Radius of Curvature & Osculating Circles

Chapter Summary (Express)

In the chapter entitled Curvature of Curves, we have learned how we can find out which way a curve is curved, that is, whether it curves upwards or downwards towards the right. This gave us no indication whatever as to how much the curve is curved, or, in other words, what is its curvature.

The curvature of a line refers to the degree of bending or deflection that occurs along a certain length of the line. For example, consider two circular paths of equal lengths but different radii (Fig. 21.1).

As you move from one point to another along these arcs, your direction changes. The change in direction, or the angle through which you turn, represents the curvature. The second path, which is part of a smaller circle, has a greater change in direction (greater angle), and therefore a greater curvature than the first. The larger the circle (i.e., the larger the radius), the less the curvature. In fact, the curvature of a circle is inversely proportional to its radius. This relationship is typically represented as

curvature = \frac{1}{radius} .

This means that as the radius of the circle increases, the curvature decreases proportionally, and vice versa.

By curvature of a line, we mean the amount of bending or deflection taking place along a certain length of the line, say along a portion of the line the length of which is one unit of length (the same unit which is used to measure the radius, whether it be one inch, one foot, or any other unit). For instance, consider two circular paths of centre $O$ and O^{\prime} and of equal lengths AB, A^{\prime} B^{\prime} (see Fig. 21.1). When passing from $A$ to $B$ along the arc $A B$ of the first one, one changes one’s direction from $A P$ to $B Q$ , since at $A$ one faces in the direction $A P$ and at $B$ one faces in the direction $B Q$ . In other words, in walking from $A$ to $B$ one unconsciously turns round through the angle $∠ P C Q$ , which is equal to the angle $∠ A O B$ . Similarly, in passing from A^\prime to B^{\prime}, along the arc A^\prime B^\prime, of equal length to $A B$ , on the second path, one turns round through the angle \angle P^{\prime} C^{\prime} Q^{\prime}, which is equal to the angle \angle A^{\prime} O^{\prime} B^{\prime}, obviously greater than the corresponding angle $∠ A O B$ . The second path bends therefore more than the first for an equal length.

This fact is expressed by saying that the curvature of the second path is greater than that of the first one. The larger the circle, the lesser the bending, that is the lesser the curvature. If the radius of the first circle is $2, 3, 4, \dots$ etc. times greater than the radius of the second, then the angle of bending or deflection along an arc of unit length will be 2, 3, 4 , ... etc. times less on the first circle than on the second, that is, it will be $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots$ etc. of the bending or deflection along the arc of same length on the second circle. In other words, the curvature of the first circle will be $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \dots$ etc. of that of the second circle. We see that, as the radius becomes 2, 3, 4, .. etc. times greater, the curvature becomes 2, 3, 4, ... etc. times smaller, and this is expressed by saying that the curvature of a circle is inversely proportional to the radius of the circle, or

$curvature = k \times \frac{1}{radius},$

where $k$ is a constant. It is agreed to take $k = 1$ , so that $curvature = \frac{1}{radius},$ always.

Illustration for A Little More About Curvature of Curves — Fig. 21.1

We can consider a straight line as an arc of circle of infinite radius. Therefore, the curvature of a straight line is zero (

\frac{1}{infinity} =

zero).

If the radius becomes indefinitely great, the curvature becomes $\frac{1}{infinity} =$ zero, since when the denominator of a fraction is indefinitely large, the value of the fraction is indefinitely small. For this reason mathematicians sometimes consider a straight line as an arc of circle of infinite radius, or zero curvature.

The curvature of a curve can vary between different points, and it may be significantly different for two points close to each other.

To determine the curvature of an arc on a curve, consider a very small arc such as

A B

(see Fig. 21.2) and draw a circle such that the arc of the circle closely matches the arc of the curve. This circle is called the osculating circle or the circle of curvature.

The curvature at a specific point on the curve is then the reciprocal of the radius of the osculating circle.

The curvature of the curve at a point can be calculated using the formula: curvature = 1/radius of curvature.

In the case of a circle, which is perfectly symmetrical and uniform, so that the curvature is the same at every point of its circumference, the above method of expressing the curvature is perfectly definite. In the case of any other curve, however, the curvature is not the same at different points, and it may differ considerably even for two points fairly close to one another. It would not then be accurate to take the amount of bending or deflection between two points as a measure of the curvature of the arc between these points, unless this arc is very small, in fact, unless it is indefinitely small.

If then we consider a very small arc such as $A B$ (see Fig. 21.2), and if we draw such a circle that an arc $A B$ of this circle coincides with the arc $A B$ of the curve more closely than would be the case with any other circle, then the curvature of this circle may be taken as the curvature of the arc $A B$ of the curve. The smaller the arc $A B$ , the easier it will be to find a circle an arc of which most nearly coincides with the arc $A B$ of the curve. When $A$ and $B$ are very near one another, so that $A B$ is so small so that the length $d s$ of the arc $A B$ is practically negligible, then the coincidence of the two arcs, of circle and of curve, may be considered as being practically perfect, and the curvature of the curve at the point $A$ (or $B$ ), being then the same as the curvature of the circle, will be expressed by the reciprocal of the radius of this circle, that is, by $\frac{1}{O A}$ , according to our way of measuring curvature, explained above.

Now, at first, you may think that, if $A B$ is very small, then the circle must be very small also. A little thinking will, however, cause you to perceive that it is by no means necessarily so, and that the circle may have any size, according to the amount of bending of the curve along this very small arc $A B$ . In fact, if the curve is almost flat at that point, the circle will be extremely large. This circle is called the circle of curvature, or the osculating circle at the point considered. Its radius is the radius of curvature of the curve at that particular point.

If you represent the arc

A B

d s

and the angle by

d θ

, the radius of curvature (

r

) is related to these quantities as

d s = r d θ or \frac{d θ}{d s} = \frac{1}{r} .

When the arc is very small, the line segment connecting the two points of the arc becomes a tangent to the curve at that point.

If the arc $A B$ is represented by $d s$ and the angle $∠ A O B$ by $d θ$ , then, if $r$ is the radius of curvature,

$d s = r d θ or \frac{d θ}{d s} = \frac{1}{r} .$

The secant $A B$ makes with the $x$ -axis the angle $θ$ , and it will be seen from the small triangle $△ A B C$ that $\frac{d y}{d x} = \tan θ$ . When $A B$ is indefinitely small, so that $B$ practically coincides with $A$ , the line $A B$ becomes a tangent to the curve at the point $A$ (or $B$ ).

Since

\frac{d y}{d x} = \tan θ

\frac{d (\frac{d y}{d x})}{d x} = \frac{d (\tan θ)}{d θ} \frac{d θ}{d x} = \sec^{2} θ \frac{d θ}{d x}

\frac{d θ}{d x} = \frac{1}{\sec^{2} θ} \frac{d^{2} y}{d x^{2}} .

It follows that \begin{align} \frac{1}{r}&=\frac{d\theta}{ds}=\frac{d\theta}{dx}\frac{dx}{ds}\\ &=\left(\frac{1}{\sec^2\theta}\theta \frac{d^2 y}{dx^2}\right)\frac{dx}{ds} \end{align} Since

\frac{d x}{d s} = \cos θ = \frac{1}{\sec θ}

(see the above figure)

\frac{1}{r} = \frac{1}{\sec^{3} θ} θ \frac{d^{2} y}{d x^{2}}

but

\sec θ = \pm \sqrt{1 + \tan^{2} θ}

. Hence

\frac{1}{r} = \pm \frac{\frac{d^{2} y}{d x^{2}}}{{(\sqrt{1 + \tan^{2} θ})}^{3}} = \pm \frac{\frac{d^{2} y}{d x^{2}}}{{1 + {(\frac{d y}{d x})}^{2}}^{\frac{3}{2}}}

Now, $\tan θ$ depends on the position of the point $A$ (or $B$ , which is supposed to nearly coincide with it), that is, it depends on $x$ , or, in other words, $\tan θ$ is "a function" of $x$ .

Differentiating with regard to $x$ to get the slope (see here), we get

$\frac{d (\frac{d y}{d x})}{d x} = \frac{d (\tan θ)}{d x}$

$\frac{d^{2} y}{d x^{2}} = \sec^{2} θ \frac{d θ}{d x} = \frac{1}{\cos^{2} θ} \frac{d θ}{d x}$

(see here);

hence

$\frac{d θ}{d x} = \cos^{2} θ \frac{d^{2} y}{d x^{2}} .$

But $\frac{d x}{d s} = \cos θ$ , and for $\frac{d θ}{d s}$ one may write $\frac{d θ}{d x} \times \frac{d x}{d s}$ therefore

$curvature = \frac{1}{r} = \frac{d θ}{d s} = \frac{d θ}{d x} \times \frac{d x}{d s} = \cos^{3} θ \frac{d^{2} y}{d x^{2}} = \frac{\frac{d^{2} y}{d x^{2}}}{\sec^{3} θ};$

but $\sec θ = \pm \sqrt{1 + \tan^{2} θ}$ ;¹ hence

$\frac{1}{r} = \pm \frac{\frac{d^{2} y}{d x^{2}}}{{(\sqrt{1 + \tan^{2} θ})}^{3}} = \pm \frac{\frac{d^{2} y}{d x^{2}}}{{1 + {(\frac{d y}{d x})}^{2}}^{\frac{3}{2}}}$

and finally,

\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\displaystyle r=\pm \frac{\left\{1+\left(\dfrac{d y}{d x}\right)^{2}\right\}^{\frac{3}{2}}}{\dfrac{d^{2} y}{d x^{2}}}.}

It is important to note that the radius $r$ must always be positive, as a negative radius would have no physical meaning. Therefore, when using the above formula, one must select the $+$ sign if the denominator $\frac{d^{2} y}{d x^{2}}$ is positive and the $-$ sign if the denominator is negative, as the numerator, being a square root, is always positive.²

It has been shown in the chapter on Curvature of Curves that if $\frac{d^{2} y}{d x^{2}}$ is positive, the curve is concave upward (also called convex), while if $\frac{d^{2} y}{d x^{2}}$ is negative, the curve is concave downward (also simply called concave). If $\frac{d^{2} y}{d x^{2}} = 0$ , the radius of curvature is infinitely great, that is, the corresponding portion of the curve is a bit of straight line. This necessarily happens whenever a curve gradually changes from being concave upward to concave downward or vice versa. The point, like $P$ in Fig. 21.3, where this occurs is called a point of inflection.

The center of the osculating circle, also known as the circle of curvature, is called the center of curvature.
Its coordinates, given by

(x_{1}, y_{1})

, result in the equation of the circle being

(x - x_{1})^{2} + (y - y_{1})^{2} = r^{2} .

