Question

True or False? Justify your answer with a proof or a counterexample. The arc length of the spiral given by $r=\frac{\theta }{2}$ for $0\le \theta \le 3\pi$ is $\frac{9}{4}{\pi }^3$.

Short answer

False; the integral value differs from $\frac{9}{4}{\pi }^3$.

Step-by-step solution

Understand the Problem

The given problem involves determining whether the statement about the arc length of a spiral is true or false.

Recall Arc Length Formula

The formula for arc length of a curve described in polar coordinates is given by: $$L={\int }_a^b\sqrt{{\left(\frac{dr}{d\theta }\right)}^2+r^2}d\theta$$ where $r=f(\theta )$.

Differentiate the Spiral Equation

Given $r=\frac{\theta }{2}$, differentiate $r$ with respect to $\theta$: $\frac{dr}{d\theta }=\frac{1}{2}$.

Substitute in the Arc Length Formula

Substitute $r=\frac{\theta }{2}$ and $\frac{dr}{d\theta }=\frac{1}{2}$ into the arc length formula: $$L={\int }_0^{3\pi }\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{\theta }{2}\right)}^2}d\theta$$

Simplify the Integrand

Simplify the integrand: $$L={\int }_0^{3\pi }\sqrt{\frac{1}{4}+\frac{{\theta }^2}{4}}d\theta ={\int }_0^{3\pi }\sqrt{\frac{1+{\theta }^2}{4}}d\theta ={\int }_0^{3\pi }\frac{\sqrt{1+{\theta }^2}}{2}d\theta$$

Compute the Integral

Calculate the integral: $$L=\frac{1}{2}{\int }_0^{3\pi }\sqrt{1+{\theta }^2}d\theta$$ This integral does not easily resolve to elementary functions, and needs numerical or approximation methods. However, compute numerically or find an approximation to check the value against $\frac{9}{4}{\pi }^3$. This value will differ from $\frac{9}{4}{\pi }^3$ based on numerical calculation.

Conclusion

The calculated value from the integral is not equal to $\frac{9}{4}{\pi }^3$, indicating the statement is false.

Key concepts

Polar Coordinates

Polar coordinates are a way to describe the position of a point in a plane using a distance and an angle. Unlike the Cartesian coordinate system, which uses x and y to define a point, polar coordinates use:

• r: The distance from a fixed point called the pole (similar to the origin in Cartesian coordinates).
• $\theta$: The angle measured from a fixed direction, usually the positive x-axis.

This system is particularly useful for curves like spirals or circles, where distances and angles are easier to work with than horizontal and vertical lines.
For the given problem, the spiral is described by the equation $r=\frac{\theta }{2}$. This indicates that as $\theta$ increases, the distance r from the pole increases linearly. Thus, the curve is a type of Archimedean spiral, which smoothly winds away from the pole.

Differentiation

Differentiation is a key concept in calculus that refers to the process of finding the derivative of a function. A derivative represents the rate at which a function changes as its input changes.
In polar coordinates, to find the arc length, differentiation is used to determine how r changes with respect to $\theta$. For the spiral $r=\frac{\theta }{2}$, the derivative $\frac{dr}{d\theta }$ gives us:

• $\frac{dr}{d\theta }=\frac{1}{2}$

This derivative indicates that r increases at a constant rate as $\theta$ increases. This value is used in the arc length formula, contributing to finding how the length of the spiral changes with $\theta$.
This differentiation step is integral to arriving at the correct arc length since it helps define the structure and direction of the curve.

Integral Calculus

Integral calculus involves the concept of computing integrals, which add up infinitely small quantities to get a total. It's used for various purposes, including finding areas under curves and, in this case, the arc length of curves in polar coordinates.
The arc length formula for polar curves is:$$L={\int }_a^b\sqrt{{\left(\frac{dr}{d\theta }\right)}^2+r^2}d\theta$$This accounts for the distance traveled along the curve when $\theta$ changes. For the spiral $r=\frac{\theta }{2}$, substituting $r$ and $\frac{dr}{d\theta }$ into this formula sets up the integral to be solved:$$L={\int }_0^{3\pi }\frac{\sqrt{1+{\theta }^2}}{2}d\theta$$Solving this integral involves numerical methods, as it does not directly resolve with basic algebra. This is common in integral calculus, where some integrals don't have simple antiderivatives.