Problem 68 - Calculus Single Variable

Question

Show that the curvature of the circle \(x^{2}+y^{2}=r^{2}\) is \(1 / r\) at all points \((x, y)\) with \(y \neq 0\).

Step-by-Step Solution

Answer

The curvature of the circle is \(1/r\) at all points where \(y \neq 0\).

Step 1: Understand the Curve's Representation

The equation given is \(x^2 + y^2 = r^2\), which represents a circle with a radius \(r\) centered at the origin \((0,0)\). We need to find the curvature of this circle.

Step 2: Calculate Derivatives from Parametric Form

First, parametrize the circle using \(x = r \cos(\theta)\) and \(y = r \sin(\theta)\). Compute the first derivatives: \(\frac{dx}{d\theta} = -r \sin(\theta)\) and \(\frac{dy}{d\theta} = r \cos(\theta)\).

Step 3: Second Derivatives for Curvature Formula

Compute the second derivatives needed for curvature: \(\frac{d^2x}{d\theta^2} = -r \cos(\theta)\) and \(\frac{d^2y}{d\theta^2} = -r \sin(\theta)\).

Step 4: Curvature Formula

The curvature \(\kappa\) of a parametric curve is \(\kappa = \frac{|x'y'' - y'x''|}{((x')^2 + (y')^2)^{3/2}}\). Substitute the parametric derivatives calculated earlier:

Step 5: Substitute Values into Curvature Formula

Substitute: \( x' = -r \sin(\theta), \ y' = r \cos(\theta), \ x'' = -r \cos(\theta), \ y'' = -r \sin(\theta)\). The numerator becomes: \(|(-r \sin(\theta))(-r \sin(\theta)) - (r \cos(\theta))(-r \cos(\theta))| = |r^2\cos^2(\theta) + r^2\sin^2(\theta)| = r^2|\cos^2(\theta) + \sin^2(\theta)| = r^2|1| = r^2\). The denominator simplifies as: \(((x')^2 + (y')^2)^{3/2} = (((-r \sin(\theta))^2 + (r \cos(\theta))^2))^{3/2} = (r^2)^{3/2} = r^{3}\).

Step 6: Simplify the Curvature Expression

Now, simplify the curvature expression: \(\kappa = \frac{r^2}{r^3} = \frac{1}{r}\). This shows that the curvature of the circle is \(\frac{1}{r}\) at all points \((x, y)\) where \(y eq 0\).

Key Concepts

Parametric Curves

Parametric curves are a useful way to represent curves by expressing the coordinates of the points on the curve as functions of a variable, often called the parameter. This approach allows a more flexible way to define a curve, especially when dealing with shapes that do not have simple algebraic equations.
For a circle, we often use trigonometric functions to describe its motion. The usual parametric form of the above circle is:

• For the x-coordinate: \(x = r \cos(\theta)\)
• For the y-coordinate: \(y = r \sin(\theta)\)

Here, \(r\) represents the radius, and \(\theta\) is a parameter that typically varies from 0 to \(2\pi\). This parameter elegantly circles all around the axis, making parametric representation powerful for capturing the whole circle without having to solve for \(y\).

Circle Equation

The circle equation is one of the fundamental geometric shapes represented in mathematics by the equation \( x^2 + y^2 = r^2 \). This simplifies to describe a circle centered at the origin

• with a radius \(r\).

When working with geometrical shapes and curves, understanding this circle equation is crucial.
In essence, if a point \((x, y)\) satisfies this equation, then it lies on the circle defined by that radius. The equation is derived directly from the Pythagorean theorem.
This highlights the sum of the squares of the coordinates

• equals the square of the radius.

This is vital when delving into concepts such as curvature, as it allows us to simplify and manipulate the equation efficiently.

Derivatives

Derivatives are a central concept in calculus, representing the rate of change of a function concerning a variable. When working with parametric curves like our circle, understanding derivatives becomes imperative.
Here's why derivatives are important:

• They help calculate important properties of curves like tangent lines, slopes, and curvatures.
• For a parametric curve, derivatives with respect to the parameter—here \(\theta\)—tell us how each coordinate (x and y) changes as \(\theta\) changes.

In our example, we first find the first derivatives:

• \(\frac{dx}{d\theta} = -r \sin(\theta)\)
• \(\frac{dy}{d\theta} = r \cos(\theta)\)

Then, the second derivatives are calculated to use in the curvature formula, providing deep insights into how steeply a curve is changing at a particular point.

Calculus

Calculus is an expansive field of mathematical study focused on change and motion, primarily using derivatives and integrals. In the context of finding the curvature of a circle, calculus becomes especially powerful.
When we calculate curvature, calculus allows us to combine derivatives to

• formulate the precise measure of how a curve bends at a specific point.

The formula used to determine the curvature \(\kappa\) of a parametric curve is as follows:
\[\kappa = \frac{|x' y'' - y' x''|}{((x')^2 + (y')^2)^{3/2}} \]Here, \(x'\) and \(y'\) are the first derivatives, and \(x''\) and \(y''\) are the second derivatives with respect to \(\theta\).
When these are plugged into the formula, it simplifies to \(\kappa = \frac{1}{r}\) for a circle, a pivotal result that strokes the brilliance of calculus by tying motion to geometry with

• precision.