By differentiating twice with respect to

x

, we can find \bbox[5px,border:1px solid black;background-color:#f2f2f2]{\displaystyle y_{1}=y+\frac{1+\left(\dfrac{d y}{d x}\right)^{2}}{\dfrac{d^{2} y}{d x^{2}}}.} Replacing this in the equation of the circle, we get \bbox[5px,border:1px solid black;background-color:#f2f2f2]{\displaystyle x_{1}=x-\frac{\dfrac{d y}{d x}\left\{1+\left(\dfrac{d y}{d x}\right)^{2}\right\}}{\dfrac{d^{2} y}{d x^{2}}}.}

The centre of the circle of curvature is called the centre of curvature. If its coordinates are $(x_{1}, y_{1})$ , then the equation of the circle is (see here)

${(x - x_{1})}^{2} + {(y - y_{1})}^{2} = r^{2}$

hence

$2 (x - x_{1}) d x + 2 (y - y_{1}) d y = 0$

and

x-x_1+\left(y-y_1\right) \frac{d y}{d x}=0.\tag{1}

Why did we differentiate? To get rid of the constant $r$ . This leaves but two unknown constants $x_{1}$ and $y_{1}$ ; differentiate again; you shall get rid of one of them. This last differentiation is not quite as easy as it seems; let us do it together; we have:

$\frac{d (x)}{d x} + \frac{d [(y - y_{1}) \frac{d y}{d x}]}{d x} = 0$

the numerator of the second term is a product; hence differentiating it gives

$(y - y_{1}) \frac{d (\frac{d y}{d x})}{d x} + \frac{d y}{d x} \frac{d (y - y_{1})}{d x} = (y - y_{1}) \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2},$

so that the result of differentiating (1) is

$1 + {(\frac{d y}{d x})}^{2} + (y - y_{1}) \frac{d^{2} y}{d x^{2}} = 0;$

from this we at once get

$y_{1} = y + \frac{1 + {(\frac{d y}{d x})}^{2}}{\frac{d^{2} y}{d x^{2}}} .$

Replacing in (1), we get

$(x - x_{1}) + {y - y - \frac{1 + {(\frac{d y}{d x})}^{2}}{\frac{d^{2} y}{d x^{2}}}} \frac{d y}{d x} = 0$

$x_{1} = x - \frac{\frac{d y}{d x} {1 + {(\frac{d y}{d x})}^{2}}}{\frac{d^{2} y}{d x^{2}}};$

$x_{1}$ and $y_{1}$ give the position of the centre of curvature. The use of these formulae will be best seen by carefully going through a few worked-out examples.

Example 21.1. Find the radius of curvature and the coordinates of the centre of curvature of the curve $y = 2 x^{2} - x + 3$ at the point $x = 0$ .

Solution

We have

\begin{align} & \frac{d y}{d x}=4 x-1,\qquad \frac{d^{2} y}{d x^{2}}=4 . \\ & r=\frac{\pm\left\{1+\left(\dfrac{d y}{d x}\right)^{2}\right\}^{\frac{3}{2}}}{\dfrac{d^{2} y}{d x^{2}}}=\frac{\left\{1+(4 x-1)^{2}\right\}^{\frac{3}{2}}}{4}, \end{align}

when $x = 0$ ; this becomes

$\frac{{1 + (- 1)^{2}}^{\frac{3}{2}}}{4} = \frac{\sqrt{8}}{4} = \frac{1}{\sqrt{2}} \approx 0.707 .$

If $(x_{1}, y_{1})$ are the coordinates of the centre f curvature then

\begin{align} & x_{1}=x-\frac{\dfrac{d y}{d x}\left\{1+\left(\dfrac{d y}{d x}\right)^{2}\right\}}{\dfrac{d^{2} y}{d x^{2}}}=x-\frac{(4 x-1)\left\{1+(4 x-1)^{2}\right\}}{4} \\ & =0-\frac{(-1)\left\{1+(-1)^{2}\right\}}{4}=\frac{1}{2} \end{align}

when $x = 0, y = 3$ , so that

$y_{1} = y + \frac{1 + {(\frac{d y}{d x})}^{2}}{\frac{d^{2} y}{d x^{2}}} = y + \frac{1 + (4 x - 1)^{2}}{4} = 3 + \frac{1 + (- 1)^{2}}{4} = 3 \frac{1}{2} .$

The curve and the circle are illustrated below. The values can be checked easily, as since when $x = 0, y = 3$ , here

$x_{1}^{2} + {(y_{1} - 3)}^{2} = r^{2} or {0.5}^{2} + {0.5}^{2} = 0.5 = {(\frac{1}{\sqrt{2}})}^{2} .$

Example 21.2. Find the radius of curvature and the position of the centre of curvature of the curve $y^{2} = m x$ ( $m > 0$ ) at the point for which $y = 0$ .

Solution

Here $y = m^{\frac{1}{2}} x^{\frac{1}{2}}$

\begin{align} & \frac{dy}{d x}=\frac{1}{2} m^{\frac{1}{2}} x^{-\frac{1}{2}}=\frac{m^{\frac{1}{2}}}{2 x^{\frac{1}{2}}} \text {, } \\[9pt] & \frac{d^{2} y}{d x^{2}}=-\frac{1}{2} \times \frac{m^{\frac{1}{2}}}{2} x^{-\frac{3}{2}}=-\frac{m^{\frac{1}{2}}}{4 x^{\frac{3}{2}}} ; \end{align}

hence

\begin{align} & r=\frac{\pm\left\{1+\left(\frac{d y}{d x}\right)^{2}\right\}^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}}=\frac{\pm\left\{1+\frac{m}{4 x}\right\}^{\frac{3}{2}}}{-\frac{m^{\frac{1}{2}}}{4 x^{\frac{5}{2}}}}=\frac{(4 x+m)^{\frac{3}{2}}}{2 m^{\frac{1}{2}}}, \end{align}

taking the $-$ sign at the numerator, so as to have $r$ positive.

Since, when $y = 0$ , $x = 0$ , we get

\begin{align} r=\frac{m^{\frac{3}{2}}}{2 m^{\frac{1}{2}}}=\frac{m}{2}. \end{align}

Also, if $(x_{1}, y_{1})$ are the coordinates of the centre,

\begin{align} x_{1}&=x-\frac{\frac{d y}{d x}\left\{1+\left(\frac{d y}{d x}\right)^{2}\right\}}{\frac{d^{2} y}{d^{2} x}}\\ &=x-\frac{\frac{m^{\frac{1}{2}}}{2 x^{\frac{1}{2}}}\left\{1+\frac{m}{4 x}\right\}}{-\frac{m^{\frac{1}{2}}}{4 x^{\frac{3}{3}}}} \\[6pt] & =x+\frac{4 x+m}{2}=3 x+\frac{m}{2} \text {, } \end{align}

when $x = 0$ , then $x_{1} = \frac{m}{2}$ . Also

\begin{align} y_{1}&=y+\frac{1+\left(\frac{d y}{d x}\right)^{2}}{\frac{d^{2} y}{d x^{2}}}\\ &=m^{\frac{1}{2}} x^{\frac{1}{2}}-\frac{1+\frac{m}{4 x}}{\frac{m^{\frac{1}{2}}}{4 x^{\frac{3}{2}}}}\\ &=-\frac{4 x^{\frac{3}{2}}}{m^{\frac{1}{2}}} \end{align}

when $x = 0, y_{1} = 0$ . Therefore, the centre of curvature is $(m / 2, 0)$ .

The following figure shows the curve $y^{2} = m x$ for $m = 1$ and its circle of curvature at $(0, 0)$ .

Example 21.3. Show that the circle is a curve of constant curvature.

Solution

If $x_{1}, y_{1}$ are the coordinates of the centre, and $R$ is the radius, the equation of the circle in rectangular coordinates is

${(x - x_{1})}^{2} + {(y - y_{1})}^{2} = R^{2};$

this is easily put into the form

$y = \sqrt{R^{2} - {(x - x_{1})}^{2}} + y_{1} = {R^{2} - {(x - x_{1})}^{2}}^{\frac{1}{2}} + y_{1} .$

To differentiate, let $R^{2} - {(x - x_{1})}^{2} = v$ ; then

$y = v^{\frac{1}{2}} + y_{1}, \frac{d y}{d v} = \frac{1}{2} v^{- \frac{1}{2}}, \frac{d v}{d x} = - 2 (x - x_{1}),$

\begin{align} \frac{d y}{d x}&=\frac{d y}{d v} \times \frac{d v}{d x}\\[6pt] &=-\frac{1}{2}\left\{R^{2}-\left(x-x_{1}\right)^{2}\right\}^{-\frac{1}{2}} \times 2\left(x-x_{1}\right) \\[6pt] &=\frac{-\left(x-x_{1}\right)}{\left\{R^{2}-\left(x-x_{1}\right)^{2}\right\}^{\frac{1}{2}}}. \end{align}

Differentiate again ; using the rule for differentiation of a fraction, we get

\begin{align} \dfrac{d^2 y}{dx^2}=\frac{\left\{R^{2}-\left(x-x_{1}\right)^{2}\right\}^{\frac{1}{2}} \times \frac{d\left\{-\left(x-x_{1}\right)\right\}}{d x}-\left\{-\left(x-x_{1}\right)\right\}\times \frac{d}{d x}\left\{R^{2}-\left(x-x_{1}\right)^{2}\right\}^{\frac{1}{2}}}{R^{2}-\left(x-x_{1}\right)^{2} } \end{align} (it is always a good plan to write out the whole expression in this way when dealing with a complicated expression); this simplifies to

\begin{align} \frac{d^{2} y}{d x^{2}} & =\frac{\left\{R^{2}-\left(x-x_{1}\right)^{2}\right\}^{\frac{1}{2}}(-1)-\frac{\left(x-x_{1}\right)^{2}}{\left\{\boldsymbol{R}^{2}-\left(x-x_{1}\right)^{2}\right\}^{\frac{1}{2}}}}{R^{2}-\left(x-x_{1}\right)^{2}} \\ & =\frac{R^{2}}{\left\{R^{2}-\left(x-x_{1}\right)^{2}\right\}^{\frac{3}{2}}} ; \end{align}

hence

$r = \frac{\pm {1 + {(\frac{d y}{d x})}^{2}}^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}} = \frac{{1 + \frac{{(x - x_{1})}^{2}}{R^{2} - {(x - x_{1})}^{2}}}^{\frac{3}{2}}}{{R^{2} - {(x - x_{1})}^{2}}^{\frac{3}{2}}} = \frac{{(R^{2})}^{\frac{3}{2}}}{R^{2}} = R$

the radius of curvature is constant and equal to the radius of the circle.

Example 21.4. Find the radius and the centre of curvature of the curve $y = x^{3} - 2 x^{2} + x - 1$ at points where $x = 0, x = 0.5$ and $x = 1.0$ . Find also the position of the point of inflection of the curve.

Solution

Here

\begin{align} & \quad \frac{d y}{d x}=3 x^{2}-4 x+1, \quad \frac{d^{2} y}{d x^{2}}=6 x-4 . \\[9pt] & r= \frac{\left\{1+\left(3 x^{2}-4 x+1\right)^{2}\right\}^{\frac{3}{2}}}{6 x-4}, \\ & x_{1}=x-\frac{\left(3 x^{2}-4 x+1\right)\left\{1+\left(3 x^{2}-4 x+1\right)^{2}\right\}}{6 x-4}, \\ & y_{1}=y+\frac{1+\left(3 x^{2}-4 x+1\right)^{2}}{6 x-4} . \end{align}

When $x = 0, y = - 1$ ,

\begin{align} r=\frac{\sqrt{8}}{4}&=\frac{1}{\sqrt{2}}= 0.7071.\\ \quad x_{1}=0+\frac{1}{2}=0.5,& \qquad y_{1}=-1-\frac{1}{2}=-1.5. \end{align}

Let’s choose two points on the curve on either side of the point $(0, - 1)$ , say points with $x$ -coordinates $- 0.1$ and $0.1$ . When $x = - 0.1$ , $y = - 1.121$ . When $x = 0.1$ , $y = - 0.919$ . If we consider the circle passing through these three points: $(0, - 1)$ , $(- 0.1, - 1.121)$ , and $(0.1, - 0.919)$ , we can determine the coordinates of its center to be $(0.5, - 1.515)$ , and its radius to be $0.718$ , ³ a very fair agreement with the circle of curvature. To improve the approximation, we can choose two other points with $x$ -coordinates closer to $x = 0$ than $0.1$ —for instance, $x = 0.01$ and $x = - 0.01$ —and repeat the calculations.

The curve $y = x^{3} - 2 x^{2} + x - 1$ , its circle of curvature at $x = 0$ , and the circle passing through $(0, - 1)$ , $(- 0.1, - 1.121)$ and $(0.1, - 0.919)$ are shown below.

