Elements of the Differential and Integral Calculus/Chapter XIV

From Wikisource

< Elements of the Differential and Integral Calculus

Elements of the Differential and Integral Calculus by William Anthony Granville
Chapter XIV

[edit] CHAPTER XIV

CIRCLE OF CURVATURE. CENTER OF CURVATURE

116. Circle of curvature.^[1] Center of curvature. If a circle be drawn through three points P₀, P₁, P₂ on a plane curve, and if P₁ and P₂ be made to approach P₀ along the curve as a limiting position, then the circle will in general approach in magnitude and position a limiting circle called the circle of curvature of the curve at the point P₀. The center of this circle is called the center of curvature.

Let the equation of the curve be

(1) $y = f (x)$ ;

and let $x 0, x 1, x 2$ be the abscissas of the points $P 0, P 1, P 2$ respectively, $(α',β')$ the coördinates of the center, and $R'$ the radius of the circle passing through the three points. Then the equation of the circle is

(x - α') 2 + (y - β') 2 = R' 2

;

and since the coordinates of the points $P 0, P 1, P 2$ must satisfy this equation, we have

(2) $\begin{cases} (x_0 - \alpha')^2 + (y_0 - \beta')^2 - R'^2 = 0, \\ (x_1 - \alpha')^2 + (y_1 - \beta')^2 - R'^2 = 0, \\ (x_2 - \alpha')^2 + (y_2 - \beta')^2 - R'^2 = 0.\end{cases}$

Now consider the function of X defined by

F (x) = (x - α') 2 + (y - β') 2 = R'2

,

in which y has been replaced by $f (x)$ from (1).

Then from equations (2) we get

F (x 0) = 0, F (x 1) = 0, F (x 2) = 0.

Hence, by Rolle's Theorem (p. 164 [§105]), $F'(x)$ must vanish for at least two values of x, one lying between $x 0$ and $x 1$ , say $x'$ , and the other lying between $x 1$ and $x 2$ say $x''$ ; that is,

F'(x') = 0, F'(x'') = 0.

Again, for the same reason, $F''(x)$ must vanish for some value of $x$ between $x'$ and $x''$ , say $x 3$ ; hence

F''(x 3) = 0.

Therefore the elements $α',β', R'$ of the circle passing through the points $P 0, P 1, P 2$ must satisfy the three equations

F (x 0) = 0, F'(x') = 0, F''(x 3) = 0

.

Now let the points $P 1$ and $P 2$ approach $P 0$ as a limiting position; then $x 1, x 2, x', x'', x 3$ will all approach $x 0$ as a limit, and the elements $α,β, R$ of the osculating circle are therefore determined by the three equations

F (x 0) = 0, F'(x 0) = 0, F''(x 0) = 0;

or, dropping the subscripts, which is the same thing,

(A) $(x - α) 2 + (y - β) 2 = R 2,$

(B) $(x - \alpha) + (y - \beta) \frac{dy}{dx} = 0,$ differentiating (A).

(C) $1 + \left( \frac{dy}{dx} \right)^2 + (y - \beta) \frac{d^2 y}{dx^2} = 0$ , differentiating (B).

Solving (B) and (C) for $x - α$ and $y - β$ , we get $\left( \frac{d^2 y}{dx^2} \ne 0 \right)$ ,

(D) $\begin{cases} x - \alpha = \frac{\frac{dy}{dx} \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]}{\frac{d^2 y}{dx^2}} \\ y - \beta = -\frac{1 + \left( \frac{dy}{dx} \right)^2}{\frac{d^2 y}{dx^2}};\end{cases}$

hence the coördinates of the center of curvature are

(E) $\alpha = x - \frac{\frac{dy}{dx} \left[1 + \left( \frac{dy}{dx} \right)^2 \right]}{\frac{d^2 y}{dx^2}}; \beta = y + \frac{1 + \left( \frac{dy}{dx} \right)^2}{\frac{d^2 y}{dx^2}}.$ $\left( \frac{d^2 y}{dx^2} \ne 0 \right)$

Substituting the values of x - α and y - β from (D) in (A), and solving for R, we get

$R = \pm \frac{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{\frac{3}{2}} }{ \frac{d^2 y}{dx^2} },$

which is identical with (42), p. 159 [§103]. Hence

Theorem. The radius of the circle of curvature equals the radius of curvature.

