Question

Use the familiar formula from geometry to find the length of the curve and then confirm using the definite integral.\(r=6 \sin \theta+8 \cos \theta\) on the interval \(0 \leq \theta \leq \pi\)

Short answer

The length of the curve is \( 10\pi \).

Step-by-step solution

Identify the Polar Curve Equation

The given polar equation is \( r = 6 \sin \theta + 8 \cos \theta \). We are tasked with finding the length of this curve from \( \theta = 0 \) to \( \theta = \pi \).

Curve Length Formula in Polar Coordinates

The formula for the length \( L \) of a curve in polar coordinates is given by: \[ L = \int_{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \ d\theta \] where \( r = f(\theta) \) and the interval is \( [a, b] \).

Compute the Derivative \( \frac{dr}{d\theta} \)

First, we find the derivative of \( r \) with respect to \( \theta \): \( \frac{dr}{d\theta} = 6 \cos \theta - 8 \sin \theta \).

Substitute into Length Formula

Substitute \( r \) and \( \frac{dr}{d\theta} \) into the formula for the length of the curve: \[ L = \int_{0}^{\pi} \sqrt{ (6 \cos \theta - 8 \sin \theta)^2 + (6 \sin \theta + 8 \cos \theta)^2 } \ d\theta \]

Simplification of the Integrand

Simplify the expression inside the integral: \( (6 \cos \theta - 8 \sin \theta)^2 + (6 \sin \theta + 8 \cos \theta)^2 = 36 \cos^2 \theta + 64 \sin^2 \theta - 96 \sin \theta \cos \theta + 36 \sin^2 \theta + 64 \cos^2 \theta + 96 \sin \theta \cos \theta = 36(\cos^2 \theta + \sin^2 \theta) + 64(\cos^2 \theta + \sin^2 \theta) \)This further simplifies to: \( 100(\cos^2 \theta + \sin^2 \theta) = 100 \).

Evaluate the Integral

Since the integrand simplifies to a constant, the length \( L \) can be evaluated as: \[ L = \int_{0}^{\pi} \sqrt{100} \, d\theta = \int_{0}^{\pi} 10 \, d\theta = 10\theta \Big|_0^{\pi} \] Evaluating this from 0 to \( \pi \) gives \( L = 10(\pi) - 10(0) = 10\pi \).

Confirm the Result

The value of the curve length is confirmed to be \( 10\pi \) using both the geometry of simplified forms and by direct integration.

Key concepts

Polar Coordinates

Polar coordinates offer a unique way to represent points in a plane, which differs from the standard Cartesian coordinate system. Instead of using x and y to fix a position, polar coordinates use the distance from a reference point (usually the origin) and an angle from a reference direction (typically the positive x-axis). This system is especially helpful in dealing with curves and shapes that have a circular or rotational symmetry. For example, the point \(r, \theta\) can be visualized as a point that is \(r\) units away from the origin at an angle \(\theta\) radians from the positive x-axis. By using equations in polar coordinates like our example \(r = 6 \sin \theta + 8 \cos \theta\), we can describe a wide range of curves easily.
Understanding how to graph and interpret these equations enhances problem-solving skills in calculus, especially when integrating over curves that don't lend themselves easily to Cartesian equations.
To convert between polar and Cartesian coordinates, utilize the relations \((x = r \cos \theta, y = r \sin \theta)\) and \(r = \sqrt{x^2 + y^2}, \theta = \arctan(y/x)\). This flexibility allows one to switch as needed for various applications.

Curve Length

The length of a curve in polar coordinates extends our understanding of distance beyond simple linear measurements. The formula for calculating the length of a curve given in polar form is \[L = \int_{a}^{b} \sqrt{\left(\frac{dr}{d\theta}\right)^2 + r^2} \, d\theta \]This formula allows us to compute the length by integrating the square root of the sum of the squares of the derivative of the radial function and the radial function itself over the desired interval.
In our example, using the polar equation \(r = 6 \sin \theta + 8 \cos \theta\), we first calculate the derivative with respect to \(\theta\), \(\frac{dr}{d\theta} = 6 \cos \theta - 8 \sin \theta\).
Substituting these into our length equation shows how calculus connects rates of change (derivatives) to measurable quantities like curve length.

• This approach is highly valuable because many curves, especially those represented in polar coordinates, cannot be easily expressed or measured using basic geometry.
• Additionally, understanding how to use calculus for these computations is a critical skill for working with more complex functions and geometries.

Definite Integral

A definite integral is a fundamental concept in calculus that allows for the accumulation of quantities, such as area under a curve or total distance covered. Definite integrals are bounded by specific upper and lower limits, providing a concrete numerical value as a result. In terms of notation, \(\int_{a}^{b} f(x) \; dx\) means finding the integral of \(f(x)\) from \(x = a\) to \(x = b\).
The process involves finding the antiderivative of \(f(x)\), evaluating it at the upper limit and lower limit, and calculating the difference. For a polar curve, such as our example \[L = \int_{0}^{\pi} \sqrt{(6 \cos \theta - 8 \sin \theta)^2 + (6 \sin \theta + 8 \cos \theta)^2} \, d\theta\]
The integral process provides the exact curve length by solving the integrand, which simplifies due to familiar trigonometric identities to 100, as seen in the example.
A definite integral helps confirm that integrals don't just compute abstract values but measure meaningful properties of practical applications.