When $x = 0.5, y = - 0.875$ , \begin{align} & r=\frac{-\left\{1+(-0.25)^{2}\right\}^{\frac{3}{2}}}{-1}= 1.095 \\ & x_{1}=0.5-\frac{-0.25 \times 1.0625}{-1}= 0.2344, \\ & y_{1}=-0.875+\frac{1.0625}{-1}= -1.938. \end{align} Considering three points $(0.4, - 0.856)$ , $(0.5, - 0.875)$ , and $(0.6, - 0.904)$ on the curve, the circle passing through them is of center $(0.2468, - 1.935)$ and of radius $1.09$ .

When $x = 1, y = - 1$ , \begin{align} & r=\frac{(1+0)^{\frac{3}{2}}}{2}=0.5, \\ & x_{1}=1-\frac{0 \times(1+0)}{2}=1, \\ & y_{1}=-1+\frac{1+0^{2}}{2}=-0.5 . \end{align} Considering three points $(0.9, - 0.919)$ , $(1, - 1)$ , and $(1.1, - 0.989)$ on the curve, the circle passing through them is of center $(0.995, - 0.495)$ and of radius $0.5051$ .

At the point of inflection $\frac{d^{2} y}{d x^{2}} = 0, 6 x - 4 = 0$ , and $x = \frac{2}{3};$ hence $y = 0.925$ (see the following figure).

Example 21.5. Find the radius and centre of curvature of the curve $y = \frac{a}{2} (e^{\frac{x}{a}} + e^{- \frac{x}{a}})$ ,⁴ at the point for which $x = 0$ . (This curve is called the catenary, as a hanging chain affects the same slope exactly.)

Solution

The equation of the curve may be written

$y = \frac{a}{2} e^{\frac{x}{a}} + \frac{a}{2} e^{- \frac{x}{a}}$ then $\frac{d y}{d x} = \frac{a}{2} \times \frac{1}{a} e^{\frac{x}{a}} - \frac{a}{2} \times \frac{1}{a} e^{- \frac{x}{a}} = \frac{1}{2} (e^{\frac{x}{a}} - e^{- \frac{x}{a}}) .$ Similarly $\frac{d^{2} y}{d x^{2}} = \frac{1}{2 a} (e^{\frac{x}{a}} + e^{- \frac{x}{a}}) = \frac{1}{2 a} \times \frac{2 y}{a} = \frac{y}{a^{2}},$

$r = \frac{{1 + \frac{1}{4} {(e^{\frac{x}{a}} - e^{- \frac{x}{a}})}^{2}}^{\frac{3}{2}}}{\frac{y}{a^{2}}} = \frac{a^{2}}{8 y} \sqrt{{(2 + e^{\frac{2 x}{a}} + e^{- \frac{2 x}{a}})}^{3},}$ since $e^{\frac{x}{a} - \frac{x}{a}} = e^{0} = 1$ , or

\begin{align} & \qquad r=\frac{a^{2}}{8 y} \sqrt{\left(2 e^{\frac{x}{a}-\frac{x}{a}}+e^{\frac{2 x}{a}}+e^{-\frac{2 x}{a}}\right)^{3}}=\frac{a^{2}}{8 y} \sqrt{\left(e^{\frac{x}{a}}+e^{-\frac{x}{a}}\right)^{6}}=\frac{y^{2}}{a}, \end{align} when $x = 0$ , $y = \frac{a}{2} (e^{0} + e^{0}) = a$ , and $\frac{d y}{d x} = \frac{1}{2} (e^{0} - e^{0}) = 0$ ; hence

$r = \frac{a^{2}}{a} = a .$

The radius of curvature at the vertex is equal to the constant $a$ .

Also

\begin{align} & x_{1}=0-\frac{0(1+0)}{\frac{1}{a}}=0, \\ & y_{1}=y+\frac{1+0}{\frac{1}{a}}=a+a=2 a . \end{align}

You are now sufficiently familiar with this type of problem to work out the following exercises by yourself. You are advised to check your answers by careful plotting of the curve and construction of the circle of curvature, as explained in Example 4 in this chapter.

Exercises

Exercise 21.1. Find the radius of curvature and the position of the centre of curvature of the curve $y = e^{x}$ at the point for which $x = 0$ .

Answer

$r = 2 \sqrt{2}$ , $x_{1} = - 2$ , $y_{1} = 3$ .

Solution

$y = e^{x} \Rightarrow \frac{d y}{d x} = e^{x} \Rightarrow \frac{d^{2} y}{d x^{2}} = e^{x}$

When $x = 0,$ $y = \frac{d y}{d x} = \frac{d^{2} y}{d x^{2}} = 1$

We just plug these numbers in the formulae for $r, x_{1}$ , and $y_{1}$

\begin{align} & r=\frac{\left(1+1^{2}\right)^{\frac{3}{2}}}{1}=2^{\frac{3}{2}}=2 \sqrt{2} \\ & x_{1}=0-\frac{1\left(1+1^{2}\right)}{1}=-2 \\ & y_{1}=1+\frac{1+1^{2}}{1}=3 \end{align}

Exercise 21.2. Find the radius and the centre of curvature of the curve $y = x (\frac{x}{2} - 1)$ at the point for which $x = 2$ .

Answer

$r = 2 \sqrt{2} \approx 2.83$ , $x_{1} = 0$ , $y_{1} = 2$ .

Solution

\begin{align} & y=x\left(\frac{x}{2}-1\right)=\frac{x^{2}}{2}-x \\ & \frac{d y}{d x}=x-1 \quad, \quad \frac{d^{2} y}{d x^{2}}=1 \end{align}

When $x = 2,$ $y = 0, \frac{d y}{d x} = 1, \frac{d^{2} y}{d x^{2}} = 1.$

We just put these numbers in the formulae for $r, x_{1}$ , and $y_{1}$

\begin{align} & r=\frac{\left(1+1^{2}\right)^{\frac{3}{2}}}{1}=2^{\frac{3}{2}}=2 \sqrt{2} \\ & x_{1}=2-\frac{1\left(1+1^{2}\right)}{1}=0 \\ & y_{1}=0+\frac{1+1^{2}}{1}=2 \end{align}

Exercise 21.3. Find the point or points of curvature unity in the curve $y = x^{2}$ .

Answer

$x \approx \pm 0.383$ , $y = 0.147$

Solution

$y = x^{2} \Rightarrow \frac{d y}{d x} = 2 x \Rightarrow \frac{d^{2} y}{d x^{2}} = 2$ Since $curvature = \frac{1}{radius} = \frac{1}{r}$ , if $curvature = 1$ , then $r = 1$ or $r = \frac{{(1 + {(\frac{d y}{d x})}^{2})}^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}} = \frac{{(1 + 4 x^{2})}^{\frac{3}{2}}}{2} = 1$

Solving the above equation for $x$ : ${(1 + 4 x^{2})}^{\frac{3}{2}} = 2$ $1 + 4 x^{2} = 2^{\frac{2}{3}}$ $4 x^{2} = 2^{\frac{2}{3}} - 1$

$x = \pm \frac{1}{2} \sqrt{2^{\frac{2}{3}} - 1} \approx \pm 0.383$

When $x \approx \pm 0.383$ , $y \approx 0.147$ .

We have solved the problem, but if we wish to draw the circle of curvature (also known as the osculating circle), we need to determine $x_{1}$ and $y_{1}$ when $x \approx \pm 0.383$ and $y \approx 0.147$ :

$x_{1} = x - \frac{2 x {[1 + (2 x)^{2}]}^{2}}{2}$ $y_{1} = x^{2} + \frac{1 + (2 x)^{2}}{2}$

When $x \approx 0.383$ and $y \approx 0.147$ : $x_{1} \approx 0.225, y_{1} \approx 0.941$

When $x \approx - 0.383$ and $y \approx 0.147$ : $x_{1} \approx - 0.225, y_{1} \approx 0.941$

The graph of $y = x^{2}$ and the two osculating circles with radii of $1$ are shown below:

Exercise 21.4. Find the radius and the centre of curvature of the curve $x y = m$ , at the point for which $x = \sqrt{m}$ .

Answer

$r = 2$ , $x_{1} = y_{1} = 2 \sqrt{m}$ .

Solution

$x y = m or y = m x^{- 1}$

$\frac{d y}{d x} = - m x^{- 2}, \frac{d^{2} y}{d x^{2}} = 2 m x^{- 3} = \frac{2 m}{x^{3}}$

When $x = \sqrt{m}$ ,

$y = \sqrt{m}, \frac{d y}{d x} = - 1, \frac{d^{2} y}{d x^{2}} = \frac{2}{\sqrt{m}}$

Hence \begin{align} & r=\frac{\left(1+(-1)^{2}\right)^{\frac{3}{2}}}{\frac{2}{\sqrt{m}}}=\frac{\sqrt{m}}{2} 2^{\frac{3}{2}}=\sqrt{2 m} \\ & x_{1}=\sqrt{m}-\frac{(-1)\left(1+(-1)^{2}\right)}{\frac{2}{\sqrt{m}}}=2 \sqrt{m} \\ & y_{1}=\sqrt{m}+\frac{1+(-1)^{2}}{\frac{2}{\sqrt{m}}}=2 \sqrt{m} \end{align}

Exercise 21.5. Find the radius and the centre of curvature of the curve $y^{2} = 4 a x$ at the point for which $x = 0$ .

Answer

$r = 2 a$ , $x_{1} = 2 a + 3 x$ , $y_{1} = - \frac{2 x^{\frac{3}{2}}}{a^{\frac{1}{2}}}$ when $x = 0$ , $x_{1} = 2 a$ , $y_{1} = 0$ .

Solution

In Example 160, we showed that if $y^{2} = m x$ , then

$r = \frac{m}{2}, x_{1} = 3 x + \frac{m}{2}, y_{1} = - \frac{4 x^{\frac{3}{2}}}{m^{\frac{1}{2}}}$

By comparison, we realize if $y^{2} = 4 a x$ , then

$r = 2 a, x_{1} = 3 x + 2 a, y_{1} = - \frac{4 x^{\frac{3}{2}}}{2 \sqrt{a}}$

When $y = 0, x = 0$ . Therefore

$r = 2 a, x_{1} = 2 a, y_{1} = 0.$

Exercise 21.6. Find the radius and the centre of curvature of the curve $y = x^{3}$ at the points for which $x = \pm 0.9$ and also $x = 0$ .

Answer

When $x = 0$ , $r = y_{1} =$ infinity, $x_{1} = 0$ .
When $x = + 0.9, r = 3 \cdot 36, x_{1} = - 2 \cdot 21, y_{1} = + 2.01$ .
When $x = - 0.9$ , $r = 3.36$ , $x_{1} = + 2.21$ , $y = - 2.01$ .

Solution

$y = x^{3} \Rightarrow \frac{d y}{d x} = 3 x^{2} \Rightarrow \frac{d^{2} y}{d x^{2}} = 6 x$

When $x = 0.9$ ,

$y = 0.729, \frac{d y}{d x} = 2.43, \frac{d^{2} y}{d x^{2}} = 5.4$

Hence

\begin{align} & r=\frac{\left(1+2.43^{2}\right)^{\frac{3}{2}}}{5.4} \approx 3.36 \\ & x_{1}=0.9-\frac{2.43\left(1+2.43^{2}\right)}{5.4} \approx-2.21 \\ & y_{1}=0.729+\frac{1+2.43^{2}}{5.4} \approx 2.01 \end{align}

When $x = - 0.9$

$y = - 0.729, \frac{d y}{d x} = 2.43, \frac{d^{2} y}{d x^{2}} = - 5.4$

Hence \begin{align} & r=-\frac{\left(1+2.43^{2}\right)^{\frac{3}{2}}}{5.4} \approx 3.36 \\ & x_{1}=-0.9-\frac{2.43\left(1+2.43^{2}\right)}{-5.4} \approx 2.21 \\ & y_{1}=-0.729+\frac{1+2.43^{2}}{-5.4} \approx-2.01 \end{align}

When $x = 0$ $y = 0, \frac{d y}{d x} = 0, \frac{d^{2} y}{d x^{2}} = 0$ Hence \begin{align} & r=\frac{\left(1+0^{2}\right)^{3 / 2}}{0}=\infty \\ & x_{1}=x-\frac{3 x^{2}\left(1+9 x^{4}\right)}{6 x}=x-\frac{1}{2} x\left(1+9 x^{4}\right) \end{align}

If we substitute $x = 0$ into the above expression for $x_{1}$ , we get $x_{1} = 0$

$y_{1} = y + \frac{1 + (3 x)^{2}}{6 x}$

If we substitute $x = 0, y = 0$ into the above equation, we get $y_{1} = 0 + \frac{1}{0} = \infty$

Exercise 21.7. Find the radius of curvature and the coordinates of the centre of curvature of the curve $y = x^{2} - x + 2$ at the two points for which $x = 0$ and $x = 1$ , respectively. Find also the maximum or minimum value of $y$ . Verify graphically all your results.