117. Second method for finding center of curvature. Here we shall make use of the definition of circle of curvature given on p. 161 [§104]. Draw a figure showing the tangent line, circle of curvature, radius of curvature, and center of curvature (α, β) corresponding to the point $P (x, y)$ on the curve. Then

α = O A = O D - A D = O D - B P = x - B P,

β = A C = A B + B C = D P + B C = y + B C .

But $B P = R sinτ, B C = R cosτ$ . Hence

(A) $α = x - R sinτ,β = y + R cosτ.$

From (29), p. 135 [§90], and (42), p. 159 [§103],

$\sin \tau = \frac{ \frac{dy}{dx} }{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{\frac{1}{2}} }, \cos \tau = \frac{1}{\left[ 1 + \left( \frac{dy}{dx} \right)^{\frac{1}{2}} \right]^{\frac{1}{2}}}, R = \frac{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^{\frac{3}{2}} }{ \frac{d^2 y}{dx^2} }$

Substituting these back in (A), we get

(50) $\alpha = x - \frac{ \frac{dy}{dx}\left[ 1 + \left( \frac{dy}{dx} \right)^2 \right] }{\frac{d^2 y}{dx^2}}; \beta = y + \frac{1 + \left( \frac{dy}{dx} \right)^2}{\frac{d^2 y}{dx^2}}.$

From (23), p. 126 [§85], we know that at a point of inflection (as Q in the next figure)

$\frac{d^2 y}{dx^2} = 0,$

Therefore, by (40), p. 157 [§102], the curvature $K = 0$ ; and from (42), p. 159 [§103], and (50), p. 180 [§117], we see that in general α, β, R increase without limit as the second derivative approaches zero. That is, if we suppose P with its tangent to move along the curve to P', at the point of inflection Q the curvature is zero, the rotation of the tangent is momentarily arrested, and as the direction of rotation changes, the center of curvature moves out indefinitely and the radius of curvature becomes infinite.

ILLUSTRATIVE EXAMPLE 1. Find the coördinates of the center of curvature of the parabola $y 2 = 4 p x$ corresponding (a) to any point on the curve; (b) to the vertex.

Solution. $\frac{dy}{dx} = \frac{2p}{y}; \frac{d^2 y}{dx^2} = -\frac{4p^2}{y^3}.$

(a) Substituting in (E), p. 179 [§116],

$\alpha = x + \frac{y^2 + 4 p^2}{y^2} \cdot \frac{2p}{y} \cdot \frac{y^3}{4 p^2} = 3x + 2p.$

$\beta = y - \frac{y^2 + 4 p^2}{y^2} \cdot \frac{y^3}{4 p^2} = -\frac{y^3}{4 p ^2}.$

Therefore $\left( 3x + 2p, -\frac{y^3}{4 p^2} \right)$ is the center of curvature corresponding to any point on the curve.

(b) $(2 p,0)$ is the center of curvature corresponding to the vertex (0, 0).

118. Center of curvature the limiting position of the intersection of normals at neighboring points. Let the equation of a curve be

(A) $y = f (x).$

The equations of the normals to the curve at two neighboring points $P 0$ and $P 1$ are^[2]

$(x_0 - X) + (y_0 - Y) \frac{dy_0}{dx_0} = 0,$

$(x_1 - X) + (y_1 - Y) \frac{dy_1}{dx_1} = 0.$

If the normals intersect at $C'(α',β')$ , the coördinates of this point must satisfy both equations, giving

(B) $\begin{cases}(x_0 - \alpha') + (y_0 - \beta) \frac{dy_0}{dx_0} = 0, \\ *x_1 - \alpha') + (y_1 - \beta') \frac{dy_1}{dx_1} = 0.\end{cases}$

Now consider the function of x defined by

$\phi(x) = (x - \alpha') + (y - \beta') \frac{dy}{dx},$

in which y has been replaced by $f (x)$ from (A).

