Problem 44 - Calculus Early Transcendentals: Pearson New International Edition

Question

, 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 evaluated as an integral: \( \int_{-3}^{3} \sqrt{\text{sech}^4 t + 4 \tanh^2 t} \, dt \).

Step 1: Identify the Formula for Arc Length

The formula for the length of a parametric curve is given by \( L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2} } \, dt \).

Step 2: Compute the Derivative \( \frac{dx}{dt} \)

Given \( x = \tanh t \), we use its derivative: \( \frac{dx}{dt} = \text{sech}^2 t \).

Step 3: Compute the Derivative \( \frac{dy}{dt} \)

Given \( y = \ln(\cosh^2 t) \), applying the chain rule gives \( \frac{dy}{dt} = 2 \cdot \frac{1}{\cosh^2 t} \cdot \sinh t \), simplifying to \( \frac{dy}{dt} = 2 \tanh t \).

Step 4: Substitute into the Arc Length Formula

Substitute \( \frac{dx}{dt} = \text{sech}^2 t \) and \( \frac{dy}{dt} = 2 \tanh t \) into the formula: \( L = \int_{-3}^{3} \sqrt{(\text{sech}^2 t)^2 + (2 \tanh t)^2} \, dt \).

Step 5: Simplify the Expression Inside the Integral

Calculate \( (\text{sech}^2 t)^2 = \text{sech}^4 t \) and \( (2 \tanh t)^2 = 4 \tanh^2 t \), leading to \( \sqrt{\text{sech}^4 t + 4 \tanh^2 t} \). Simplify using trigonometric identities.

Step 6: Evaluate the Integral

Evaluate \( L = \int_{-3}^{3} \sqrt{\text{sech}^4 t + 4 \tanh^2 t} \, dt \). Exact simplification often involves recognizing \( \text{sech}^2 t + \tanh^2 t = 1 \), therefore simplifying further calculations ensuring proper understanding of standard integration techniques if applied.

Key Concepts

Arc Length Formula

To find the length of a parametric curve, we use the arc length formula. This is given by:\[L = \int_{a}^{b} \sqrt{\left( \frac{dx}{dt} \right)^{2} + \left( \frac{dy}{dt} \right)^{2} } \, dt\]This formula applies to curves defined by parametric equations where each variable, typically \(x\) and \(y\), is expressed in terms of a third variable, often \(t\).

Make sure to:

• Identify the correct limits of integration based on the given interval for \(t\).
• Compute the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) accurately.
• Substitute these derivatives into the formula and perform the integration over the specified interval.

Such computations allow us to measure the true length of even complex curves, which a simple distance formula cannot achieve.

Hyperbolic Functions

Hyperbolic functions, like trigonometric functions, are often used in the context of calculus, especially when dealing with parametric curves. Some of the most common hyperbolic functions are \(\sinh t\), \(\cosh t\), and \(\tanh t\), each defined as follows:

• \(\sinh t = \frac{e^t - e^{-t}}{2}\)
• \(\cosh t = \frac{e^t + e^{-t}}{2}\)
• \(\tanh t = \frac{\sinh t}{\cosh t}\)

These functions share several properties with trigonometric ones, such as identities and derivatives.

For instance:

• \(\text{sech}^2 t + \tanh^2 t = 1\)
• \(\frac{d}{dt}\sinh t = \cosh t\)
• \(\frac{d}{dt}\cosh t = \sinh t\)
• \(\frac{d}{dt}\tanh t = \text{sech}^2 t\)

These properties make them essential tools in calculus, especially when defining or analyzing the components of a curve along specific intervals.

Derivative Computation

Taking derivatives is a central task when working with parametric curves, as it allows us to explore and use the properties of functions within these curves.

For example, if given the parametric equation \(x = \tanh t\), we compute its derivative as follows:

• \(\frac{dx}{dt} = \text{sech}^2 t\)

You compute \(\frac{dy}{dt}\) similarly, often using the chain rule. For \(y = \ln(\cosh^2 t)\):

• Apply the chain rule: calculate the derivative of the outer function \(\ln u\) as \(\frac{1}{u}\), and multiply by the derivative of the inner function.
• Simplify resulting in \(\frac{dy}{dt} = 2 \tanh t\).

Remembering these steps and using proper calculus techniques facilitates solving even seemingly complex problems.

Trigonometric Identities

Trigonometric identities are valuable in simplifying expressions and solving integrals when working with parametric and hyperbolic functions.

For instance, identify crucial identities like:

• \(\sin^2 \theta + \cos^2 \theta = 1\)
• \(1 + \tan^2 \theta = \sec^2 \theta\)
• \(\text{sech}^2 t + \tanh^2 t = 1\)

These identities can simplify the integrands of functions we encounter in calculus.

Used properly, they reduce complex expressions into more manageable forms. By substituting or reducing terms using these identities, integrals that result in terms like \(\sqrt{\text{sech}^4 t + 4 \tanh^2 t}\) become easier to evaluate, making the solution not only accurate but also efficient.