Answer

When $x = 0$ , $r = \sqrt{2} \approx 1.41$ , $x_{1} = 1, y_{1} = 3$ .
When $x = 1$ , $r = \sqrt{2} \approx 1.41$ , $x_{1} = 0, y_{1} = 3$ .
Minimum $= 1.75$ .

Solution

$y = x^{2} - x + 2$

Then

$\frac{d y}{d x} = 2 x - 1 \frac{d^{2} y}{d x^{2}} = 2$

When $x = 0$ , $y = 2, \frac{d y}{d x} = - 1, \frac{d^{2} y}{d x^{2}} = 2$

\begin{align} & r=\frac{\left(1+(-1)^{2}\right)^{\frac{3}{2}}}{2}=\sqrt{2} \approx 1.41 \\ & x_{1}=0-\frac{(-1)\left(1+(-1)^{2}\right)}{2}=1 \\ & y_{1}=2+\frac{1+(-1)^{2}}{2}=3 \end{align}

When $x = 1$ $y = 2, \frac{d y}{d x} = 1, \frac{d^{2} y}{d x^{2}} = 2$ Hence, \begin{align} & r=\frac{\left(1+1^{2}\right)^{\frac{3}{2}}}{2}=\sqrt{2} \approx 1.41 \\ & x_{1}=1-\frac{1\left(1+1^{2}\right)}{2}=0 \\ & y_{1}=2+\frac{1+1^{2}}{2}=3 \end{align}

To find the maximum or minimum value of $y$

$\frac{d y}{d x} = 2 x - 1 = 0 \Rightarrow x = \frac{1}{2} = 0.5$

Since $\frac{d^{2} y}{d x^{2}} = 2 > 0$ , the curve is concave upward and the value of $y$ at $x = 0.5$ , which is $y = {0.5}^{2} - 0.5 + 2 = 1.75$ is the minimum value of $y$ .

Exercise 21.8. Find the radius of curvature and the coordinates of the centre of curvature of the curve $y = x^{3} - x - 1$ at the points for which $x = - 2, x = 0$ , and $x = 1$ .

Answer

For $x = - 2$ , $r \approx 112.3$ , $x_{1} \approx 109.8$ , $y_{1} \approx - 17.2$ .
For $x = 0$ , $r = x_{1} = y_{1} =$ infinity.
For $x = 1$ , $r \approx 1.86$ , $x_{1} \approx - 0.67$ , $y_{1} \approx - 0.17$ .

Solution

$y = x^{3} - x - 1 \Rightarrow \frac{d y}{d x} = 3 x^{2} - 1 \Rightarrow \frac{d^{2} y}{d x^{2}} = 6 x$

When $x = - 2$ , $y = - 7, \frac{d y}{d x} = 11, \frac{d^{2} y}{d x} = - 12$

\begin{align} & r=-\frac{\left(1+11^{2}\right)^{\frac{3}{2}}}{-12} \approx 112.3 \\ & x_{1}=-2-\frac{11\left(1+11^{2}\right)}{-12} \approx 109.8 \\ & y_{1}=-7+\frac{1+11^{2}}{-12} \approx-17.2 \end{align}

When $x = 0$ , $y = - 1, \frac{d y}{d x} = - 1, \frac{d^{2} y}{d x^{2}} = 0$

Hence, \begin{align} & r=\frac{\left(1+(-1)^{2}\right)^{\frac{3}{2}}}{0}=\infty \\ & x_{1}=0-\frac{(-1)\left(1+(-1)^{2}\right)}{0}=\infty \\ & y_{1}=-1+\frac{1+(-1)^{2}}{0}=\infty \end{align}

When $x = 1$ $y = - 1, \frac{d y}{d x} = 2, \frac{d^{2} y}{d x^{2}} = 6$

\begin{align} & r=\frac{\left(1+2^{2}\right)^{\frac{3}{2}}}{6} \approx 1.86 \\ & x_{1}=1-\frac{2\left(1+2^{2}\right)}{6}=-\frac{2}{3} \approx-0.67\\ &y_{1}=-1+\frac{1+2}{6}=-\frac{1}{6} \approx-0.17 \end{align}

Exercise 21.9. Find the coordinates of the point or points of inflection of the curve $y = x^{3} + x^{2} + 1$ .

Answer

$x = - \frac{1}{3} \approx - 0.33$ , $y = \frac{29}{27} \approx + 1.08$

Solution

$y = x^{3} + x^{2} + 1$

$\frac{d y}{d x} = 3 x^{2} + 2 x, \frac{d^{2} y}{d x^{2}} = 6 x + 2$

To find the point(s) of inflection

$\begin{matrix} \frac{d^{2} y}{d x^{2}} = 6 x + 2 = 0 \\ x = - \frac{1}{3} \end{matrix}$

If $x < \frac{- 1}{3}, \frac{d^{2} y}{d x^{2}} < 0$ and if $x > - \frac{1}{3}, \frac{d^{2} y}{d x^{2}} > 0$ .

Therefore, the direction of concavity changes at $x = - \frac{1}{3}$ .

When $x = - \frac{1}{3}, y = \frac{29}{27} \approx 1.07$ .

Hence, $(- \frac{1}{3}, \frac{29}{27})$ is the point of inflection.

Exercise 21.10. Find the radius of curvature and the coordinates of the centre of curvature of the curve $y = {(4 x - x^{2} - 3)}^{\frac{1}{2}}$ at the points for which $x = 1.2$ , $x = 2$ , and $x = 2.5$ . What is this curve?

Answer

$r = 1$ , $x = 2$ , $y = 0$ for all points. A circle.

Solution

$y = {(4 x - x^{2} - 3)}^{\frac{1}{2}}$

\begin{align} \frac{d y}{d x} & =\frac{1}{2}(4-2 x)\left(4 x-x^{2}-3\right)^{-\frac{1}{2}}\\ &=(2-x)\left(4 x-x^{2}-3\right)^{-\frac{1}{2}} \end{align}

\begin{align} \frac{d^{2} y}{d x^{2}} & =-\left(4 x-x^{2}-3\right)^{-\frac{1}{2}}-(2-x)^{2}\left(4 x-x^{2}-3\right)^{\frac{-3}{2}} \\ & =\frac{-\left(4 x-x^{2}-3\right)-(2-x)^{2}}{\left(4 x-x^{2}-3\right)^{\frac{3}{2}}} \\ & =\frac{-1}{\left(4 x-x^{2}-3\right)^{\frac{3}{2}}} \end{align}

Instead of calculating $r$ , $x_{1}$ , and $y_{1}$ for the given points, we find the general formulae for them in this case:

\begin{align} & r=\frac{\left(1+\left(\frac{d y}{d x}\right)^{2}\right)^{2}}{\frac{d^{2} y}{d x^{2}}} \\ & =-\frac{\left(1+\frac{(2-x)^{2}}{4 x-x^{2}-3}\right)^{\frac{3}{2}}}{\frac{-1}{\left(4 x-x^{2}-3\right)^{\frac{3}{2}}}} \\ & =\frac{\left(\frac{4 x-x^{2}-3+\left(4-4 x-x^{2}\right)}{4 x-x^{2}-3}\right)^{\frac{3}{2}}}{\frac{1}{\left(4 x-x^{2}-3\right)^{\frac{3}{2}}}}\\ & =1 \end{align} \begin{align} x_{1}&=x-\frac{\frac{2-x}{\sqrt{4 x-x^{2}-3}}\left(1+\frac{(2-x)^{2}}{4 x-x^{2}-3}\right)}{\frac{-1}{\left(4 x-x^{2}-3\right)^{\frac{3}{2}}}} \\ &=x+(2-x)\\ &=2 \end{align} \begin{align} y_{1}&=y+\frac{1+\frac{(2-x)^{2}}{4 x-x^{2}-3}}{\frac{-1}{\left(4 x-x^{2}-3\right)^{\frac{3}{2}}}}\\ & =y+\frac{\frac{1}{4 x-x^{2}-3}}{\frac{-1}{\left(4 x-x^{2}-3\right)^{\frac{3}{2}}}} \\ & =y-\sqrt{4 x-x^{2}-3} \\ & =0 \end{align}

Therefore, for all points $r = 1, x_{1} = 2, y_{1} = 0$ . This is the equation of a circle. To see that note

$\begin{matrix} y = {(4 x - x^{2} - 3)}^{\frac{1}{2}} \\ y^{2} = 4 x - x^{2} - 3 \\ x^{2} - 4 x + y^{2} = - 3 \\ x^{2} - 4 x + 4 + y^{2} = 1 \\ (x - 2)^{2} + y^{2} = 1 \end{matrix}$

The last one is the equation of a circle of radius 1 and of center $(2, 1)$ .

Exercise 21.11. Find the radius and the centre of curvature of the curve $y = x^{3} - 3 x^{2} + 2 x + 1$ at the points for which $x = 0, x = + 1.5$ . Find also the position of the point of inflection.

Answer

When $x = 0$ , $r \approx 1.86, x_{1} \approx 1.67, y_{1} \approx 0.17$ .
When $x = 1.5$ , $r \approx 0.365$ , $x_{1} \approx 1.59$ , $y_{1} \approx 0.98$ .
$x = 1$ , $y = 1$ for zero curvature.

Solution

$y = x^{3} - 3 x^{2} + 2 x + 1$

$\frac{d y}{d x} = 3 x^{2} - 6 x + 1, \frac{d^{2} y}{d x^{2}} = 6 x - 6$

When $x = 0$ $y = 0, \frac{d y}{d x} = 2, \frac{d^{2} y}{d x^{2}} = - 6$

Hence, \begin{align} & r=-\frac{\left(1+2^{2}\right)^{\frac{3}{2}}}{-6}=\frac{5^{\frac{3}{2}}}{6}=\frac{5 \sqrt{5}}{6} \approx 1.86 \\ & x_{1}=0-\frac{2\left(1+2^{2}\right)}{-6}=\frac{10}{6}=\frac{5}{3} \approx 1.67 \\ & y_{1}=1+\frac{1+2^{2}}{-6}=\frac{1}{6} \approx 0.17 \end{align}

When $x = 1.5$ ,

$y = \frac{5}{8}, \frac{d y}{d x} = - \frac{1}{4} \frac{d^{2} y}{d x^{2}} = 3$

\begin{align} & r=\frac{\left(1+\frac{1}{16}\right)^{\frac{3}{2}}}{3}=\frac{17 \sqrt{17}}{192} \approx 0.365 \\ & x_{1}=1.5-\frac{-\frac{1}{4}\left(1+\frac{1}{16}\right)}{3}=\frac{305}{192} \approx 1.59 \\ & y_{1}=\frac{5}{8}+\frac{1+\frac{1}{16}}{3}=\frac{47}{48} \approx 0.98 \end{align}

At $x = 1$ , the second derivative $\frac{d^{2} y}{d x^{2}} = 6 x - 6$ changes from negative to positive, indicating a change in concavity from downward to upward. When $x = 1$ , the corresponding $y$ value is $y = 1$ . Therefore, the point $(1, 1)$ is a point of inflection on the curve.

