Problem 30 - Thomas Calculus

Question

Find the lengths of the curves. $$ \begin{array}{l}{x=\ln (\sec t+\tan t)-\sin t} \\ {y=\cos t, \quad 0 \leq t \leq \pi / 3}\end{array} $$

Step-by-Step Solution

Answer

The length of the curve is \(\ln(2)\).

Step 1: Understand the Arc Length Formula

To find the length of a curve given in parametric form, we use the arc length formula: \[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]where \(x = f(t)\) and \(y = g(t)\). Here, we have \(x = \ln(\sec t + \tan t) - \sin t\) and \(y = \cos t\).

Step 2: Differentiate x and y with respect to t

Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\):- \(x = \ln(\sec t + \tan t) - \sin t\) Using the derivative of \(\ln u\), we get: \(\frac{d}{dt}[\ln(\sec t + \tan t)] = \frac{1}{\sec t + \tan t} \left(\sec t \tan t + \sec^2 t\right) = \sec t\) Thus, \(\frac{dx}{dt} = \sec t - \cos t\).- For \(y = \cos t\), \(\frac{dy}{dt} = -\sin t\).

Step 3: Substitute into the Arc Length Formula

Substitute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) into the arc length formula:\[L = \int_{0}^{\pi/3} \sqrt{(\sec t - \cos t)^2 + (-\sin t)^2} \, dt\]

Step 4: Simplify the Integrand

Simplify the expression under the square root:\[(\sec t - \cos t)^2 = \sec^2 t - 2\sec t \cos t + \cos^2 t = 1 + \tan^2 t - 2 + \cos^2 t = \tan^2 t\]This simplifies to:\[L = \int_{0}^{\pi/3} \sqrt{\tan^2 t + \sin^2 t} \, dt\]Recognizing trigonometric identities can lead to further simplifications.

Step 5: Further Simplify and Integrate

Combine the integrand using identities:Notice \(\tan^2 t + \sin^2 t = \frac{\sin^2 t}{\cos^2 t} + \sin^2 t = \sin^2 t (1 + \frac{1}{\cos^2 t}) = \sin^2 t \sec^2 t\) which simplifies further:\[L = \int_{0}^{\pi/3} \sqrt{\sin^2 t \sec^2 t} \, dt = \int_{0}^{\pi/3} |\sin t \sec t| \, dt\]In the range \([0, \pi/3]\), \(\sin t\) and \(\sec t\) are positive, so:\[L = \int_{0}^{\pi/3} \sin t \sec t \, dt = \int_{0}^{\pi/3} \tan t \, dt\]

Step 6: Evaluate the Integral

Evaluate the integral:\[L = \int_{0}^{\pi/3} \tan t \, dt = [-\ln|\cos t|]_{0}^{\pi/3}\]Substitute the limits:\[L = -\ln|\cos(\pi/3)| + \ln|\cos(0)| = -\ln\left(\frac{1}{2}\right) + 0 = \ln(2)\]

Step 7: Conclusion

The length of the curve from \(t = 0\) to \(t = \pi/3\) is \(\ln(2)\).

Key Concepts

Parametric Equations

Parametric equations are a way of defining a curve using a set of equations, with each equation representing a different dimension of the curve and using a common variable, usually denoted as \(t\).
In our problem, we are given the parametric equations \(x = \ln(\sec t + \tan t) - \sin t\) and \(y = \cos t\). This means that as the value of \(t\) changes, the values of \(x\) and \(y\) together describe the movement along a curve from one point to another.
Parametric equations are useful in situations where coordinates alone don't fully describe the position, because they can easily model motion and time-dependent processes. In our exercise, \(t\) varies from 0 to \(\pi/3\), which gives us a specific segment of the curve. To find the length of this segment, we make use of calculus, specifically the arc length formula for parametric equations.

Trigonometric Functions

Trigonometric functions play a crucial role in many mathematical problems, especially those involving curves and cycles such as sine, cosine, and tangent. In this problem, trigonometric functions are used in both the definitions of \(x\) and \(y\). Understanding their properties and identities is essential for simplifying expressions and solving integrals.
For example, in the derivatives we calculated in the solution, the derivatives of trigonometric functions were vital. The derivative of \(\cos t\) is \(-\sin t\), and knowing this helps us compute easier steps in our solutions.

• \(\sec t\) is the reciprocal of the cosine function, so \(\sec t = 1/\cos t\).
• \(\tan t\) is \(\sin t/\cos t\), which also appears in many calculations.

The exercise uses the arc length formula which incorporates these functions. Also, to simplify the expressions further, certain identities such as \(\tan^2 t + 1 = \sec^2 t\) are used which are crucial for making integration possible.

Integration Techniques

Integration is a fundamental concept in calculus and helps to determine the area under a curve or, in this case, the length of a curve. The exercise involves definite integration from 0 to \(\pi/3\). The formula for the arc length of a parametric curve uses integration to accumulate the entire length of the curve described by the parameters \(x(t)\) and \(y(t)\).
Once we have the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), they are used to find \(\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}\). This integrand is simplified using trigonometric identities. Understanding how to manipulate these forms is essential for efficient problem-solving.
The key was simplifying \(\int_{0}^{\pi/3} \tan t \, dt\), which is a standard integral resulting in \(-\ln|\cos t|\). Substituting the limits of integration gives the length of the curve. Recognizing when a standard form appears in an integrand allows us to solve it more efficiently, emphasizing the importance of mastering a variety of integration techniques.