Question

Show that the curvature is greatest at the endpoints of the major axis, and is least at the endpoints of the minor axis, for the ellipse given by \(x^{2}+4 y^{2}=4\).

Short answer

The curvature is greatest at the endpoints of the major axis (\(x=2\), \(x=-2\)) and is least at the endpoints of the minor axis (\(x=0\)).

Step-by-step solution

Rewrite Ellipse Equation

Express the ellipse equation \(x^{2}+4y^{2}=4\) as \(y=\sqrt{1-\frac{x^{2}}{4}}\).

Compute First Derivative

Differentiate implicitly to find the first derivative \(y'(x) = -\frac{x}{4\sqrt{1-\frac{x^{2}}{4}}}\).

Compute Second Derivative

Differentiate \(y'(x)\) again to find the second derivative \(y''(x) = -\frac{1}{4\sqrt{1-\frac{x^{2}}{4}}}+\frac{x^{2}}{16(1-\frac{x^{2}}{4})^\frac{3}{2}}\).

Compute curvature of the ellipse at any point \((x,y)\)

The curvature is given by \(\kappa(x,y) = \frac{{|y''(x)|}}{{(1+(y'(x)^{2}))^\frac{3}{2}}}\)

Obtain Endpoint Curvatures

Substitute the endpoints of the major axis \((+2,0)\) and \((-2,0)\), and the endpoints of the minor axis \((0,+1)\) and \((0,-1)\), to calculate \(\kappa\) at these points. Substitute \(x=2\) and \(x=-2\) to get the endpoints of the major axis and \(x=0\) for the endpoints of the minor axis. The curvature \(\kappa\) at these points are \(\kappa (2,0) = \kappa (-2,0) = \frac{1}{4}\) and \(\kappa (0,+1) = \kappa (0,-1) = \frac{1}{2}\).

Conclusion

Therefore, we have now established that the greatest curvature occurs at the endpoints of the major axis, i.e., \(x=2\) or \(x=-2\), and least curvature occurs at the endpoints of the minor axis, i.e., \(x=0\).

Key concepts

Curvature Calculation

Curvature refers to the amount by which a curve deviates from being straight or flat. In the context of an ellipse, which is a closed, smooth, and symmetric curve, calculating curvature helps us understand how the shape bends at different points. The standard formula for curvature \(\kappa\) at a given point on a curve described by a function y(x) is:
\[ \kappa(x,y) = \frac{{|y''(x)|}}{{(1+(y'(x)^{2}))^{\frac{3}{2}}}} \]
This equation involves the absolute value of the second derivative of y with respect to x, denoted by \(y''(x)\) and the square of the first derivative \(y'(x)\). Essentially, the curvature quantifies how rapidly the curve changes direction at a particular point. Curvature is greater where the curve is more sharply bent. For example, in the case of our ellipse \(x^2 + 4y^2 = 4\), we use calculus techniques to calculate \(y'(x)\) and \(y''(x)\) and then apply them into the curvature formula to find the curvature at any point on the ellipse.

Major Axis and Minor Axis

An ellipse has two main axes - the longest diameter is known as the major axis and the shortest as the minor axis. These axes intersect at the center of the ellipse and are perpendicular to each other. The endpoints of the major axis are called vertices, and for the ellipse given by the equation \(x^2 + 4y^2 = 4\), the vertices are located at \(\pm2, 0\) since if we set \(y=0\) in our equation and solve for \(x\), we get \(x=\pm2\). Similarly, the endpoints of the minor axis can be found by setting \(x=0\) and solving for \(y\), resulting in \(y=\pm1\). The lengths of the major and minor axes play a crucial role in determining the overall shape and properties of the ellipse, including how its curvature varies at different points. For instance, an ellipse's curvature is known to be greatest at the endpoints of the major axis and least at the endpoints of the minor axis.

Implicit Differentiation

Implicit differentiation is a method used in calculus to find the derivative of a function that is not explicitly solved for one variable in terms of the other (that is, not expressed as \(y=f(x)\)). For an ellipse, which generally has an equation involving both \(x\) and \(y\) in a single expression, implicit differentiation is an efficient way to compute the derivatives needed for curvature.

To apply implicit differentiation to the given ellipse equation \(x^2 + 4y^2 = 4\), we differentiate both sides of the equation with respect to \(x\), treating \(y\) as a function of \(x\) (that is, \(y=y(x)\)). This process yields the first derivative \(y'(x)\) and, upon differentiating again, the second derivative \(y''(x)\). These derivatives are essential ingredients in the curvature formula. Additionally, implicit differentiation sometimes simplifies calculations because it allows us to work directly with the given equation without first needing to solve for \(y\) explicitly.

Second Derivative Test for Curvature

The second derivative test for curvature is a valuable calculus tool to determine how the curvature of a curve changes. This test involves examining the sign of the second derivative of a function at a given point. A positive second derivative indicates the curve is concave up, implying that it bends like a cup that can hold water, and a negative second derivative indicates that the curve is concave down, resembling an arch.

For the ellipse \(x^2 + 4y^2 = 4\), the second derivatives at the endpoints of the axes reveal the extremal curvatures. For the major axis endpoints \(\pm2, 0\), \(y''(x)\) is positive, suggesting that these points are where the ellipse has the greatest concavity and therefore the greatest curvature. Conversely, at the endpoints of the minor axis \(0, \pm1\), \(y''(x)\) is negative, indicating the points of least curvature. Thus, the second derivative provides direct insight into the curvature's behavior along the ellipse and is instrumental in confirming where the curvature is maximized and minimized.