Problem 46 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 35-46, find the length of the parametric curve defined over the given interval. $$ x=\sin t-t \cos t, y=\cos t+t \sin t ; \frac{\pi}{4} \leq t \leq \frac{\pi}{2} $$

Step-by-Step Solution

Answer

The length of the curve is approximately \( 0.7101 \).

Step 1: Identify the Parametric Equations

Given the parametric equations \( x = \sin t - t \cos t \) and \( y = \cos t + t \sin t \), we'll use these to find the derivatives with respect to \( t \).

Step 2: Find Derivatives of x and y

To find the length of the curve, we need \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).\[ \frac{dx}{dt} = \cos t + t\sin t \] \[ \frac{dy}{dt} = -\sin t + t\cos t \]

Step 3: Determine the Formula for Curve Length

The length \( L \) of a parametric curve from \( t = a \) to \( t = b \) is given by:\[ L = \int_{a}^{b} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt \] In this case, \( a = \frac{\pi}{4} \) and \( b = \frac{\pi}{2} \).

Step 4: Substitute the Derivatives into the Formula

Substitute \( \frac{dx}{dt} = \cos t + t\sin t \) and \( \frac{dy}{dt} = -\sin t + t\cos t \) into the length formula:\[ L = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{ \left( \cos t + t\sin t \right)^2 + \left( -\sin t + t\cos t \right)^2 } \, dt \]

Step 5: Simplify the Integrand Inside the Root

Calculate inside the square root:\( \left( \cos t + t\sin t \right)^2 = \cos^2 t + 2t \sin t \cos t + t^2 \sin^2 t \) \( \left( -\sin t + t\cos t \right)^2 = \sin^2 t - 2t \sin t \cos t + t^2 \cos^2 t \) Combine them:\( \cos^2 t + 2t \sin t \cos t + t^2 \sin^2 t + \sin^2 t - 2t \sin t \cos t + t^2 \cos^2 t = 1 + t^2 \)

Step 6: Evaluate the Integral

Now the expression inside the integral is simplified:\[ L = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{1 + t^2} \, dt \] This integral should be evaluated using standard integration techniques such as substitution or numerical methods.

Step 7: Final Step: Solution

Evaluating the integral for \( L = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{1 + t^2} \, dt \), we get a numerical result: approximately \( \frac{1}{2} \left( \sqrt{5} - \sqrt{2} + \ln{\left(\sqrt{2} + 1\right)} - \ln{\left( \sqrt{5} + 2 \right)} \right) \).

Key Concepts

Derivatives of Parametric Equations

Parametric equations are used to define a set of relationships between two or more variables using a third variable, often denoted as \( t \). In the given exercise, we have two parametric equations: \( x = \sin t - t \cos t \) and \( y = \cos t + t \sin t \).
To understand how the curve behaves, we need to find how \( x \) and \( y \) change with respect to \( t \). This involves calculating their derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).

• For \( x = \sin t - t \cos t \), differentiation gives: \( \frac{dx}{dt} = \cos t + t\sin t - \cos t = \cos t + t\sin t \).
• For \( y = \cos t + t \sin t \), differentiation yields: \( \frac{dy}{dt} = -\sin t + \sin t + t\cos t = -\sin t + t\cos t \).

This step is crucial because it prepares us for finding the length of the curve by using these derivatives within the length formula.

Integral of Square Root Function

The process of finding the length of a parametric curve involves solving an integral with a square root function. In mathematical terms, this length \( L \) from \( t = a \) to \( t = b \) is given by:
\[ L = \int_{a}^{b} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt \]
Substituting \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) from our earlier derivations gives us:
\[ L = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{ \left( \cos t + t\sin t \right)^2 + \left( -\sin t + t\cos t \right)^2 } \, dt \]
This integral is essential as it encompasses both the rate of change of \( x \) and \( y \) with respect to \( t \). Solving this integral usually requires techniques like substitution or numerical methods, as integrals involving square roots often cannot be solved by basic methods.

Simplifying Integrands

Simplifying the integrand before solving the integral can make calculations easier and more straightforward. Here, our task was to simplify:
\[ \sqrt{ \left( \cos t + t\sin t \right)^2 + \left( -\sin t + t\cos t \right)^2 } \]
When expanded, these become:

• \( \left( \cos t + t \sin t \right)^2 = \cos^2 t + 2t \sin t \cos t + t^2 \sin^2 t \)
• \( \left( -\sin t + t \cos t \right)^2 = \sin^2 t - 2t \sin t \cos t + t^2 \cos^2 t \)

Adding these together simplifies to \( \cos^2 t + \sin^2 t + t^2(\sin^2 t + \cos^2 t) \), which reduces to \( 1 + t^2 \) because \( \cos^2 t + \sin^2 t = 1 \).
This simplification allows us to rewrite the integral as \( \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{1 + t^2} \, dt \) and then proceed with evaluation. Understanding how to simplify these expressions is a key component in effectively solving integrals involving parametric curves.