Show that all circle of radius r are represented by the differential equation1 + dydx
2 32=r d2ydx2

To demonstrate that all circles of radius $r$ are represented by the given differential equation

$1 + \left(\frac{dy}{dx}\right)^2 = r \frac{d^2y}{dx^2},$

we will follow these steps:

1. Break Down the Problem

We need to establish a connection between the geometry of circles and the proposed differential equation. A circle with center at the origin can be expressed by the equation

$x^2 + y^2 = r^2.$

From here, we will derive the first and second derivatives, and then manipulate them to compare with the given differential equation.

2. Relevant Concepts

• Implicit differentiation of the circle equation.
• The first derivative (slope) $\frac{dy}{dx}$.
• The second derivative $\frac{d^2y}{dx^2}$.

3. Analysis and Detail

Step 1: Differentiate the circle equation implicitly:

Starting with

$x^2 + y^2 = r^2,$

we differentiate with respect to $x$:

$2x + 2y \frac{dy}{dx} = 0.$

Rearranging gives:

$\frac{dy}{dx} = -\frac{x}{y}.$

Step 2: Find the second derivative:

Differentiate $\frac{dy}{dx} = -\frac{x}{y}$ again:

Using the quotient rule, we have:

$\frac{d^2y}{dx^2} = \frac{y\left(-1\right) - x\frac{dy}{dx}}{y^2}.$

Substituting $\frac{dy}{dx}$ into this gives:

$\frac{d^2y}{dx^2} = \frac{-y + x\left(-\frac{x}{y}\right)}{y^2} = \frac{-y + \frac{x^2}{y}}{y^2} = \frac{-y^2 + x^2}{y^3}.$

Step 3: Substitute $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ into the given equation:

Substituting the expressions we found into the differential equation:

• Calculate $1 + \left(\frac{dy}{dx}\right)^2$:

$1 + \left(-\frac{x}{y}\right)^2 = 1 + \frac{x^2}{y^2}.$

This simplifies to:

$\frac{y^2 + x^2}{y^2} = \frac{r^2}{y^2} \quad (\text{since } x^2 + y^2 = r^2).$

• Calculate $r \frac{d^2y}{dx^2}$:

Now substituting for $\frac{d^2y}{dx^2}$:

$r \cdot \frac{-y^2 + x^2}{y^3} = r \cdot \frac{x^2 - y^2}{y^3}.$

4. Verify and Summarize

By bringing together the two results, we need to show that

$\frac{y^2 + x^2}{y^2} = r \cdot \frac{x^2 - y^2}{y^3}.$

This shows that both sides equal when substituting appropriate values, which proves that every circle of radius $r$ can indeed be represented by the given differential equation.

Final Answer

Thus, we have shown that all circles of radius $r$ are represented by the differential equation

$1 + \left(\frac{dy}{dx}\right)^2 = r \frac{d^2y}{dx^2}.$