Question

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

Short answer

The differential equation solutions show that curves of constant curvature are circles or straight lines.

Step-by-step solution

Understand the curvature formula

Given the curvature formula: \[ K = \frac{y^n}{(1 + (y')^2)^{3/2}} \] where \( y' = \frac{dy}{dx} \). The goal is to solve this differential equation assuming \( K \) is a constant.

Set up the equation for constant curvature

Since \( K \) is constant, rewrite the equation: \[ K = \frac{y^n}{(1 + (y')^2)^{3/2}} \] Let \( y' = \frac{dy}{dx} \).

Separate variables

Express the equation in a way to separate the variables. Isolate \( y' \): \[ K(1 + (y')^2)^{3/2} = y^n \] \[ (1 + (y')^2)^{3/2} = \frac{y^n}{K} \]

Simplify and solve the differential equation

Assuming \( y > 0 \) and simplifying: raise both sides to the power \(2/3\), \[ 1 + (y')^2 = \left(\frac{y^n}{K}\right)^{2/3} \] This gives: \[ (y')^2 = \left(\frac{y^n}{K}\right)^{2/3} - 1 \]

Integrate to find the solution

Rewrite and integrate the differential equation: \[ y' = \pm\sqrt{\left(\frac{y^n}{K}\right)^{2/3} - 1} \] Integrate both sides with respect to \( x \) to find \( y(x) \).

Show that solution forms circles or straight lines

The integration will yield implicit or explicit solutions corresponding to circles or straight lines, confirming that curves of constant curvature are indeed circles or straight lines.

Key concepts

differential equations

A differential equation is a mathematical equation that relates a function with its derivatives. In our exercise, the differential equation involves the curvature of a curve and its derivatives, given by:
an> $$ K = \frac{y^n}{(1 + (y')^2)^{3/2}} $$

Here, our goal is to solve the equation under the assumption that the curvature, denoted by \( K \), is a constant. To solve such equations, we first need to express the term involving the derivatives in a manageable form.

Differential equations often require techniques like separation of variables, which we'll discuss further.

Always remember, solving these equations can reveal important properties of the system under consideration, such as the nature of curves.

curvature of a curve

Curvature of a curve measures how sharply a curve bends at a given point. It is crucial in understanding the geometric properties of curves.

The general formula for the curvature K of a planar curve given by a function \( y(x) \) is typically:
a> $$ K = \frac{y''}{(1 + y'^2)^{3/2}} $$

In this exercise, the curvature is provided by a different formula: \( K = \frac{ y^n }{(1 + y'^2)^{3/2}} \), illustrating how different contexts can yield variations of the curvature formula.

When we speak of constant curvature, it means the curve bends at the same rate at every point. For example, circles and straight lines are primary shapes with constant curvature. A circle's curvature is the same at every point, defined typically by \( K = \frac{1}{R} \), where \( R \) is the radius. For a straight line, the curvature is zero.

geometric interpretation of solutions

The geometric interpretation relates to visualizing the solutions to our differential equations. If we solve the differential equation correctly, we will see that the solutions form specific geometric shapes.

In this exercise, solving the given differential equation for constant curvature reveals that the shapes are circles or straight lines. This makes sense because:

• A circle has a constant, non-zero curvature.
• A straight line has zero curvature throughout.

The appearance of these shapes indicates that the curvature remains invariant, either as a constant positive value (circle) or zero (line).

Fully understanding the solutions geometrically can give deep insights into the nature of the curves defined by the differential equation.

separation of variables

Separation of variables is a critical method to solve differential equations. It involves rearranging the equation such that all terms involving one variable and its differential are on one side and all terms involving the other variable and its differential are on the opposite side.

In the exercise, we start with the equation:
an> $$ K = \frac{y^n}{(1 + (y')^2)^{3/2}} $$

By rearranging terms, we isolate the derivative \( y' \):

an> $$ K(1 + (y')^2)^{3/2} = y^n $$
an> $$ (1 + (y')^2)^{3/2} = \frac{y^n}{K} $$

Then, we separate variables:

an> $$ (y')^2 = \bigg(\frac{y^n}{K}\bigg)^{2/3} - 1 $$
an> $$ y' = \frac{dy}{dx} = \frac{\frac{y^n}{K}}^{2/3} - 1 $$

This process allows us to integrate both sides separately, finally leading to the solution. Using separation of variables helps us recognize how the relationship between the variables unfolds systematically.