals provide powerful means to find exact arc lengths or areas.