Then equations (B) show that

φ(x 0) = 0,φ(x 1) = 0.

But then, by Rolle's Theorem (p. 164 [§105]), $φ'(x)$ must vanish for some value of x between $x 0$ and $x 1$ say x'. Therefore $α'$ and $β'$ are determined by the two equations

φ(x 0) = 0,φ'(x') = 0.

If now $P 1$ approaches $P 0$ as a limiting position, then x' approaches $x 0$ , giving

φ(x 0) = 0,φ'(x 0) = 0;

and $C'(α',β')$ will approach as a limiting position the center of curvature $C (α,β)$ corresponding to $P 0$ on the curve. For if we drop the subscripts and write the last two equations in the form

$(x - \alpha') + (y - \beta') \frac{dy}{dx} = 0,$

$1 + \left( \frac{dy}{dx} \right)^2 + (y - \beta') \frac{d^2 y}{dx^2} = 0,$

it is evident that solving for $α'$ and $β'$ will give the same results as solving (B) and (C), p. 179, for $α$ and $β$ . Hence

Theorem. The center of curvature C corresponding to a point P on a curve is the limiting position of the intersection of the normal to the curve at P with a neighboring normal.

119. Evolutes. The locus of the centers of curvature of a given curve is called the evolute of that curve. Consider the circle of curvature corresponding to a point P on a curve. If P moves along the given curve, we may suppose the corresponding circle of curvature to roll along the curve with it, its radius varying so as to be always equal to the radius of curvature of the curve at the point P. The curve $C C 7$ described by the center of the circle is the evolute of $P P 7$ It is instructive to make an approximate construction of the evolute of a curve by estimating (from the shape of the curve) the lengths of the radii of curvature at different points on the curve and then drawing them in and drawing the locus of the centers of curvature.

Formula (E), p. 179 [§116], gives the coordinates of any point $(α,β)$ on the evolute expressed in terms of the cöordinates of the corresponding point $(x, y)$ of the given curve. But y is a function of x; therefore

$\alpha = x - \frac{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right] \frac{dy}{dx} }{ \frac{d^2 y}{dx^2} }, \beta = y + \frac{ 1 + \left( \frac{dy}{dx} \right)^2 }{ \frac{d^2 y}{dx^2} }$

give us at once the parametric equations of the evolute in terms of the parameter x.

To find the ordinary rectangular equation of the evolute we eliminate x between the two expressions. No general process of elimination can be given that will apply in all cases, the method to be adopted depending on the form of the given equation. In a large number of cases, however, the student can find the rectangular equation of the evolute by taking the following steps:

General directions for finding the equation of the evolute in rectangular coördinates.

FIRST STEP. Find α and β from (50), p. 180 [§117].

SECOND STEP. Solve the two resulting equations for x and y in terms of α and β.

THIRD STEP. Substitute these values of x and y in the given equation. This gives a relation between the variables α and β which is the equation of the evolute.

ILLUSTRATIVE EXAMPLE 1. Find the equation of the evolute of the parabola $y 2 = 4 p x$ .

Solution.	$\frac{dy}{dx}$	$=\ \frac{2p}{y}, \frac{d^2 y}{dx^2} = -\frac{4p^2}{y^3}.$
First step.	$α$	$= 3x + 2p, \beta = -\frac{y^3}{4p^2}.$
Second step.	$x$	$= \frac{\alpha - 2p}{3}, y = -(4p^2 \beta)^{\frac{1}{3}}.$
Third step	$(4p^2 \beta)^{\frac{2}{3}}$	$= 4p \left( \frac{\alpha - 2p}{3} \right);$
or,	$p β 2$	$\frac{4}{27} (\alpha - 2p)^3.$

Remembering that α denotes the abscissa and β the ordinate of a rectangular system of coordinates, we see that the evolute of the parabola AOB is the semi cubical parabola DC'E; the centers of curvature for $O, P, P 1, P 2$ being at $C', C, C 1, C 2$ respectively.

