Problem 35 - Essential Calculus

Question

Find the length of the curve over the given interval. $$ \begin{array}{ll} \text { Polar Equation } & \text { Interval } \\ \hline r=1+\sin \theta a \cos \theta & 0 \leq \theta \leq 2 \pi \end{array} $$

Step-by-Step Solution

Answer

The length of the curve over the given interval is \( L = \int_0^{2\pi} \sqrt{1 + \sin^2\theta} d\theta \)

Step 1: Find the derivative of r with respect to \( \theta \)

The derivative of \( r = 1 + \sin \theta \cos \theta \) with respect to \( \theta \) is obtained by applying the product rule. Thus, \( \frac{{dr}}{{d\theta}} = \sin^2\theta - \cos^2\theta \).

Step 2: Substitute r and \( \frac{{dr}}{{d\theta}} \) into the formula

Substitute \( r = 1 + \sin \theta \cos \theta \) and \( \frac{{dr}}{{d\theta}} = \sin^2\theta - \cos^2\theta \) into the formula for the length of a polar curve, which gives \( L = \int_0^{2\pi} \sqrt{(1 + 2\sin \theta \cos \theta + \sin^2 \theta \cos^2 \theta) + (\sin^2\theta - \cos^2\theta)^2} d\theta \).

Step 3: Simplify the integral

Simplify the integral to get a more manageable expression. After simplification, the integral to be evaluated becomes \( L = \int_0^{2\pi} \sqrt{1 + \sin^2\theta} d\theta \).

Step 4: Evaluate the integral

Evaluate the integral from \( 0 \) to \( 2\pi \) of \( \sqrt{1 + \sin^2\theta} \). Unfortunately, this integral cannot be expressed in terms of elementary functions, so the answer will be presented in its integral form.

Key Concepts

Polar Coordinates

Polar coordinates provide a different way of describing points in a plane compared to the Cartesian coordinate system. Instead of using a grid of x and y values, polar coordinates use a combination of distance and direction to specify a point.
The position of a point is defined by two values:

• \( r \) - the radial distance from the point to the origin, also known as the pole.
• \( \theta \) - the angle between the positive x-axis and the line from the origin to the point.

In many cases, using polar coordinates can simplify the process of solving problems involving circles and spirals, as it naturalizes the representation of these curves. When calculating the length of a curve in polar coordinates, the formula is generally \[L = \int_{a}^{b} \sqrt{\left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta.\] Understanding this formula and how to apply it is crucial when working with polar curves.

Integral Calculus

Integral calculus is a branch of calculus focused on accumulation and areas under or between curves. In this context, it is used to determine the length of curves.
The integral is a mathematical representation that summarizes infinite limits of summation. The process involves finding the antiderivative, which can sometimes lead to complicated expressions.

• Definite integrals are represented by the integral symbol with upper and lower limits, indicating the range over which the function is integrated.
• Indefinite integrals lack such boundaries, focusing instead on the general form of antiderivatives.

When calculating the length of a curve in polar coordinates, integration helps accumulate the 'length units' along the curve's path. A strong grasp of integral calculus, especially in finding and simplifying integrals, is essential for understanding curve lengths in polar coordinates.

Product Rule

The product rule is a fundamental tool in calculus used for finding the derivative of the product of two functions. It is essential in scenarios where two differentiable functions are multiplied together.
The formula is given by: \[(uv)' = u'v + uv'.\]Here, \( u \) and \( v \) are functions of the same variable, and \( u' \) and \( v' \) are their respective derivatives. The essence is that the derivative of the product is not simply the product of their derivatives but includes additional terms for thorough accuracy.

• This rule becomes incredibly useful for deriving the functions involved in polar equations, which often include trigonometric components like \( \sin \theta \) and \( \cos \theta \).
• Applying the product rule appropriately allows for accurate integration in the process of determining polar curve lengths.

Mastering the product rule is critical, as it allows students to differentiate more complex functions derived from basic trigonometric identities and apply these derivatives effectively in calculus operations.