Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the length of each curve. (a) \(y=\int_{\pi / 6}^{x} \sqrt{64 \sin ^{2} u \cos ^{4} u-1} d u, \frac{\pi}{6} \leq x \leq \frac{\pi}{3}\) (b) \(x=a \cos t+a t \sin t, y=a \sin t-a t \cos t,-1 \leq t \leq 1\)

Step-by-Step Solution

Answer

(a) \( \frac{1}{3} (3\sqrt{3} - 1) \); (b) \( a \).

Step 1: Understand the formula for arc length

To find the length of a curve defined by a function, we use the formula for the arc length: \[ L = \int_{a}^{b} \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]For parametric curves, we use the formula: \[ L = \int_{t_1}^{t_2} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \].

Step 2: Solve part (a) with arc length formula

For part \(a\), the function given is:\[ y = \int_{\pi / 6}^{x} \sqrt{64 \sin^2 u \cos^4 u - 1} \, du \]Since \( y = F(x) \) where \( F'(x) = \sqrt{64 \sin^2 x \cos^4 x - 1} \), the derivative \( \frac{dy}{dx} = \sqrt{64 \sin^2 x \cos^4 x - 1} \). The formula for the arc length becomes:\[ L = \int_{\pi/6}^{\pi/3} \sqrt{1 + \left( \sqrt{64 \sin^2 x \cos^4 x - 1} \right)^2} \, dx \]Simplify the expression under the square root:\[ L = \int_{\pi/6}^{\pi/3} \sqrt{1 + 64 \sin^2 x \cos^4 x - 1} \, dx = \int_{\pi/6}^{\pi/3} \sqrt{64 \sin^2 x \cos^4 x} \, dx \]\[ L = \int_{\pi/6}^{\pi/3} 8 \sin x \cos^2 x \, dx \]Use the substitution \( u = \cos x \), \( du = -\sin x \, dx \) to evaluate the integral.

Step 3: Evaluate the integral for part (a)

Substitute the limits and evaluate:\[ L = \int_{\sqrt{3}/2}^{\frac{1}{2}} -8u^2 du \]\[ L = -8 \left[ \frac{u^3}{3} \right]_{\sqrt{3}/2}^{1/2} \]Calculate:\[ L = -8 \left( \frac{1}{24} - \frac{3\sqrt{3}}{24} \right) = -8 \left( \frac{1 - 3\sqrt{3}}{24} \right) \]\[ L = \frac{8(3\sqrt{3} - 1)}{24} = \frac{1}{3} (3\sqrt{3} - 1) \]Hence, the length of the curve for part (a) is \( \frac{1}{3} (3\sqrt{3} - 1) \).

Step 4: Solve part (b) by finding derivatives

For part (b), the parametric equations are given by:\[ x = a \cos t + a t \sin t, \quad y = a \sin t - a t \cos t \]First, find the derivatives:\[ \frac{dx}{dt} = -a \sin t + a \sin t + a t \cos t = a t \cos t \]\[ \frac{dy}{dt} = a \cos t - a \cos t - a t \sin t = -a t \sin t \].

Step 5: Substitute derivatives into the length formula for part (b)

Use the parametric arc length formula:\[ L = \int_{-1}^{1} \sqrt{(a t \cos t)^2 + (-a t \sin t)^2} \, dt \]\[ L = \int_{-1}^{1} \sqrt{a^2 t^2 \cos^2 t + a^2 t^2 \sin^2 t} \, dt \]\[ L = \int_{-1}^{1} \sqrt{a^2 t^2 (\cos^2 t + \sin^2 t)} \, dt \]\[ L = \int_{-1}^{1} \sqrt{a^2 t^2} \, dt \]\[ L = a \int_{-1}^{1} |t| \, dt \]Split the integral based on the absolute value of \( t \):\[ L = a \left( \int_{-1}^{0} -t \, dt + \int_{0}^{1} t \, dt \right) \].

Step 6: Evaluate the integral for part (b)

Calculate the integral:\[ L = a \left( -\left[ \frac{t^2}{2} \right]_{-1}^{0} + \left[ \frac{t^2}{2} \right]_{0}^{1} \right) \]\[ L = a \left( -\left( 0 - \frac{1}{2} \right) + \left( \frac{1}{2} - 0 \right) \right) \]\[ L = a \left( \frac{1}{2} + \frac{1}{2} \right) = a \]Thus, the length of the curve in part (b) is \( a \).

Key Concepts

Parametric Curves

Parametric curves are a special way to represent curves using parameters. Most commonly, we use a variable, typically denoted as \( t \), to express both the \( x \) and \( y \) coordinates as functions of \( t \). This approach offers great flexibility because it doesn’t confine us to a simple \( y = f(x) \) style. Instead, each value of \( t \) corresponds to a specific point on the curve, which can be useful when describing more complex shapes.
For example, in the exercise, part (b) is defined by parameterizing \( x \) and \( y \) coordinates with respect to \( t \): \( x = a \cos t + a t \sin t \) and \( y = a \sin t - a t \cos t \). This allows easy manipulation and integration, especially when curves have loops or multiple parts.

• Useful for complex curves
• Allows separate representation of \( x \) and \( y \)
• Parametric equations help solve intricate mathematical problems

Derivatives

Derivatives are fundamental in calculus, representing the rate at which a function changes. When dealing with parametric curves, derivatives help determine the slope of the tangent line to the curve at any given point. For a parametric function defined in terms of \( t \), we calculate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). These derivatives are used in various calculations, including finding the arc length of a parametric curve.
In the solution for part (b), derivatives provided the essential tools needed to apply the arc length formula. We found:

• \( \frac{dx}{dt} = a t \cos t \)
• \( \frac{dy}{dt} = -a t \sin t \)

These derivatives were crucial for evaluating the curve's geometry and further simplifying the arc length integration process.

Integration

Integration is a core method in calculus for finding the total accumulation of quantities. In terms of curves, it helps find areas under curves or, as in our exercise, the total arc length. The arc length formula for curves in Cartesian coordinates is modified for parametric equations.
The general formula for arc length \( L \) of a parametric curve \( x = f(t) \) and \( y = g(t) \) over an interval \( [t_1, t_2] \) is:
\[ L = \int_{t_1}^{t_2} \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \]
Part (a) involved using integration to find the arc length by simplifying the integral \( \int \sqrt{64 \sin^2 x \cos^4 x} \, dx \) with trigonometric substitution ultimately.

• Integration synthesizes continuous information
• Essential for calculating arc lengths and areas
• Involves definite integrals for specific intervals

Trigonometric Substitution

Trigonometric substitution is an integration technique used to simplify integrands containing square roots, particularly when they involve expressions like \( \sin x \) or \( \cos x \). This method involves substituting trigonometric functions to transform a complex integral into a more manageable form.
In exercise part (a), the substitution \( u = \cos x \), \( du = -\sin x \, dx \), was employed to simplify the expression within the integral. Such substitutions convert the problem into a standard integral form, making it easier to evaluate.
This technique is invaluable in analyzing curves where frequent trigonometric identities play a role, helping break down integrals into simpler parts.

• Great for handling difficult integrals
• Works well with expressions involving square roots
• Widely used in arc length calculations when dealing with trigonometric functions