ILLUSTRATIVE EXAMPLE 2. Find the equation of the evolute of the ellipse $b 2 a 2 + a 2 y 2 = a 2 b 2$ .

Solution.	$\frac{dy}{dx}$	$= -\frac{b^2 x}{a^2 y}, \frac{d^2 y}{dx^2} = -\frac{b^4}{a^2 y^3}.$
First step.	$α$	$= \frac{(a^2 - b^2)x^3}{a^4},$
	$β$	$= -\frac{(a^2 - b^2)y^3}{b^4}.$
Second step.	$x$	$= \left( \frac{a^4 \alpha}{a^2 - b^2} \right)^{frac{1}{3}}$
	$y$	$= -\left( \frac{b^4 \beta}{a^2 - b^2} \right)^{\frac{1}{3}}$

Third step. $(a\alpha)^{\frac{2}{3}} + (b\beta)^{\frac{2}{3}} = (a^2 - b^2)^{\frac{2}{3}}$ , the equation of the evolute $E H E' H'$ of the ellipse $A B A' B', E, E', H', H$ are the centers of curvature corresponding to the points $A, A', B, B',$ on the curve, and $C, C', C''$ correspond to the points $P, P', P''$ .

When the equations of the curve are given in parametric form, we proceed to find $\frac{dy}{dx}$ and $\frac{d^2 y}{dx^2}$ , as on p. 160 [§103], from

(A)	$\frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }, \frac{d^2 y}{dx^2} = \frac{ \frac{dx}{dt} \frac{d^2 y}{dt^2} - \frac{dy}{dt} \frac{d^2 x}{dt^2} }{ \left( \frac{dx}{dt} \right)^3 }$

and then substitute the results in formulas (50), p. 180 [§117]. This gives the parametric equations of the evolute in terms of the same parameter that occurs in the given equations.

ILLUSTRATIVE EXAMPLE 3. The parametric equations of a curve are

(B)	$x = \frac{t^2 + 1}{4}, y = \frac{t^3}{6}.$

Find the equation of the evolute in parametric form, plot the curve and the evolute, find the radius of curvature at the point where t = 1, and draw the corresponding circle of curvature.

Solution.	$\frac{dx}{dt} = \frac{t}{2}, \frac{d^2 x}{dt^2} = \frac{1}{2},$
	$\frac{dy}{dt} = \frac{t^2}{2}, \frac{d^2 y}{dt^2} = t.$

Substituting in above formulas (A) and then in (50), p. 180 [§117], gives

(C)	$\alpha = \frac{1 - t^2 - 2t^4}{4}, \beta = \frac{4t^3 + 3t}{6},$

the parametric equations of the evolute. Assuming values of the parameter t, we calculate $x, y;α,β$ from (B) and (C); and tabulate the results as follows:

Now plot the curve and its evolute.

t	x	y	&aplha;	β
3	$\frac{5}{2}$	$\frac{9}{2}$
2	$\frac{5}{4}$	$\frac{4}{3}$	$-\frac{35}{4}$	$\frac{19}{3}$
$\frac{3}{2}$	$\frac{13}{16}$	$\frac{9}{16}$	$-\frac{91}{32}$	$3$
1	$\frac{1}{2}$	$\frac{1}{6}$	$-\frac{1}{2}$	$\frac{7}{6}$
0	$\frac{1}{4}$	0	$\frac{1}{4}$	0
$\frac{1}{2}$	$-\frac{1}{6}$	$-\frac{1}{2}$	$-\frac{7}{6}$
$-\frac{3}{2}$	$\frac{13}{16}$	$-\frac{9}{16}$	$-\frac{91}{32}$	$- 3$
$\frac{5}{4}$	$-\frac{4}{3}$	$-\frac{35}{4}$	$-\frac{19}{3}$
$\frac{5}{2}$	$-\frac{9}{2}$

The point $(\frac{1}{4}, 0)$ is common to the given curve and its evolute. The given curve (semi cubical parabola) lies entirely to the right and the evolute entirely to the left of $x = \frac{1}{4}$ .

