The curvature of a curve in the \((x, y)\) plane is $$K=y^{\prime \prime}\left(1+y^{\prime 2}\right)^{-3 / 2}$$. With \(K=\) const., solve this differential equation to show that curves of constant curvature are circles (or straight lines).

Differential equations are mathematical equations involving an unknown function and its derivatives. In this exercise, we are dealing with a second-order differential equation. This means the equation involves the second derivative of a function. The given equation for curvature involves functions and their rates of change.

In general, differential equations help us describe various physical phenomena. In this case, the equation describes the curvature of a curve in the \(x, y\) plane. Solving differential equations often involves finding an unknown function that satisfies the given relationship with its derivatives.

Curvature measures how sharply a curve bends at a given point. For a curve in the \(x, y\) plane, the curvature of a curve at any point is given by the formula: \[ K = \frac{y''}{(1+y'^2)^{3/2}} \]. Here, \(y'\) represents the first derivative of \(y\) with respect to \(x\), and \(y''\) is the second derivative.

If the curvature \(K\) is constant, it means the curve bends in the same way everywhere. Common examples of such curves are circles, which have a constant non-zero curvature, and straight lines, which have zero curvature.

In this problem, we assume \(K\) is constant and derive solutions that satisfy this condition. This leads us to the conclusion that curves with constant curvature must be circles or straight lines.

Geometric interpretation helps us understand the shapes and properties of curves from a visual and geometric perspective. For curves with constant curvature, we are looking at shapes that bend uniformly.

When solving the differential equation for constant curvature, we find that the solutions are circles or straight lines. This is because:

• A circle has constant non-zero curvature, meaning every point on the circle bends equally.
• A straight line has zero curvature, meaning it doesn't bend at all.

Thus, the geometric interpretation aligns with our mathematical solutions, confirming that constant curvature curves are indeed circles or lines.

Integration techniques are essential for solving differential equations as they help find antiderivatives of functions. In this problem, we use integration to solve for the first derivative and then find the function itself.

The process involves:

• Recognizing the separation of variables to isolate the second derivative, \[ y'' = K(1 + y'^2)^{3/2} \]
• Integrating both sides with respect to \(x\) to solve for \(y'\)

By integrating the expression for \(y'\), we encounter an integration constant, \(C\), representing the family of curves that satisfy the constant curvature condition. Finally, further integration yields functions representing the general shape of the curve, seen as circles or lines based on the integration results.