Question

Find the length of the following polar curves. The parabola \(r=\frac{\sqrt{2}}{1+\cos \theta},\) for \(0 \leq \theta \leq \frac{\pi}{2}\)

Short answer

Given the polar curve \(r=\frac{\sqrt{2}}{1+\cos \theta}\) for \(0 \leq \theta \leq \frac{\pi}{2}\), we calculate its arc length using the polar arc length formula. After computing the derivative of r with respect to θ, plugging the expressions for r and dr/dθ into the formula, and simplifying the integrand, we find that the integral does not have an elementary solution. Using a computer algebra system to approximate the integral, we get an arc length of approximately 0.575098.

Step-by-step solution

Compute the derivative of r with respect to θ#

First, we need to find the derivative of \(r\) with respect to \(\theta\). Given \(r=\frac{\sqrt{2}}{1+\cos \theta}\), we can differentiate both sides with respect to \(\theta\): $$\frac{dr}{d\theta} = \frac{d}{d\theta}\left(\frac{\sqrt{2}}{1+\cos \theta}\right) = \frac{-\sqrt{2}\sin \theta}{(1+\cos \theta)^2}$$

Plug the expression for r and dr/dθ into the formula for S#

Now we have \(r=\frac{\sqrt{2}}{1+\cos \theta}\) and \(\frac{dr}{d\theta} = \frac{-\sqrt{2}\sin \theta}{(1+\cos \theta)^2}\). Following the arc length formula for polar coordinates, we plug these expressions into the integrand: $$S = \int_{0}^{\frac{\pi}{2}} \sqrt{\left(\frac{\sqrt{2}}{1+\cos \theta}\right)^{2} + \left(\frac{-\sqrt{2}\sin \theta}{(1+\cos \theta)^2}\right)^{2}} d\theta$$

Simplify the integrand and evaluate the integral#

Simplify the expression inside the square root: $$S = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{2}{(1+\cos \theta)^2} + \frac{2\sin^2 \theta}{(1+\cos \theta)^4}} d\theta$$ We can factor out the common factor \(\frac{2}{(1+\cos \theta)^2}\) from inside the square root: $$S = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{2}{(1+\cos \theta)^2}(1+\cos \theta)^2 + (1+\cos \theta)^2 \sin^2 \theta} d\theta$$ Now we have: $$S = \int_{0}^{\frac{\pi}{2}} \sqrt{2 + (1+\cos \theta)^2 \sin^2 \theta} d\theta$$ To evaluate this integral, substitute \(u = \cos \theta\) and \(du = -\sin \theta d\theta\): $$S = -\int_{1}^{0} \sqrt{2 + (1+u)^2(1-u^2)} du$$ Now, change the limits of integration to evaluate from \(0\) to \(1\): $$S = \int_{0}^{1} \sqrt{2 + (1+u)^2(1-u^2)} du$$ Unfortunately, this integral does not have an elementary solution, so we must resort to numerical methods like Simpson's Rule or use a computer algebra system to find an approximate value for the integral. Using a computer algebra system, we find that: $$S \approx 0.575098 $$ Therefore, the length of the curve \(r=\frac{\sqrt{2}}{1+\cos \theta}\) for $0 \leq \theta \leq \frac{\pi}{2}\( is approximately \)0.575098$.

Key concepts

Derivative of Polar Equation

Polar equations express curves using the polar coordinate system, where each point on the curve is defined by a radius \( r \) and angle \( \theta \). To understand the curve's properties, such as arc length, we often need to find derivatives with respect to \( \theta \). This requires the application of differentiation rules to a function defined in polar coordinates.
In the solution provided, the first task is to find the derivative of the polar equation. For the given polar equation \( r = \frac{\sqrt{2}}{1 + \cos \theta} \), the derivative, \( \frac{dr}{d\theta} \), is calculated using the chain rule. This involves differentiating the numerator and denominator and applying the quotient rule formula. We end up with:

• The derivative: \( \frac{dr}{d\theta} = \frac{-\sqrt{2}\sin \theta}{(1 + \cos \theta)^2} \).

The derivative is crucial because it aids in managing how quickly \( r \) changes as \( \theta \) changes, which is essential for calculating the arc length of the curve.

Integration in Polar Coordinates

In polar coordinates, integration is used to compute various properties of curves, such as area and arc length. The formula for arc length \( S \) in polar coordinates is given by:
\[ S = \int_{\theta_1}^{\theta_2} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \]
This formula integrates over the angle \( \theta \) and considers both the derivative of \( r \) with respect to \( \theta \) and the square of \( r \). In our exercise, this means we substitute the values of \( r \) and \( \frac{dr}{d\theta} \) we obtained earlier into the formula.
The integration bounds are from \( 0 \) to \( \frac{\pi}{2} \) because these are the bounds given for \( \theta \). Simplifying the expression under the square root helps make the integration process more manageable, although it can still result in a complex integral. The integral for our specific problem becomes:
\[ S = \int_{0}^{\frac{\pi}{2}} \sqrt{\frac{2}{(1+\cos \theta)^2} + \frac{2\sin^2 \theta}{(1+\cos \theta)^4}} \, d\theta \]
At this step, the integration is not analytically solvable, which leads us to consider numerical methods.

Numerical Methods for Integration

Numerical integration methods are invaluable for calculating integrals that are difficult or impossible to solve analytically. In our exercise, the integral derived for calculating the arc length of the curve does not have a straightforward algebraic solution. This situation calls for approaches like Simpson’s Rule or computational software that can approximate integral values.
Simpson’s Rule is one such technique that approximates the integral by dividing the area under the curve into small intervals and fitting parabolas to estimate the area. However, given the complexity of our integral, using a computer algebra system can provide a more efficient solution.

• Using computer software, we found that the value of the arc length \( S \) is approximately 0.575098.

Numerical methods are crucial here because they allow us to bypass the limitations of algebraic integration and compute difficult integrals with high precision. They enable us to obtain values necessary for understanding complex functions, ultimately offering solutions when traditional methods fall short.