Problem 44 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 35-46, find the length of the parametric curve defined over the given interval. $$ x=\tanh t, y=\ln \left(\cosh ^{2} t\right) ;-3 \leq t \leq 3 $$

Step-by-Step Solution

Answer

The length of the curve is approximately 1.99 units.

Step 1: Recall the Formula for the Length of a Parametric Curve

The formula to find the length of a parametric curve \( (x(t), y(t)) \) over an interval \([a,b]\) is given by:\[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]

Step 2: Calculate Derivative of x with respect to t

Given \( x = \tanh t \), we need to find \( \frac{dx}{dt} \). The derivative of \( \tanh t \) is \( \text{sech}^2 t \). So, \( \frac{dx}{dt} = \text{sech}^2 t \).

Step 3: Calculate Derivative of y with respect to t

Given \( y = \ln (\cosh^2 t) \), apply the chain rule to find \( \frac{dy}{dt} \).First find: \( \frac{d}{dt} (\ln (\cosh^2 t)) = \frac{1}{\cosh^2 t} \cdot \frac{d}{dt} (\cosh^2 t) \).Using the derivative \( \frac{d}{dt} (\cosh^2 t) = 2 \cosh t \cdot \sinh t \) which simplifies to \( \sinh(2t) \).Finally, \( \frac{dy}{dt} = \frac{\sinh(2t)}{\cosh^2 t} = 2 \tanh t \).

Step 4: Setup the Integral for the Curve Length

Substitute the derivatives calculated into the curve length formula:\[ L = \int_{-3}^{3} \sqrt{\left(\text{sech}^2 t \right)^2 + \left(2 \tanh t \right)^2} \, dt \]

Step 5: Simplify the Integrand

Simplify the expression within the integral:\[ \left(\text{sech}^2 t \right)^2 + \left(2 \tanh t \right)^2 = \text{sech}^4 t + 4 \tanh^2 t \]Using \( \tanh^2 t = 1 - \text{sech}^2 t \):\[ = \text{sech}^4 t + 4(1 - \text{sech}^2 t) = \text{sech}^4 t + 4 - 4 \text{sech}^2 t \]

Step 6: Solve the Integral

Using simplification and properties of hyperbolic functions, \[ L = \int_{-3}^{3} \text{sech}^2 t \, dt \]Solving this gives:\[ \int_{-3}^{3} \text{sech}^2 t \, dt = \left. \tanh t \right|_{-3}^{3} \]Since \( \tanh(3) \approx 0.995 \) and \( \tanh(-3) \approx -0.995 \) the length is:\[ L = 0.995 - (-0.995) = 1.99 \]

Step 7: Conclude the Calculation

Thus, after performing the integration, the length of the parametric curve \( x = \tanh t, y = \ln (\cosh^2 t) \) over the interval \([-3, 3]\) is approximately 1.99.

Key Concepts

Hyperbolic Functions in Calculus

Hyperbolic functions are analogous to trigonometric functions but are based on hyperbolas rather than circles. In this problem, we encounter two common hyperbolic functions: \( \tanh t \) and \( \cosh t \).
\( \tanh t \) stands for hyperbolic tangent, and \( \cosh t \) is the hyperbolic cosine function. These functions have properties that are useful in calculus, such as derivatives and integrals, which are crucial for solving problems involving parametric curves.

• The derivative of \( \tanh t \) is \( \text{sech}^2 t \), where \( \text{sech} t = \frac{1}{\cosh t} \).
• The function \( \cosh t \) itself is defined by \( \frac{e^t + e^{-t}}{2} \).
• Hyperbolic identities, such as \( \tanh^2 t = 1 - \text{sech}^2 t \), mimic trigonometric identities like \( \sin^2 \theta + \cos^2 \theta = 1 \).

Understanding these properties is key to managing the algebra involved in problems using hyperbolic functions.

Integration in Calculus

Integration is a fundamental process in calculus. It reverses differentiation and is used to calculate areas, volumes, and in this case, lengths of curves. When dealing with parametric equations, the integration takes a slightly different form:
The length of a parametric curve \((x(t), y(t))\) is determined using:\[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]

• This formula accounts for changes in both the x and y directions as the parameter \( t \) varies.
• Replacing the derivatives calculated, you simplify the integrand.
• Sometimes, hyperbolic identities or algebraic manipulation are required to make the integral easier to solve.
• For example, using the identity \( \tanh^2 t = 1 - \text{sech}^2 t \) simplified our integral.

Successful integration, in this case, provided the curve's length, leading to a final result after evaluating the integral's limits.

Derivative Calculation

Calculating derivatives is a key skill in calculus, and it plays a critical role when working with parametric curves. Derivatives tell us the rate of change of a function with respect to a variable. In this exercise:
Two derivatives are essential to find:

• The derivative of \( x = \tanh t \) is \( \text{sech}^2 t \).
• The derivative of \( y = \ln(\cosh^2 t) \) is found using the chain rule. First, \( \ln(\cosh^2 t) \) becomes \( \frac{1}{\cosh^2 t} \times \sinh(2t) \) simplifying to \( 2 \tanh t \).
• To find these, recognize the connection between hyperbolic functions and their derivatives. The chain rule helps when differentiating composite functions like \( \ln(\cosh^2 t) \).

By determining these derivatives, we can substitute back into the length formula, enabling calculation of the curved arc's precise length over the given interval.