The circle of curvature at $A (\frac{1}{2}, \frac{1}{6})$ , where $t = 1$ , will have its center at $A' (-\frac{1}{2}, -\frac{7}{6})$ on the evolute and radius = $A A'$ . To verify our work find radius of curvature at A. From (42), p. 159 [§103], we get

$R = \frac{t(1 + t^2)^{\frac{3}{2}}}{2} = \sqrt{2}, \text{when} t 1.$

This should equal the distance

$AA' = \sqrt{(\frac{1}{2} + \frac{1}{2})^2 +(\frac{1}{6} - \frac{7}{6})^2} = \sqrt{2}$ .

ILLUSTRATIVE EXAMPLE 4. Find the parametric equations of the evolute of the cycloid,

(C)	$\begin{cases} x = a(t - \sin t) \\ y = a(1 - \cos t).\end{cases}$

Solution. As in ILLUSTRATIVE EXAMPLE 2, p. 160 [§103], we get
	$\frac{dy}{dx} = \frac{\sin t}{1 - \cos t}, \frac{d^2 y}{dx^2} = -\frac{1}{\alpha (1 - \cos t)^2}.$
Substituting these results in formulas (50), p. 180 [§117], we get
(D)	$\begin{cases} \alpha = a(t + \sin t), \\ \beta = -a(1 - \cos t).\end{cases}$ Ans.

NOTE. If we eliminate t between equations (D), there results the rectangular equation of the evolute $O O' Q v$ referred to the axes $O'α$ and $O'β$ . The coördinates of O with respect to these axes are $( - π a, - 2 a)$ . Let us transform equations (D) to the new set of axes OX and OY. Then

α = x - π a,β = y - 2 a, t = t' - π.

Substituting in (D) and reducing, the equations of the evolute become

(E) $\begin{cases} x = a(t' - \sin t'), \\ y = a(1 - \cos t').\end{cases}$

Since (E) and (C) are identical in form, we have:

The evolute of a cycloid is itself a cycloid whose generating circle equals that of the given cycloid.

120. Properties of the evolute. From (A), p. 180 [§117],

(A) $α = x - R sinτ,β = y + R cosτ.$

Let us choose as independent variable the lengths of the arc on the given curve; then $x, y, R, T,α,β$ are functions of s. Differentiating (A) with respect to s gives

(B) $\frac{d\alpha}{ds} = \frac{dx}{ds} - R \cos \tau \frac{d\tau}{ds} - \sin \tau \frac{dR}{ds},$

(C) $\frac{d\beta}{ds} = \frac{dy}{ds} - R \sin \tau \frac{d\tau}{ds} + \cos \tau \frac{dR}{ds}.$

But $\frac{dx}{ds} = \cos \tau, \frac{dy}{ds} = \sin \tau$ , from (26), p. 134 [§90]; and $\frac{d\tau}{ds} = \frac{1}{R}$ , from (38) and (39), p. 156 [§100–1].

Substituting in (B) and (C), we obtain

(D) $\frac{d\alpha}{ds} = \cos \tau - R \cos \tau \cdot \frac{1}{R} - \sin \tau \frac{dR}{ds} = - \sin \tau \frac{dR}{ds},$

(E) $\frac{d\beta}{ds} = \sin \tau - R \sin \tau \cdot \frac{1}{R} + \cos \tau \frac{dR}{ds} = \cos \tau \frac{dR}{ds}.$

Dividing (E) by (D) gives

(F) $\frac{d\beta}{d\alpha} = - \cot \tau = - \frac{1}{\tan \tau} = - \frac{1}{\frac{dy}{dx}}.$

But	$\frac{d\beta}{d\alpha} = \tan \tau$ = slope of tangent to the evolute at C, and
	$\frac{dy}{dx} = \tan \tau$ = slope of tangent to the given curve at the corresponding point $P (x, y)$ .

Substituting the last two results in (F), we get

$\tan \tau' = - \frac{1}{\tan \tau}.$

Since the slope of one tangent is the negative reciprocal of the slope of the other, they are perpendicular. But a line perpendicular to the tangent at P is a normal to the curve. Hence

A normal to the given curve is a tangent to its evolute.