Exercise 21.12. Find the radius and centre of curvature of the curve $y = \sin θ$ at the points for which $θ = \frac{π}{4}$ and $θ = \frac{π}{2}$ . Find the position of the point of inflection.

Answer

When $θ = \frac{π}{2}$ , $r = 1$ , $θ_{1} = \frac{π}{2}$ , $y_{1} = 0$ .
When $θ = \frac{π}{4}$ , $r \approx 2.598$ , $θ_{1} \approx 2.285$ , $y_{1} \approx - 1.41$

Solution

$y = \sin θ$

$\frac{d y}{d θ} = \cos θ, \frac{d^{2} y}{d θ^{2}} = - \sin θ$

Notice that in this exercise, we are not using polar coordinates. We are using the regular Cartesian coordinates, but instead of $x_{1}$ the independent variable is denoted by $θ$ . Therefore,

\begin{align} & r= \pm \frac{\left(1+\left(\frac{d y}{d \theta}\right)^{2}\right)^{\frac{3}{2}}}{\frac{d^{2} y}{d \theta^{2}}} \\ & \theta_{1}=\theta-\frac{\frac{d y}{d \theta}\left[1+\left(\frac{d y}{d \theta}\right)^{2}\right]}{\frac{d^{2} y}{d \theta^{2}}} \\ & y_{1}=y+\frac{1+\left(\frac{d y}{d \theta}\right)^{2}}{\frac{d^{2} y}{d \theta^{2}}} \end{align}

When $θ = \frac{π}{4}$

$y = \frac{1}{\sqrt{2}}, \frac{d y}{d θ} = \frac{1}{\sqrt{2}}, \frac{d^{2} y}{d θ^{2}} = - \frac{1}{\sqrt{2}}$ Hence \begin{align} & r=-\frac{\left(1+\frac{1}{2}\right)^{\frac{3}{2}}}{\frac{-1}{\sqrt{2}}}=\frac{3 \sqrt{3}}{2} \approx 2.598\\ & \theta_{1}=\frac{\pi}{4}-\frac{\frac{1}{\sqrt{2}}\left(1+\frac{1}{2}\right)}{-\frac{1}{\sqrt{2}}}=\frac{\pi+6}{4} \approx 2.285 \\ & y_{1}=\frac{1}{\sqrt{2}}+\frac{1+\frac{1}{2}}{-\frac{1}{\sqrt{2}}}=-\sqrt{2} \approx-1.414 \end{align}

When $θ = \frac{π}{2}$ , $y = 1, \frac{d y}{d θ} = 0, \frac{d^{2} y}{d θ^{2}} = - 1.$ Hence, \begin{align} & r=-\frac{(1+0)^{\frac{3}{2}}}{-1}=1 \\ & \theta_{1}=\frac{\pi}{2}-\frac{0\left(1+0^{2}\right)}{-1}=\frac{\pi}{2} \\ & y_{1}=1+\frac{1+0^{2}}{-1}=0 \end{align}

Exercise 21.13. Draw a circle of radius 3 , the centre of which has for its coordinates $x = 1$ , $y = 0$ . Deduce the equation of such a circle from first principles (see here). Find by calculation the radius of curvature and the coordinates of the centre of curvature for several suitable points, as accurately as possible, and verify that you get the known values.

Solution

The equation of a circle of radius $R$ and center $(x_{0}, y_{0})$ is

${(x - x_{0})}^{2} + {(y - y_{0})}^{2} = R^{2}$

Therefore, in this case

$(x - 1)^{2} + y^{2} = R^{2}, (R = 3)$

Differentiate with respect to $x$

$2 (x - 1) + 2 y \frac{d y}{d x} = 0$

Therefore

$\frac{d y}{d x} = - \frac{x - 1}{y} .$

Differentiating again using the quotient rule yields

\begin{align} \frac{d^{2} y}{d x^{2}} & =-\frac{y-\frac{d y}{d x}(x-1)}{y^{2}} \\ & =-\frac{y+\frac{(x-1)^{2}}{y}}{y^{2}} \\ & =-\frac{y^{2}+(x-1)^{2}}{y^{3}} \\ & =-\frac{R^{2}}{y^{3}} \end{align}

Radius of curvature:

\begin{align} r & =\frac{\left[1+\left(\dfrac{d y}{d x}\right)^{2}\right]^{\frac{3}{2}}}{\dfrac{d^{2} y}{d x^{2}}} \\ & =-\frac{\left[1+\dfrac{(x-1)^{2}}{y^{2}}\right]^{\frac{3}{2}}}{-\dfrac{R^{2}}{y^{3}}} \\ & =\frac{\left(\dfrac{R^{2}}{y^{2}}\right)^{\frac{3}{2}}}{\dfrac{R^{2}}{y^{2}}}\\ &=\frac{\dfrac{R^{3}}{y^2}}{\dfrac{R^{2}}{y^{2}}}=R \end{align}

Centre of curvature:

\begin{align} x_{1} & =x-\frac{-\frac{x-1}{y}\left(1+\frac{(x-1)^{2}}{y^{2}}\right)}{-\frac{R^{2}}{y^{3}}} \\ & =x-\frac{\frac{x-1}{y^{3}}\left((x-1)^{2}+y^{2}\right)}{\frac{R^{2}}{y^{3}}} \\ & =x-\frac{(x-1) R^{2}}{R^{2}}\\ & =x-(x-1)=1. \end{align} \begin{align} y_{1} & =y+\frac{1+\frac{(x-1)^{2}}{y^{2}}}{-\frac{R^{2}}{y^{3}}} \\ & =y-\frac{y^{2}+(x-1)^{2}}{y^{2}} \\ & =y-\frac{\dfrac{R^{2}}{y^{3}}}{\dfrac{R^{2}}{y^{2}}} \\ & =y-y=0. \end{align} Hence the center of curvature is $(1, 0)$ .

Exercise 21.14. Find the radius and centre of curvature of the curve $y = \cos θ$ at the points for which $θ = 0$ , $θ = \frac{π}{4},$ and $θ = \frac{π}{2}$ .

Answer

When $θ = 0$ , $r = 1$ , $θ_{1} = 0$ , $y_{1} = 0$ .
When $θ = \frac{π}{4}$ , $r \approx 2.598$ , $θ_{1} \approx 0.7146$ , $y_{1} \approx - 1.41$ .
When $θ = \frac{π}{2}$ , $r = θ_{1} = y_{1} =$ infinity.

Solution

$y = \cos θ$

Again, similar to exercise 12, we are using the Cartesian coordinates, but the independent variable is denoted by $θ$ , instead of $x$

$y = \cos θ \Rightarrow \frac{d y}{d θ} = - \sin θ \Rightarrow \frac{d^{2} y}{d θ^{2}} = - \cos θ$

When $θ = 0$ $y = 1, \frac{d y}{d x} = 0, \frac{d^{2} y}{d θ^{2}} = - 1.$ Hence, \begin{align} & r=-\frac{\left(1+0^{2}\right)^{\frac{3}{2}}}{-1}=1 \\ & \theta_{1}=0-\frac{0\left(1+0^{2}\right)}{-1}=0 \\ & y_{1}=1+\frac{1+0^{2}}{-1}=0 \end{align}

When $θ = \frac{π}{4}$ ,

$y = \frac{1}{\sqrt{2}}, \frac{d y}{d x} = - \frac{1}{\sqrt{2}}, \frac{d^{2} y}{d x^{2}} = - \frac{1}{\sqrt{2}} .$ Hence, \begin{align} & r=-\frac{\left(1+\frac{1}{2}\right)^{\frac{3}{2}}}{\frac{-1}{\sqrt{2}}}=\frac{3 \sqrt{3}}{2} \approx 2.598 \\ & \theta_{1}=\frac{\pi}{4}-\frac{-\frac{1}{\sqrt{2}}\left(1+\frac{1}{2}\right)}{-\frac{1}{\sqrt{2}}}=\frac{\pi}{4}-\frac{3}{2} \approx-0.715\\ & y_{1}=\frac{1}{\sqrt{2}}+\frac{1+\frac{1}{2}}{-\frac{1}{\sqrt{2}}}=-\sqrt{2} \approx-1.414 \end{align}

When $θ = \frac{π}{2}$ , $y = 0, \frac{d y}{d θ} = - 1, \frac{d^{2} y}{d θ^{2}} = 0.$ Hence, \begin{align} & r=\frac{(1+1)^{\frac{3}{2}}}{0}=\infty \\ & \theta_{1}=\frac{\pi}{2}-\frac{-1(1+1)}{0}=\infty\\ & y_{1}=0+\frac{1+1}{0}=\infty \end{align}

Exercise 21.15. Find the radius of curvature and the centre of curvature of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ at the points for which $x = 0$ and at the points for which $y = 0$ .

Answer

$r^{\cdot} = \frac{{(a^{4} y^{2} + b^{4} x^{2})}^{\frac{3}{2}}}{a^{4} b^{4}}$ , where $x = 0$ , $r = \frac{a^{2}}{b}$ , $x_{1} = 0$ , $y_{1} = \frac{b^{2} - a^{2}}{b}$ .

Solution

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

Multiplying both sides by $a^{2} b^{2}$ , we get $b^{2} x^{2} + a^{2} y^{2} = a^{2} b^{2}$

Differentiating both sides of $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with respect to $x$ , we obtain \begin{align} & 2 b^{2} x+2 a^{2} y \frac{d y}{d x}=0 \\ & \frac{d y}{d x}=-\frac{b^{2}}{a^{2}} \frac{x}{y} \end{align}

Differentiating the last equation with respect to $x$ using the quotient rule, we get

\begin{align} \frac{d^{2} y}{d x^{2}} & =-\frac{b^{2}}{a^{2}} \frac{y-\frac{d y}{d x} x}{y^{2}} \\ & =-\frac{b^{2}}{a^{2}} \frac{y+\frac{b^{2}}{a^{2}} \frac{x}{y}}{y^{2}} \\ & =-\frac{b^{2}}{a^{2}} \frac{a^{2} y^{2}+b^{2} x^{2}}{a^{2} y^{3}} \\ & =-\frac{b^{2}}{a^{2}} \frac{a^{2} b^{2}}{a^{2} y^{3}}=-\frac{b^{4}}{a^{2} y^{3}} \end{align}

Hence \begin{align} r & = \pm \frac{\left(1+\left(\frac{d y}{d x}\right)^{2}\right)^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}} \\ & =\frac{\left(1+\frac{b^{4}}{a^{4}} \frac{x^{2}}{y^{2}}\right)^{\frac{3}{2}}}{\frac{b^{4}}{a^{2} y^{3}}}=\frac{\left(\frac{a^{4} y^{2}+b^{4} x^{2}}{a^{4} y^{2}}\right)^{\frac{3}{2}}}{\frac{b^{4}}{a^{2} y^{2}}} \\ & =\frac{\left(a^{4} y^{2}+b^{4} x^{2}\right)^{\frac{3}{2}}}{a^{4} b^{4}} \end{align}

\begin{align} x_{1} & =x-\frac{-\frac{a^{2}}{\bar{y}}\left(1+\frac{a}{a^{4}} \frac{x}{y^{2}}\right)}{-\frac{b^{4}}{a^{2} y^{3}}} \\ & =x-\frac{\frac{b^{2}}{a^{6}} \frac{x}{y^{3}}\left(a^{4} y^{2}+b^{4} x^{2}\right)}{\frac{b^{4}}{a^{2} y^{3}}} \\ & =x-\frac{x}{a^{4} b^{2}}\left(a^{4} y^{2}+b^{4} x^{2}\right) \end{align}

\begin{align} y_{1} & =y+\frac{1+\frac{b^{4}}{a^{4}} \frac{x^{2}}{y^{2}}}{-\frac{b^{4}}{a^{2} y^{3}}} \\ & =y-\frac{y}{a^{2} b^{4}}\left(a^{4} y^{2}+b^{4} x^{2}\right) \\ & =y-\frac{\frac{1}{a^{4} y^{2}}\left(a^{4} y^{2}+b^{4} x^{2}\right)}{a^{4} y^{3}} \end{align}

