Problem 18 - Thomas Calculus

Question

Show that the ellipse \(x=a \cos t, y=b \sin t, a > b >0,\) has its largest curvature on its major axis and its smallest curvature on its minor axis. (As in Exercise \(17,\) the same is true for any ellipse.)

Step-by-Step Solution

Answer

The largest curvature of the ellipse is on the major axis, and the smallest is on the minor axis.

Step 1: Parametrize the Ellipse

The given parameterization of the ellipse is \(x = a \cos t\) and \(y = b \sin t\). Here, \(a\) represents the semi-major axis, and \(b\) represents the semi-minor axis. Since \(a > b > 0\), the major axis is horizontal.

Step 2: Compute First Derivatives

Find the first derivatives with respect to \(t\): \( \frac{dx}{dt} = -a \sin t \) and \( \frac{dy}{dt} = b \cos t \). These represent the tangent vector to the curve.

Step 3: Compute Second Derivatives

Find the second derivatives with respect to \(t\): \( \frac{d^2x}{dt^2} = -a \cos t \) and \( \frac{d^2y}{dt^2} = -b \sin t \). These second derivatives are needed to find the curvature.

Step 4: Use Curvature Formula

The curvature \(k\) of a parametrized curve \((x(t), y(t))\) is given by the formula \(k = \frac{|\frac{dx}{dt} \cdot \frac{d^2y}{dt^2} - \frac{dy}{dt} \cdot \frac{d^2x}{dt^2}|}{\left( \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 \right)^{3/2}}\).

Step 5: Plug in Derivatives into the Curvature Formula

Substitute the derivatives: \(k = \frac{|(-a \sin t)(-b \sin t) - (b \cos t)(-a \cos t)|}{\left( (-a \sin t)^2 + (b \cos t)^2 \right)^{3/2}} = \frac{|ab|}{\left( a^2 \sin^2 t + b^2 \cos^2 t \right)^{3/2}}\).

Step 6: Evaluate Curvature on Major and Minor Axes

On the major axis, \(t = 0\) or \(t = \pi\), thus \(k_{major} = \frac{ab}{(b^2)^{3/2}} = \frac{a}{b^2}\). On the minor axis, \(t = \frac{\pi}{2}\) or \(t = \frac{3\pi}{2}\), thus \(k_{minor} = \frac{ab}{(a^2)^{3/2}} = \frac{b}{a^2}\).

Step 7: Compare Curvature Values

Since \(a > b > 0\), it follows that \(\frac{a}{b^2} > \frac{b}{a^2}\). This proves that the curvature is greatest on the major axis and smallest on the minor axis.

Key Concepts

Parametrization of Curves

To understand the curvature of an ellipse, we first need to discuss the parametrization of curves. An ellipse can be represented in parametric form using a set of equations. For an ellipse with a semi-major axis of length \(a\) and a semi-minor axis of length \(b\), the parametrization is given by:

• \(x = a \cos t\)
• \(y = b \sin t\)

Here, \(t\) is a parameter that varies over an interval, typically between 0 and \(2\pi\), and traces the entire curve of the ellipse as \(t\) changes. This parameterization helps in simplifying complex geometric calculations on curves by transforming them into functions of a single variable \(t\). For the context of this problem, it's important to note that \(a > b > 0\), indicating that the ellipse has its major axis horizontal. Understand that this way of defining ellipses allows us to find derivatives easily, which are key in determining curvature.

First and Second Derivatives

The next step in calculating the curvature of an ellipse is determining the first and second derivatives of its parametric equations. Here's why derivatives are crucial:First derivatives, \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), capture the rate of change of the ellipse coordinates with respect to the parameter \(t\), essentially describing the tangent vector of the curve at any point. For the ellipse:

• \( \frac{dx}{dt} = -a \sin t\)
• \( \frac{dy}{dt} = b \cos t\)

Second derivatives, \( \frac{d^2x}{dt^2} \) and \( \frac{d^2y}{dt^2} \), reveal how the tangent vector itself changes, offering insight into the curve's concavity:

• \( \frac{d^2x}{dt^2} = -a \cos t\)
• \( \frac{d^2y}{dt^2} = -b \sin t\)

These derivatives are vital components when applied to the curvature formula, bridging the connection between pure geometry and calculus.

Curvature Calculation

Curvature is a measure of how sharply a curve bends at a given point. For a parametrized curve like our ellipse, the curvature \( k \) is calculated using a specific formula:\[k = \frac{| \frac{dx}{dt} \cdot \frac{d^2y}{dt^2} - \frac{dy}{dt} \cdot \frac{d^2x}{dt^2} |}{\left( \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 \right)^{3/2}}\]Now, by substituting the known derivatives:

• The numerator becomes \(|-a \sin t \cdot -b \sin t - b \cos t \cdot -a \cos t| = |ab \sin^2 t + ab \cos^2 t| = |ab|\)
• The denominator transforms into \(\left( a^2 \sin^2 t + b^2 \cos^2 t \right)^{3/2}\)

This results in the curvature function: \[k = \frac{|ab|}{\left( a^2 \sin^2 t + b^2 \cos^2 t \right)^{3/2}}\]By evaluating this function at specific values of \(t\), representing points on the major and minor axes, we observe that the curvature is largest at points \( t = 0 \) or \( t = \pi \) (major axis) resulting in \( k_{major} = \frac{a}{b^2} \), and smallest at \( t = \frac{\pi}{2} \) or \( t = \frac{3\pi}{2} \) (minor axis) yielding \( k_{minor} = \frac{b}{a^2} \). Therefore, since \( a > b > 0 \), the inequality \( \frac{a}{b^2} > \frac{b}{a^2} \) confirms the curvature is indeed greatest on the major axis.