Miscellaneous · Q15: Circle $(x-a)^2+(y-b)^2=c^2$⁠: $\frac{[1+y'^2]^{3/2}}{y''}=\mp c$

Basics you need

Solution

1. Where is $y$ a differentiable function of $x$⁠?

   Solving the circle equation for $y$ gives two functions on $a - c < x < a + c$⁠:

   $$\begin{aligned} \text{upper half: } y &= b + \sqrt{c^2 - (x-a)^2}, \\ \text{lower half: } y &= b - \sqrt{c^2 - (x-a)^2}. \end{aligned}$$

   Each is differentiable on the open interval (the quantity under the root is positive there). At the two end points $(a \pm c, b)$⁠, where $y = b$⁠, the tangent is vertical and $\frac{dy}{dx}$ does not exist. So throughout we work at points with $y \ne b$⁠.

   The expression in the question contains $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$⁠, so it only makes sense where $y$ is a twice-differentiable function of $x$⁠. A circle is not the graph of a single function; it is two semicircles glued at $(a \pm c, b)$⁠. Knowing this in advance tells us where the formulas below are valid, and it will explain the sign in Step 5.
   $\sqrt{u}$ is differentiable for $u > 0$⁠; chain rule, Theorem 4. Basics: composite functions

2. First derivative by implicit differentiation

   Differentiate $(x - a)^2 + (y - b)^2 = c^2$ with respect to $x$⁠:

   $$2(x - a) + 2(y - b)\frac{dy}{dx} = 0,$$ $$\frac{dy}{dx} = -\frac{x - a}{y - b} \qquad (y \ne b).$$
   $a$⁠, $b$⁠, $c$ are constants, so $c^2$ differentiates to 0. For $(y-b)^2$ the outer function is "square" and the inner is $y - b$⁠, whose derivative is $\frac{dy}{dx}$⁠: that is where the factor $\frac{dy}{dx}$ comes from. Dividing by $2(y - b)$ needs $y \ne b$⁠, as in Step 1.
   Implicit differentiation (Section 5.3.2) with the chain rule (Basic 1). Lesson: implicit differentiation

3. Second derivative with the quotient rule

   Differentiate $\frac{dy}{dx} = -\frac{x - a}{y - b}$ again, with $u = x - a$ ($u' = 1$⁠) and $v = y - b$ ($v' = y'$⁠):

   $$\begin{aligned} \frac{d^2y}{dx^2} &= -\frac{1\cdot(y - b) - (x - a)\,y'}{(y - b)^2}. \end{aligned}$$

   Substitute $y' = -\frac{x-a}{y-b}$⁠, so $(x - a)y' = -\frac{(x-a)^2}{y-b}$⁠:

   $$\begin{aligned} \frac{d^2y}{dx^2} &= -\frac{(y - b) + \dfrac{(x-a)^2}{y-b}}{(y-b)^2} \\ &= -\frac{(y-b)^2 + (x-a)^2}{(y-b)^3} \\ &= -\frac{c^2}{(y - b)^3}. \end{aligned}$$
   Multiplying the top and bottom by $y - b$ clears the small fraction. Then the circle equation replaces $(x-a)^2 + (y-b)^2$ by $c^2$⁠. This is how $x - a$ disappears: we use the fact that the point lies on the circle.
   Second order derivative (Section 5.7); quotient rule (rule (3), Section 5.3). Lesson: second order derivative

4. Compute $1 + \left(\frac{dy}{dx}\right)^2$

   $$\begin{aligned} 1 + \left(\frac{dy}{dx}\right)^2 &= 1 + \frac{(x-a)^2}{(y-b)^2} \\ &= \frac{(y-b)^2 + (x-a)^2}{(y-b)^2} \\ &= \frac{c^2}{(y-b)^2}. \end{aligned}$$

   Raising to the power $\frac32$ (Basic 4, with $c > 0$⁠):

   $$\left[1 + \left(\frac{dy}{dx}\right)^2\right]^{\frac32} = \frac{c^3}{|y - b|^3}.$$
   The power $\frac32$ means "square root, then cube". The square root of $(y-b)^2$ is $|y - b|$⁠, not $y - b$⁠: a square root is never negative. The left side here is positive, so the right side must be too, which $|y-b|^3$ guarantees.
   $\left(\frac{p}{q}\right)^{3/2} = \frac{p^{3/2}}{q^{3/2}}$ for $p, q > 0$⁠; $\left(c^2\right)^{3/2} = c^3$ since $c > 0$⁠.

5. Form the ratio

   $$\begin{aligned} \frac{\left[1 + (y')^2\right]^{3/2}}{y''} &= \frac{c^3}{|y-b|^3}\cdot\frac{(y-b)^3}{-c^2} \\ &= -c\,\frac{(y - b)^3}{|y - b|^3}. \end{aligned}$$

   The fraction $\frac{(y-b)^3}{|y-b|^3}$ is $+1$ when $y > b$ and $-1$ when $y < b$⁠. Therefore, calling the given expression $K$⁠,

   $$K = \begin{cases} -c, & y > b \ \text{(upper half)}, \\ +c, & y < b \ \text{(lower half)}. \end{cases}$$

   On each semicircle, that is for each of the two functions of Step 1, the expression is constant, and in both cases it involves only the radius $c$⁠, not $a$ or $b$⁠. This is what was to be proved.

   Both pieces of the answer are numbers built from $c$ alone. Changing $a$ or $b$ only slides the circle around the plane, which cannot change how sharply it bends, so the result had to be independent of $a$ and $b$⁠. The size of the ratio, $c$⁠, is what later courses call the radius of curvature (not defined in this book): a circle curves the same amount everywhere, and its radius of curvature is its own radius. The sign records which way the curve bends (the sign of $y''$⁠).
   $\frac{p^3}{|p|^3} = \pm 1$ according to the sign of $p$ (Basic 4).

6. Confirm on the upper semicircle directly

   Write $w = c^2 - (x - a)^2 > 0$⁠, so $y = b + \sqrt{w}$ and $w' = -2(x-a)$⁠. Then

   $$\begin{aligned} y' &= \frac{w'}{2\sqrt w} = -\frac{x - a}{\sqrt w}, \\ y'' &= -\frac{\sqrt w - (x-a)\cdot\frac{-(x-a)}{\sqrt w}}{w} \\ &= -\frac{w + (x-a)^2}{w^{3/2}} = -\frac{c^2}{w^{3/2}}. \end{aligned}$$

   Also $1 + (y')^2 = \frac{w + (x-a)^2}{w} = \frac{c^2}{w}$⁠, so $[1 + (y')^2]^{3/2} = \frac{c^3}{w^{3/2}}$⁠, and the ratio is $\frac{c^3}{w^{3/2}}\cdot\frac{w^{3/2}}{-c^2} = -c$⁠. ✓ The same calculation with $y = b - \sqrt w$ flips the signs of $y'$ and $y''$ and gives $+c$⁠.

   Here no absolute values are needed, because $\sqrt w$ is positive by definition; the sign comes out automatically. It agrees with Step 5, since $y - b = \sqrt w > 0$ on the upper half.
   Chain rule for $\sqrt{w}$ (Theorem 4) and the quotient rule. Lesson: chain rule

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