When $x = 0$ , then $y = + b$ or $y = - b$

When $x = 0$ and $y = b$

$r = \frac{1}{a^{4} b^{4}} {(a^{4} b^{2})}^{\frac{3}{2}} = \frac{a^{2}}{b}$

\begin{align} & x_{1}=0-0=0 \\ & y_{1}=b-\frac{b}{a^{2} b^{4}}\left(a^{4} b^{2}\right)=b-\frac{a^{2}}{b}=\frac{b^{2}-a^{2}}{b} \end{align}

When $x = 0$ and $y = - b$

\begin{align} & y=-b \\ & r=\frac{a^{2}}{b}, \quad x_{1}=0 \quad y_{1}=-b+\frac{a^{2}}{b}=\frac{a^{2}-b^{2}}{b} \end{align}

When $y = 0$ , then $x = a$ or $x = - a$

When $x = a$ and $y = 0$

\begin{align} & r=\frac{1}{a^{4} b^{4}}\left(b^{4} a^{2}\right)^{\frac{3}{2}}=\frac{b^{6} a^{3}}{a^{4} b^{4}}=\frac{b^{2}}{a} \\ & x_{1}=a-\frac{a}{a^{4} b^{2}}\left(0+b^{4} a^{2}\right)=a-\frac{b^{2}}{a}=\frac{a^{2}-b^{2}}{a} \\ & y_{1}=0-0=0 \end{align}

When $x = - a$ and $y = 0$

$r = \frac{b^{2}}{a}, x_{1} = \frac{b^{2} - a^{2}}{a}, y_{1} = 0$

Depending on $θ$ , this can be $+ \sqrt{1 + \tan^{2} θ}$ or $- \sqrt{1 + \tan^{2} θ}$ .↩︎

Alternatively, one may write $r = \frac{{1 + {(\frac{d y}{d x})}^{2}}^{\frac{3}{2}}}{| \frac{d^{2} y}{d x^{2}} |},$ where $| \frac{d^{2} y}{d x^{2}} |$ denotes the absolute value of $\frac{d^{2} y}{d x^{2}}$ . For the definition of the absolute value, see here.↩︎

Let $C$ with coordinates $(a, b)$ be the centre of this circle whose radius is $r$ . Then the equation of this circle is

$(a - x)^{2} + (b - y)^{2} = r^{2}$

Since the point $(0, - 1)$ lies on this circle, its coordinates must satisfy the equation of this circle. Thus,

a^2+(b+1)^2=r^2 \tag{1}

Similarly, since $(- 0.1, - 1.121)$ and $(0.1, - 0.919)$ satisfy this equation too, we have

(a+0.1)^2+(b+1.121)^2=r^2 \tag{2}

(a-0.1)^2+(b+0.919)^2=r^2 \tag{3}

If we subtract Eq. (1) from (2) and (1) from (3) and use $A^{2} - B^{2} = (A - B) (A + B)$ , we get \left\{ \begin{align} &0.1(2a+0.1)+0.121 (2b+2.121)=0\\ &-0.1(2a-0.1)-0.081(2b+1.919)=0 \end{align} \right. or \left\{ \begin{align} &0.2a+0.241b=-0.266641\\ &-0.2a-0.162 b=0.145439 \end{align} \right. Solving these two equations for two unknowns $a$ and $b$ , we can find the coordinates of the centre: $a = 0.499974, b = - 1.51502$ From Eq. (1), the radius of this circle is $r = \sqrt{(0.499974 - 0)^{2} + (- 1.51502 + 1)^{2}} = 0.717788$ ↩︎

Notice that $y = \frac{a}{2} (e^{\frac{x}{a}} + e^{- \frac{x}{a}}) = a \cosh \frac{x}{a}$ . For the definitions of the hyperbolic functions see here.↩︎

Full Chapter

By curvature of a line, we mean the amount of bending or deflection taking place along a certain length of the line, say along a portion of the line the length of which is one unit of length (the same unit which is used to measure the radius, whether it be one inch, one foot, or any other unit). For instance, consider two circular paths of centre $O$ and O^{\prime} and of equal lengths AB, A^{\prime} B^{\prime} (see the following figure). When passing from $A$ to $B$ along the arc $A B$ of the first one, one changes one’s direction from $A P$ to $B Q$ , since at $A$ one faces in the direction $A P$ and at $B$ one faces in the direction $B Q$ . In other words, in walking from $A$ to $B$ one unconsciously turns round through the angle $∠ P C Q$ , which is equal to the angle $∠ A O B$ . Similarly, in passing from A^\prime to B^{\prime}, along the arc A^\prime B^\prime, of equal length to $A B$ , on the second path, one turns round through the angle \angle P^{\prime} C^{\prime} Q^{\prime}, which is equal to the angle \angle A^{\prime} O^{\prime} B^{\prime}, obviously greater than the corresponding angle $∠ A O B$ . The second path bends therefore more than the first for an equal length.

$curvature = k \times \frac{1}{radius},$

where $k$ is a constant. It is agreed to take $k = 1$ , so that $curvature = \frac{1}{radius},$ always.

If then we consider a very small arc such as $A B$ (see the next figure ), and if we draw such a circle that an arc $A B$ of this circle coincides with the arc $A B$ of the curve more closely than would be the case with any other circle, then the curvature of this circle may be taken as the curvature of the arc $A B$ of the curve. The smaller the arc $A B$ , the easier it will be to find a circle an arc of which most nearly coincides with the arc $A B$ of the curve. When $A$ and $B$ are very near one another, so that $A B$ is so small so that the length $d s$ of the arc $A B$ is practically negligible, then the coincidence of the two arcs, of circle and of curve, may be considered as being practically perfect, and the curvature of the curve at the point $A$ (or $B$ ), being then the same as the curvature of the circle, will be expressed by the reciprocal of the radius of this circle, that is, by $\frac{1}{O A}$ , according to our way of measuring curvature, explained above.

If the arc $A B$ is represented by $d s$ and the angle $∠ A O B$ by $d θ$ , then, if $r$ is the radius of curvature,

$d s = r d θ or \frac{d θ}{d s} = \frac{1}{r} .$

Now, $\tan θ$ depends on the position of the point $A$ (or $B$ , which is supposed to nearly coincide with it), that is, it depends on $x$ , or, in other words, $\tan θ$ is "a function" of $x$ .

Differentiating with regard to $x$ to get the slope (see here), we get

$\frac{d (\frac{d y}{d x})}{d x} = \frac{d (\tan θ)}{d x}$

$\frac{d^{2} y}{d x^{2}} = \sec^{2} θ \frac{d θ}{d x} = \frac{1}{\cos^{2} θ} \frac{d θ}{d x}$

(see here);

hence

$\frac{d θ}{d x} = \cos^{2} θ \frac{d^{2} y}{d x^{2}} .$

But $\frac{d x}{d s} = \cos θ$ , and for $\frac{d θ}{d s}$ one may write $\frac{d θ}{d x} \times \frac{d x}{d s}$ therefore

$curvature = \frac{1}{r} = \frac{d θ}{d s} = \frac{d θ}{d x} \times \frac{d x}{d s} = \cos^{3} θ \frac{d^{2} y}{d x^{2}} = \frac{\frac{d^{2} y}{d x^{2}}}{\sec^{3} θ};$

but $\sec θ = \pm \sqrt{1 + \tan^{2} θ}$ ;¹ hence

$\frac{1}{r} = \pm \frac{\frac{d^{2} y}{d x^{2}}}{{(\sqrt{1 + \tan^{2} θ})}^{3}} = \pm \frac{\frac{d^{2} y}{d x^{2}}}{{1 + {(\frac{d y}{d x})}^{2}}^{\frac{3}{2}}}$

and finally,

\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\displaystyle r=\pm \frac{\left\{1+\left(\dfrac{d y}{d x}\right)^{2}\right\}^{\frac{3}{2}}}{\dfrac{d^{2} y}{d x^{2}}}.}

It has been shown in the chapter on Curvature of Curves that if $\frac{d^{2} y}{d x^{2}}$ is positive, the curve is concave upward (also called convex), while if $\frac{d^{2} y}{d x^{2}}$ is negative, the curve is concave downward (also simply called concave). If $\frac{d^{2} y}{d x^{2}} = 0$ , the radius of curvature is infinitely great, that is, the corresponding portion of the curve is a bit of straight line. This necessarily happens whenever a curve gradually changes from being concave upward to concave downward or vice versa. The point, like $P$ in the following figure, where this occurs is called a point of inflection.

The centre of the circle of curvature is called the centre of curvature. If its coordinates are $(x_{1}, y_{1})$ , then the equation of the circle is (see here)

${(x - x_{1})}^{2} + {(y - y_{1})}^{2} = r^{2}$

hence

$2 (x - x_{1}) d x + 2 (y - y_{1}) d y = 0$

and

x-x_1+\left(y-y_1\right) \frac{d y}{d x}=0.\tag{1}

$\frac{d (x)}{d x} + \frac{d [(y - y_{1}) \frac{d y}{d x}]}{d x} = 0$

the numerator of the second term is a product; hence differentiating it gives

$(y - y_{1}) \frac{d (\frac{d y}{d x})}{d x} + \frac{d y}{d x} \frac{d (y - y_{1})}{d x} = (y - y_{1}) \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2},$

so that the result of differentiating (1) is

$1 + {(\frac{d y}{d x})}^{2} + (y - y_{1}) \frac{d^{2} y}{d x^{2}} = 0;$

from this we at once get

\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\displaystyle y_{1}=y+\frac{1+\left(\dfrac{d y}{d x}\right)^{2}}{\dfrac{d^{2} y}{d x^{2}}}.}

Replacing in (1), we get

$(x - x_{1}) + {y - y - \frac{1 + {(\frac{d y}{d x})}^{2}}{\frac{d^{2} y}{d x^{2}}}} \frac{d y}{d x} = 0$

\bbox[5px,border:1px solid black;background-color:#f2f2f2]{\displaystyle x_{1}=x-\frac{\dfrac{d y}{d x}\left\{1+\left(\dfrac{d y}{d x}\right)^{2}\right\}}{\dfrac{d^{2} y}{d x^{2}}};}

$x_{1}$ and $y_{1}$ give the position of the centre of curvature. The use of these formulae will be best seen by carefully going through a few worked-out examples.

Example 1.

Example 21.1. Find the radius of curvature and the coordinates of the centre of curvature of the curve $y = 2 x^{2} - x + 3$ at the point $x = 0$ .

Solution.

We have

when $x = 0$ ; this becomes

\frac{{1 + (- 1)^{2}}^{\frac{3}{2}}}{4} = \frac{\sqrt{8}}{4} = \frac{1}{\sqrt{2}} \approx 0.707 .

If $(x_{1}, y_{1})$ are the coordinates of the centre f curvature then

when $x = 0, y = 3$ , so that

y_{1} = y + \frac{1 + {(\frac{d y}{d x})}^{2}}{\frac{d^{2} y}{d x^{2}}} = y + \frac{1 + (4 x - 1)^{2}}{4} = 3 + \frac{1 + (- 1)^{2}}{4} = 3 \frac{1}{2} .