Again, squaring equations (D) and (E) and adding, we get

(G) $\left( \frac{d\alpha}{ds} \right)^2 + \left( \frac{d\beta}{ds} \right)^2 = \left( \frac{dR}{ds} \right)^2.$

But if s' = length of arc of the evolute, the left-hand member of (G) is precisely the square of $\frac{ds'}{ds}$ (from (34), p. 141 [§94], where $t = s, s = s', x = α, y = β$ ). Hence (G) asserts that

$\left( \frac{ds'}{ds} \right)^2 = \left( \frac{dR}{ds} \right)^2, \text{ or } \frac{ds'}{ds} = \plusmn \frac{dR}{ds}.$

That is, the radius of curvature of the given curve increases or decreases as fast as the arc of the evolute increases. In our figure this means that

$P_1 C_1 - PC = \text{arc}\ CC_1.$

The length of an arc of the evolute is equal to the difference between the radii of curvature of the given curve which are tangent to this arc at its extremities.

Thus in Illustrative Example 4, p. 186 [§118], we observe that if we fold $Q v P v ( = 4 a)$ over to the left on the evolute, $P v$ will reach to O', and we have:

The length of one arc of the cycloid (as OO'Q^v) is eight times the length of the radius of the generating circle.

121. Involutes and their mechanical construction. Let a flexible ruler be bent in the form of the curve $C 1 C 9$ the evolute of the curve. $P 1 P 9$ , and suppose a string of length $R 9$ , with one end fastened at $C 9$ , to be wrapped around the ruler (or curve). It is clear from the results of the last section that when the string is unwound and kept taut, the free end will describe the curve $P 1 P 9$ . Hence the name evolute.

The curve $P 1 P 9$ is said to be an involute of $C 1 C 9$ . Obviously any point on the string will describe an involute, so that a given curve has an infinite number of involutes but only one evolute.

The involutes $P 1 P 9, P 1' P 9', P 1'' P 9''$ are called parallel curves since the distance between any two of them measured along their common normals is constant.

The student should observe how the parabola and ellipse on pp. 183, 184 [§119] may be constructed in this way from their evolutes.

EXAMPLES

Find the coördinates of the center of curvature and the equation of the evolute of each of the following curves. Draw the curve and its evolute, and draw at least one circle of curvature.

1.	The hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^1} = 1.$	Ans.	$\alpha = \frac{(a^2 + b^2)x^3}{a^4}, \beta = -\frac{(a^2 + b^2)y^3}{b^4};$
evolute $(a\alpha)^{\frac{2}{3}} - (b\beta)^{\frac{2}{3}} = (a^2 + b^2)^{\frac{2}{3}}.$
2.	The hypocycloid $x^{\frac{2}{3}} + y^{\frac{2}{3}} = a^{\frac{2}{3}}.$		$\alpha = x + 3x^{\frac{1}{3}} y^{\frac{2}{3}}, \beta = y + 3 x^{\frac{2}{3}} y^{\frac{1}{3}};$
evolute $(\alpha + \beta)^{\frac{2}{3}} + (\alpha - \beta)^{\frac{2}{3}} = 2a^{\frac{2}{3}}.$
3.	Find the coordinates of the center of curvature of the cubical parabola $y 3 = a 2 x .$
		Ans.	$\alpha = \frac{a^4 + 15y^4}{6a^2 y}, \beta = \frac{a^4y - 9y^5}{2a^4}.$

4. Show that in the parabola $x^{\frac{1}{2}} + y^{\frac{1}{2}} = a^{\frac{1}{2}}$ we have the relation $α + β = 3(x + y).$

5. Given the equation of the equilateral hyperbola $2 x y = a 2$ show that

$\alpha + \beta = \frac{(y + x)^3}{a^2}, \alpha - \beta = \frac{(y - x)^3}{a^2}.$

From this derive the equation of the evolute $(\alpha + \beta)^{\frac{2}{3}} - (\alpha - \beta)^{\frac{2}{3}} = 2 a^{\frac{2}{3}}.$

Find the parametric equations of the evolutes of the following curves in terms of the parameter t. Draw the curve and its evolute, and draw at least one circle of curvature.