The curve and the circle are illustrated below. The values can be checked easily, as since when $x = 0, y = 3$ , here

x_{1}^{2} + {(y_{1} - 3)}^{2} = r^{2} or {0.5}^{2} + {0.5}^{2} = 0.5 = {(\frac{1}{\sqrt{2}})}^{2} .

Example 2.

Example 21.2. Find the radius of curvature and the position of the centre of curvature of the curve $y^{2} = m x$ ( $m > 0$ ) at the point for which $y = 0$ .

Solution.

Here $y = m^{\frac{1}{2}} x^{\frac{1}{2}}$

hence

taking the $-$ sign at the numerator, so as to have $r$ positive.

Since, when $y = 0$ , $x = 0$ , we get

\begin{align} r=\frac{m^{\frac{3}{2}}}{2 m^{\frac{1}{2}}}=\frac{m}{2}. \end{align}

Also, if $(x_{1}, y_{1})$ are the coordinates of the centre,

when $x = 0$ , then $x_{1} = \frac{m}{2}$ . Also

when $x = 0, y_{1} = 0$ . Therefore, the centre of curvature is $(m / 2, 0)$ .

The following figure shows the curve $y^{2} = m x$ for $m = 1$ and its circle of curvature at $(0, 0)$ .

Example 3.

Example 21.3. Show that the circle is a curve of constant curvature.

Solution.

If $x_{1}, y_{1}$ are the coordinates of the centre, and $R$ is the radius, the equation of the circle in rectangular coordinates is

{(x - x_{1})}^{2} + {(y - y_{1})}^{2} = R^{2};

this is easily put into the form

y = \sqrt{R^{2} - {(x - x_{1})}^{2}} + y_{1} = {R^{2} - {(x - x_{1})}^{2}}^{\frac{1}{2}} + y_{1} .

To differentiate, let $R^{2} - {(x - x_{1})}^{2} = v$ ; then

y = v^{\frac{1}{2}} + y_{1}, \frac{d y}{d v} = \frac{1}{2} v^{- \frac{1}{2}}, \frac{d v}{d x} = - 2 (x - x_{1}),

Differentiate again ; using the rule for differentiation of a fraction, we get

(it is always a good plan to write out the whole expression in this way when dealing with a complicated expression); this simplifies to

hence

r = \frac{\pm {1 + {(\frac{d y}{d x})}^{2}}^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}} = \frac{{1 + \frac{{(x - x_{1})}^{2}}{R^{2} - {(x - x_{1})}^{2}}}^{\frac{3}{2}}}{{R^{2} - {(x - x_{1})}^{2}}^{\frac{3}{2}}} = \frac{{(R^{2})}^{\frac{3}{2}}}{R^{2}} = R

the radius of curvature is constant and equal to the radius of the circle.

Example 4.

Solution.

Here

When $x = 0, y = - 1$ ,

\begin{align} r=\frac{\sqrt{8}}{4}&=\frac{1}{\sqrt{2}}= 0.7071.\\ \quad x_{1}=0+\frac{1}{2}=0.5,& \qquad y_{1}=-1-\frac{1}{2}=-1.5. \end{align}

The curve $y = x^{3} - 2 x^{2} + x - 1$ , its circle of curvature at $x = 0$ , and the circle passing through $(0, - 1)$ , $(- 0.1, - 1.121)$ and $(0.1, - 0.919)$ are shown below.

At the point of inflection $\frac{d^{2} y}{d x^{2}} = 0, 6 x - 4 = 0$ , and $x = \frac{2}{3};$ hence $y = 0.925$ (see Fig. 21.4).

Example 5.

Solution. The equation of the curve may be written

y = \frac{a}{2} e^{\frac{x}{a}} + \frac{a}{2} e^{- \frac{x}{a}}

then (see these examples),

\frac{d y}{d x} = \frac{a}{2} \times \frac{1}{a} e^{\frac{x}{a}} - \frac{a}{2} \times \frac{1}{a} e^{- \frac{x}{a}} = \frac{1}{2} (e^{\frac{x}{a}} - e^{- \frac{x}{a}}) .

Similarly

\frac{d^{2} y}{d x^{2}} = \frac{1}{2 a} (e^{\frac{x}{a}} + e^{- \frac{x}{a}}) = \frac{1}{2 a} \times \frac{2 y}{a} = \frac{y}{a^{2}},

r = \frac{{1 + \frac{1}{4} {(e^{\frac{x}{a}} - e^{- \frac{x}{a}})}^{2}}^{\frac{3}{2}}}{\frac{y}{a^{2}}} = \frac{a^{2}}{8 y} \sqrt{{(2 + e^{\frac{2 x}{a}} + e^{- \frac{2 x}{a}})}^{3},}

since $e^{\frac{x}{a} - \frac{x}{a}} = e^{0} = 1$ , or

when $x = 0$ , $y = \frac{a}{2} (e^{0} + e^{0}) = a$ , and $\frac{d y}{d x} = \frac{1}{2} (e^{0} - e^{0}) = 0$ ; hence

r = \frac{a^{2}}{a} = a .

The radius of curvature at the vertex is equal to the constant $a$ .

Also

\begin{align} & x_{1}=0-\frac{0(1+0)}{\frac{1}{a}}=0, \\ & y_{1}=y+\frac{1+0}{\frac{1}{a}}=a+a=2 a . \end{align}

Exercises

Exercise 1.

Exercise 21.1. Find the radius of curvature and the position of the centre of curvature of the curve $y = e^{x}$ at the point for which $x = 0$ .

Answer

$r = 2 \sqrt{2}$ , $x_{1} = - 2$ , $y_{1} = 3$ .

Solution

$y = e^{x} \Rightarrow \frac{d y}{d x} = e^{x} \Rightarrow \frac{d^{2} y}{d x^{2}} = e^{x}$

When $x = 0,$ $y = \frac{d y}{d x} = \frac{d^{2} y}{d x^{2}} = 1$

We just plug these numbers in the formulae for $r, x_{1}$ , and $y_{1}$

\begin{align} & r=\frac{\left(1+1^{2}\right)^{\frac{3}{2}}}{1}=2^{\frac{3}{2}}=2 \sqrt{2} \\ & x_{1}=0-\frac{1\left(1+1^{2}\right)}{1}=-2 \\ & y_{1}=1+\frac{1+1^{2}}{1}=3 \end{align}

Exercise 2.

Exercise 21.2. Find the radius and the centre of curvature of the curve $y = x (\frac{x}{2} - 1)$ at the point for which $x = 2$ .

Answer

$r = 2 \sqrt{2} \approx 2.83$ , $x_{1} = 0$ , $y_{1} = 2$ .

Solution

\begin{align} & y=x\left(\frac{x}{2}-1\right)=\frac{x^{2}}{2}-x \\ & \frac{d y}{d x}=x-1 \quad, \quad \frac{d^{2} y}{d x^{2}}=1 \end{align}

When $x = 2,$ $y = 0, \frac{d y}{d x} = 1, \frac{d^{2} y}{d x^{2}} = 1.$

We just put these numbers in the formulae for $r, x_{1}$ , and $y_{1}$

\begin{align} & r=\frac{\left(1+1^{2}\right)^{\frac{3}{2}}}{1}=2^{\frac{3}{2}}=2 \sqrt{2} \\ & x_{1}=2-\frac{1\left(1+1^{2}\right)}{1}=0 \\ & y_{1}=0+\frac{1+1^{2}}{1}=2 \end{align}

Exercise 3.

Exercise 21.3. Find the point or points of curvature unity in the curve $y = x^{2}$ .

Answer

$x \approx \pm 0.383$ , $y = 0.147$

Solution

Solving the above equation for $x$ : ${(1 + 4 x^{2})}^{\frac{3}{2}} = 2$ $1 + 4 x^{2} = 2^{\frac{2}{3}}$ $4 x^{2} = 2^{\frac{2}{3}} - 1$

$x = \pm \frac{1}{2} \sqrt{2^{\frac{2}{3}} - 1} \approx \pm 0.383$

When $x \approx \pm 0.383$ , $y \approx 0.147$ .

$x_{1} = x - \frac{2 x {[1 + (2 x)^{2}]}^{2}}{2}$ $y_{1} = x^{2} + \frac{1 + (2 x)^{2}}{2}$

When $x \approx 0.383$ and $y \approx 0.147$ : $x_{1} \approx 0.225, y_{1} \approx 0.941$

When $x \approx - 0.383$ and $y \approx 0.147$ : $x_{1} \approx - 0.225, y_{1} \approx 0.941$

The graph of $y = x^{2}$ and the two osculating circles with radii of $1$ are shown below:

Exercise 4.

Exercise 21.4. Find the radius and the centre of curvature of the curve $x y = m$ , at the point for which $x = \sqrt{m}$ .

Answer

$r = 2$ , $x_{1} = y_{1} = 2 \sqrt{m}$ .

Solution

$x y = m or y = m x^{- 1}$

$\frac{d y}{d x} = - m x^{- 2}, \frac{d^{2} y}{d x^{2}} = 2 m x^{- 3} = \frac{2 m}{x^{3}}$

When $x = \sqrt{m}$ ,

$y = \sqrt{m}, \frac{d y}{d x} = - 1, \frac{d^{2} y}{d x^{2}} = \frac{2}{\sqrt{m}}$

Exercise 5.

Exercise 21.5. Find the radius and the centre of curvature of the curve $y^{2} = 4 a x$ at the point for which $x = 0$ .

Answer

$r = 2 a$ , $x_{1} = 2 a + 3 x$ , $y_{1} = - \frac{2 x^{\frac{3}{2}}}{a^{\frac{1}{2}}}$ when $x = 0$ , $x_{1} = 2 a$ , $y_{1} = 0$ .

Solution

In Example 160, we showed that if $y^{2} = m x$ , then

$r = \frac{m}{2}, x_{1} = 3 x + \frac{m}{2}, y_{1} = - \frac{4 x^{\frac{3}{2}}}{m^{\frac{1}{2}}}$

By comparison, we realize if $y^{2} = 4 a x$ , then

$r = 2 a, x_{1} = 3 x + 2 a, y_{1} = - \frac{4 x^{\frac{3}{2}}}{2 \sqrt{a}}$

When $y = 0, x = 0$ . Therefore

$r = 2 a, x_{1} = 2 a, y_{1} = 0.$

Exercise 6.

Exercise 21.6. Find the radius and the centre of curvature of the curve $y = x^{3}$ at the points for which $x = \pm 0.9$ and also $x = 0$ .

Answer

When $x = 0$ , $r = y_{1} =$ infinity, $x_{1} = 0$ .
When $x = + 0.9, r = 3 \cdot 36, x_{1} = - 2 \cdot 21, y_{1} = + 2.01$ .
When $x = - 0.9$ , $r = 3.36$ , $x_{1} = + 2.21$ , $y = - 2.01$ .

Solution

$y = x^{3} \Rightarrow \frac{d y}{d x} = 3 x^{2} \Rightarrow \frac{d^{2} y}{d x^{2}} = 6 x$

When $x = 0.9$ ,

$y = 0.729, \frac{d y}{d x} = 2.43, \frac{d^{2} y}{d x^{2}} = 5.4$

Hence

When $x = - 0.9$

$y = - 0.729, \frac{d y}{d x} = 2.43, \frac{d^{2} y}{d x^{2}} = - 5.4$

If we substitute $x = 0$ into the above expression for $x_{1}$ , we get $x_{1} = 0$

$y_{1} = y + \frac{1 + (3 x)^{2}}{6 x}$

If we substitute $x = 0, y = 0$ into the above equation, we get $y_{1} = 0 + \frac{1}{0} = \infty$

Exercise 7.

Answer

When $x = 0$ , $r = \sqrt{2} \approx 1.41$ , $x_{1} = 1, y_{1} = 3$ .
When $x = 1$ , $r = \sqrt{2} \approx 1.41$ , $x_{1} = 0, y_{1} = 3$ .
Minimum $= 1.75$ .

Solution

$y = x^{2} - x + 2$

Then

$\frac{d y}{d x} = 2 x - 1 \frac{d^{2} y}{d x^{2}} = 2$

When $x = 0$ , $y = 2, \frac{d y}{d x} = - 1, \frac{d^{2} y}{d x^{2}} = 2$

\begin{align} & r=\frac{\left(1+(-1)^{2}\right)^{\frac{3}{2}}}{2}=\sqrt{2} \approx 1.41 \\ & x_{1}=0-\frac{(-1)\left(1+(-1)^{2}\right)}{2}=1 \\ & y_{1}=2+\frac{1+(-1)^{2}}{2}=3 \end{align}

To find the maximum or minimum value of $y$

$\frac{d y}{d x} = 2 x - 1 = 0 \Rightarrow x = \frac{1}{2} = 0.5$

Since $\frac{d^{2} y}{d x^{2}} = 2 > 0$ , the curve is concave upward and the value of $y$ at $x = 0.5$ , which is $y = {0.5}^{2} - 0.5 + 2 = 1.75$ is the minimum value of $y$ .

Exercise 8.

Exercise 21.8. Find the radius of curvature and the coordinates of the centre of curvature of the curve $y = x^{3} - x - 1$ at the points for which $x = - 2, x = 0$ , and $x = 1$ .

Answer

Solution

$y = x^{3} - x - 1 \Rightarrow \frac{d y}{d x} = 3 x^{2} - 1 \Rightarrow \frac{d^{2} y}{d x^{2}} = 6 x$

When $x = - 2$ , $y = - 7, \frac{d y}{d x} = 11, \frac{d^{2} y}{d x} = - 12$

When $x = 0$ , $y = - 1, \frac{d y}{d x} = - 1, \frac{d^{2} y}{d x^{2}} = 0$

Hence, \begin{align} & r=\frac{\left(1+(-1)^{2}\right)^{\frac{3}{2}}}{0}=\infty \\ & x_{1}=0-\frac{(-1)\left(1+(-1)^{2}\right)}{0}=\infty \\ & y_{1}=-1+\frac{1+(-1)^{2}}{0}=\infty \end{align}

When $x = 1$ $y = - 1, \frac{d y}{d x} = 2, \frac{d^{2} y}{d x^{2}} = 6$

Exercise 9.

Exercise 21.9. Find the coordinates of the point or points of inflection of the curve $y = x^{3} + x^{2} + 1$ .

Answer

$x = - \frac{1}{3} \approx - 0.33$ , $y = \frac{29}{27} \approx + 1.08$

Solution

$y = x^{3} + x^{2} + 1$

$\frac{d y}{d x} = 3 x^{2} + 2 x, \frac{d^{2} y}{d x^{2}} = 6 x + 2$

To find the point(s) of inflection

$\begin{matrix} \frac{d^{2} y}{d x^{2}} = 6 x + 2 = 0 \\ x = - \frac{1}{3} \end{matrix}$

If $x < \frac{- 1}{3}, \frac{d^{2} y}{d x^{2}} < 0$ and if $x > - \frac{1}{3}, \frac{d^{2} y}{d x^{2}} > 0$ .

Therefore, the direction of concavity changes at $x = - \frac{1}{3}$ .

When $x = - \frac{1}{3}, y = \frac{29}{27} \approx 1.07$ .

Hence, $(- \frac{1}{3}, \frac{29}{27})$ is the point of inflection.

Exercise 10.

Answer

$r = 1$ , $x = 2$ , $y = 0$ for all points. A circle.

Solution

$y = {(4 x - x^{2} - 3)}^{\frac{1}{2}}$

\begin{align} \frac{d y}{d x} & =\frac{1}{2}(4-2 x)\left(4 x-x^{2}-3\right)^{-\frac{1}{2}}\\ &=(2-x)\left(4 x-x^{2}-3\right)^{-\frac{1}{2}} \end{align}

Instead of calculating $r$ , $x_{1}$ , and $y_{1}$ for the given points, we find the general formulae for them in this case:

Therefore, for all points $r = 1, x_{1} = 2, y_{1} = 0$ . This is the equation of a circle. To see that note

$\begin{matrix} y = {(4 x - x^{2} - 3)}^{\frac{1}{2}} \\ y^{2} = 4 x - x^{2} - 3 \\ x^{2} - 4 x + y^{2} = - 3 \\ x^{2} - 4 x + 4 + y^{2} = 1 \\ (x - 2)^{2} + y^{2} = 1 \end{matrix}$

The last one is the equation of a circle of radius 1 and of center $(2, 1)$ .

Exercise 11.

Answer

When $x = 0$ , $r \approx 1.86, x_{1} \approx 1.67, y_{1} \approx 0.17$ .
When $x = 1.5$ , $r \approx 0.365$ , $x_{1} \approx 1.59$ , $y_{1} \approx 0.98$ .
$x = 1$ , $y = 1$ for zero curvature.

Solution

$y = x^{3} - 3 x^{2} + 2 x + 1$

$\frac{d y}{d x} = 3 x^{2} - 6 x + 1, \frac{d^{2} y}{d x^{2}} = 6 x - 6$

When $x = 0$ $y = 0, \frac{d y}{d x} = 2, \frac{d^{2} y}{d x^{2}} = - 6$

When $x = 1.5$ ,

$y = \frac{5}{8}, \frac{d y}{d x} = - \frac{1}{4} \frac{d^{2} y}{d x^{2}} = 3$

Exercise 12.

Answer

When $θ = \frac{π}{2}$ , $r = 1$ , $θ_{1} = \frac{π}{2}$ , $y_{1} = 0$ .
When $θ = \frac{π}{4}$ , $r \approx 2.598$ , $θ_{1} \approx 2.285$ , $y_{1} \approx - 1.41$

Solution

$y = \sin θ$

$\frac{d y}{d θ} = \cos θ, \frac{d^{2} y}{d θ^{2}} = - \sin θ$

Notice that in this exercise, we are not using polar coordinates. We are using the regular Cartesian coordinates, but instead of $x_{1}$ the independent variable is denoted by $θ$ . Therefore,

When $θ = \frac{π}{4}$

Exercise 13.

Solution

The equation of a circle of radius $R$ and center $(x_{0}, y_{0})$ is

${(x - x_{0})}^{2} + {(y - y_{0})}^{2} = R^{2}$

Therefore, in this case

$(x - 1)^{2} + y^{2} = R^{2}, (R = 3)$

Differentiate with respect to $x$

$2 (x - 1) + 2 y \frac{d y}{d x} = 0$

Therefore

$\frac{d y}{d x} = - \frac{x - 1}{y} .$

Differentiating again using the quotient rule yields

\begin{align} \frac{d^{2} y}{d x^{2}} & =-\frac{y-\frac{d y}{d x}(x-1)}{y^{2}} \\ & =-\frac{y+\frac{(x-1)^{2}}{y}}{y^{2}} \\ & =-\frac{y^{2}+(x-1)^{2}}{y^{3}} \\ & =-\frac{R^{2}}{y^{3}} \end{align}

Radius of curvature:

Centre of curvature:

Exercise 14.

Exercise 21.14. Find the radius and centre of curvature of the curve $y = \cos θ$ at the points for which $θ = 0$ , $θ = \frac{π}{4},$ and $θ = \frac{π}{2}$ .

Answer

Solution

$y = \cos θ$

Again, similar to exercise 12, we are using the Cartesian coordinates, but the independent variable is denoted by $θ$ , instead of $x$

$y = \cos θ \Rightarrow \frac{d y}{d θ} = - \sin θ \Rightarrow \frac{d^{2} y}{d θ^{2}} = - \cos θ$

When $θ = \frac{π}{4}$ ,

Exercise 15.

Answer

$r^{\cdot} = \frac{{(a^{4} y^{2} + b^{4} x^{2})}^{\frac{3}{2}}}{a^{4} b^{4}}$ , where $x = 0$ , $r = \frac{a^{2}}{b}$ , $x_{1} = 0$ , $y_{1} = \frac{b^{2} - a^{2}}{b}$ .

Solution

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

Multiplying both sides by $a^{2} b^{2}$ , we get $b^{2} x^{2} + a^{2} y^{2} = a^{2} b^{2}$

Differentiating the last equation with respect to $x$ using the quotient rule, we get

When $x = 0$ , then $y = + b$ or $y = - b$

When $x = 0$ and $y = b$

$r = \frac{1}{a^{4} b^{4}} {(a^{4} b^{2})}^{\frac{3}{2}} = \frac{a^{2}}{b}$

\begin{align} & x_{1}=0-0=0 \\ & y_{1}=b-\frac{b}{a^{2} b^{4}}\left(a^{4} b^{2}\right)=b-\frac{a^{2}}{b}=\frac{b^{2}-a^{2}}{b} \end{align}

When $x = 0$ and $y = - b$

\begin{align} & y=-b \\ & r=\frac{a^{2}}{b}, \quad x_{1}=0 \quad y_{1}=-b+\frac{a^{2}}{b}=\frac{a^{2}-b^{2}}{b} \end{align}

When $y = 0$ , then $x = a$ or $x = - a$

When $x = a$ and $y = 0$

When $x = - a$ and $y = 0$

$r = \frac{b^{2}}{a}, x_{1} = \frac{b^{2} - a^{2}}{a}, y_{1} = 0$

1. Depending on $θ$ , this can be $+ \sqrt{1 + \tan^{2} θ}$ or $- \sqrt{1 + \tan^{2} θ}$ .↩︎

2. Alternatively, one may write $r = \frac{{1 + {(\frac{d y}{d x})}^{2}}^{\frac{3}{2}}}{| \frac{d^{2} y}{d x^{2}} |},$ where $| \frac{d^{2} y}{d x^{2}} |$ denotes the absolute value of $\frac{d^{2} y}{d x^{2}}$ . For the definition of the absolute value, see page .↩︎

3. Let $C$ with coordinates $(a, b)$ be the centre of this circle whose radius is $r$ . Then the equation of this circle is

$(a - x)^{2} + (b - y)^{2} = r^{2}$

Since the point $(0, - 1)$ lies on this circle, its coordinates must satisfy the equation of this circle. Thus,

$\begin{matrix} (1) & a^{2} + (b + 1)^{2} = r^{2} \end{matrix}$

Similarly, since $(- 0.1, - 1.121)$ and $(0.1, - 0.919)$ satisfy this equation too, we have

$\begin{matrix} (2) & (a + 0.1)^{2} + (b + 1.121)^{2} = r^{2} \end{matrix}$

$\begin{matrix} (3) & (a - 0.1)^{2} + (b + 0.919)^{2} = r^{2} \end{matrix}$

If we subtract Eq. (1) from (2) and (1) from (3) and use $A^{2} - B^{2} = (A - B) (A + B)$ , we get ${\begin{aligned} 0.1 (2 a + 0.1) + 0.121 (2 b + 2.121) = 0 \\ - 0.1 (2 a - 0.1) - 0.081 (2 b + 1.919) = 0 \end{aligned}$ or ${\begin{aligned} 0.2 a + 0.241 b = - 0.266641 \\ - 0.2 a - 0.162 b = 0.145439 \end{aligned}$ Solving these two equations for two unknowns $a$ and $b$ , we can find the coordinates of the centre: $a = 0.499974, b = - 1.51502$ From Eq. (1), the radius of this circle is $r = \sqrt{(0.499974 - 0)^{2} + (- 1.51502 + 1)^{2}} = 0.717788$ ↩︎

4. Notice that $y = \frac{a}{2} (e^{\frac{x}{a}} + e^{- \frac{x}{a}}) = a \cosh \frac{x}{a}$ . For the definitions of the hyperbolic functions see here.↩︎

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