6. The hypocycloid	$\begin{cases} x = a \cos^3 t, \\ y = a \sin^3 t. \end{cases}$	Ans.	$\begin{cases} \alpha = a \cos^3 t + 3a \cos t \sin^2 t, \\ \beta = 3a \cos^2 t \sin t + a \sin^3 t. \end{cases}$
7. The curve	$\begin{cases} x = 3t^2, \\ y = 3t - t^3. \end{cases}$		$\begin{cases} \alpha = \frac{3}{2} ( 1 + 2t^2 - t^4 ), \\ \beta = -4 t^3. \end{cases}$
8. The curve	$\begin{cases} x = a(\cos t + t \sin t), \\ y = a(\sin t - t \cos t). \end{cases}$	Ans.	$\begin{cases} \alpha = a \cos t, \\ \beta = a \sin t. \end{cases}$
9. The curve	$\begin{cases} x = 3t, \\ y = t^2 -6. \end{cases}$		$\begin{cases} \alpha = -\frac{4}{3} t^3, \\ \beta = 3t^2 - \frac{3}{2}. \end{cases}$
10. The curve	$\begin{cases} x = 6 - t^2 \\ y = 2t. \end{cases}$		$\begin{cases} \alpha = 4 - 3t^2, \beta = -2t^3. \end{cases}$
11. The curve	$\begin{cases} x = 2t, \\ y = t^2 - 2. \end{cases}$		$\begin{cases} \alpha = -2 t^3, \\ \beta = 3t^2. \end{cases}$
12. The curve	$\begin{cases} x = 4t, \\ y = 3 + t^2. \end{cases}$		$\begin{cases} \alpha = -t^3, \\ \beta = 11 + 3t^2. \end{cases}$
13. The curve	$\begin{cases} x = 9 - t^2, \\ y = 2t. \end{cases}$		$\begin{cases} \alpha = 7 - 3t^2, \beta = -2t^3. \end{cases}$
14. The curve	$\begin{cases} x = 2t, \\ y = \frac{1}{3}t^3. \end{cases}$		$\begin{cases} \alpha = \frac{4t - t^5}{4}. \\ \beta = \frac{12 + 5t^4}{6t}. \end{cases}$
15. The curve	$\begin{cases} x = \frac{1}{3} t^3, \\ y = t^2. \end{cases}$		$\begin{cases} \alpha = \frac{4t^3 + 12t}{3} \\ \beta = -\frac{2t^2 + t^4}{2}. \end{cases}$
16. The curve	$\begin{cases} x = 2t, \\ y = \frac{3}{t}. \end{cases}$		$\begin{cases} \alpha = \frac{12t^4 + 9}{4t^3} \\ \beta = \frac{27 + 4t^4}{6t}. \end{cases}$

17. $x = 4 - t 2, y = 2 t .$	22. $x = t, y = t 3 .$
18. $x = 2 t, y = 16 - t 2 .$	23. $x = sin t, y = 3cos t .$
19. $x = t, y = sin t .$	24. $x = 1 - cos t, y = t - sin t .$
20. $x = \frac{4}{t}, y = 3t.$	25. $x = cos 4 t, y = sin 4 t .$
21. $x = t^2, y = \frac{1}{6} t^3.$	26. $x = a sec t, y = b tan t .$

↑ Sometimes called the osculating circle. The circle of curvature was defined from another point of view on p. 161. [§104]
↑ From (2), p. 77 [§65], X and Y being the variable coordinates.

[0] Sometimes called the osculating circle. The circle of curvature was defined from another point of view on p. 161. [§104]

[1] From (2), p. 77 [§65], X and Y being the variable coordinates.

[1]

[2]

Elements of the Differential and Integral Calculus/Chapter XIV

From Wikisource

[edit] CHAPTER XIV

Views

Personal tools

Search

Navigation

Toolbox

Print/export