chengyewang/TexOCR-eval huggingface dataset:
4. Find the volume generated by revolving about $OY$ the area bounded by the coordinate axes and the parabola $x^{2/3}+y^{2/3}=a^{2/3}$.

5. Find the volume generated by revolving about the $x$-axis the area bounded by the catenary $y=\frac{a}{2}\left(e^{x/a}+e^{-x/a}\right)$, the $x$-axis and the lines $x=\pm a$.

6. Find the volume generated by revolving one arch of the sine curve $y=\sin x$ about $OX$.

7. A cone has its vertex on the surface of a sphere and its axis coincides with a diameter of the sphere. Find the common volume.

8. Find the volume generated by revolving about the $y$-axis, the part of the parabola $y^2=4ax$ cut off by the line $x=a$.

9. Find the volume generated by revolving about $x=a$ the part of the parabola $y^2=4ax$ cut off by the line $x=a$.

10. Find the volume generated by revolving about $y=-2a$ the part of the parabola $y^2=4ax$ cut off by the line $x=a$.

11. Find the volume generated by revolving one arch of the cycloid

$x=a(\phi-\sin\phi), \qquad y=a(1-\cos\phi)$

about the $x$-axis.

12. Find the volume generated by revolving the curve

$x=a\cos^3\phi, \qquad y=a\sin^3\phi$

about the $y$-axis.

13. Find the volume generated by revolving the cardioid $r=a(1+\cos\theta)$ about the initial line.

14. Find the volume generated by revolving the cardioid $r=a(1+\cos\theta)$ about the line $x=-\frac{a}{4}$.

15. Find the volume generated by revolving the ellipse

$x^2+xy+y^2=3$

about the $x$-axis.

16. Find the volume generated by revolving about the line $y=x$ the part of the parabola $x^{1/3}+y^{1/3}=a^{1/3}$ cut off by the line $x+y=a$.

\textbf{29. Volume of a Solid with Given Area of Section.} — Divide the solid into slices by parallel planes. Let $X$ be the area of section at distance $x$ from a fixed point. The plate $PQRS$ with lateral surface perpendicular to $PQR$ has the volume

$PQR\cdot \Delta x = X\,\Delta x$.
215. Moment of inertia of a right circular cylinder about a line tangent to its base.

216. Moment of inertia of a segment cut from a sphere by a plane, about a diameter parallel to that plane.

217. Center of gravity of a right circular cone whose density varies as the distance from the center of the base.

218. Volume bounded by the $xy$-plane, the cylinder $x^2 + y^2 = 2ax$ and the cone $z^2 = x^2 + y^2$.

219. Find the attraction of a uniform wire of length $l$ and mass $M$ on a particle of unit mass at distance $c$ from the wire in the perpendicular at one end.

220. Find the attraction of a right circular cylinder on a particle at the middle of its base.

221. Show that the attraction of a homogeneous shell bounded by two concentric spherical surfaces on a particle in the enclosed space is zero.

\section{Chapter X}

Solve the following differential equations:

222. $y\,dx + (x - xy)\,dy = 0$.

223. $\sin x \sin y\,dx + \cos x \cos y\,dy = 0$.

224. $(2xy - y^2 + 6x^2)\,dx + (3y^2 + x^2 - 2xy)\,dy = 0$.

225. $x\,\frac{dy}{dx} + y = x^3 y$.

226. $x\,\frac{dy}{dx} + y = \cot x$.

227. $x\,dy - (y + e^{1/x})\,dx = 0$.

228. $(1 + x^2)\,dy + (xy + x)\,dx = 0$.

229. $x\,dx + y\,dy = x\,dy - y\,dx$.

230. $(\sin x + y)\,dy + (y\cos x - x^2)\,dx = 0$.

231. $y(e^x + 2)\,dx + (e^x + 2x)\,dy = 0$.

232. $(xy^2 - x)\,dx + (y + xy)\,dy = 0$.

233. $(1 + x^2)\,\frac{dy}{dx} + xy = 2y$.

234. $x\,dy - y\,dx = \sqrt{x^2 + y^2}\,dx$.

235. $(x - y)\,dx + x\,dy = 0$.

236. $x\,dy - y\,dx = x\sqrt{x^2 + y^2}\,dx$.

237. $e^{x+y}\,dy + (1 + e^y)\,dx = 0$.

238. $(2x + 3y - 1)\,dx + (4x + 6y - 5)\,dy = 0$.

239. $(3y^2 + 3xy + x^2)\,dx = (x^2 + 2xy)\,dy$.

240. $(1 + x^2)\,dy + (xy - x^2)\,dx = 0$.
Using V, with $u = x^2 + 1$, $v = x^2 - 1$,
\[
 dy = \frac{(x^2-1)\,d(x^2+1) - (x^2+1)\,d(x^2-1)}{(x^2-1)^2}
\]
\[
= \frac{(x^2-1)2x\,dx - (x^2+1)2x\,dx}{(x^2-1)^2}
\]
\[
= -\frac{4x\,dx}{(x^2-1)^2}.
\]

\textbf{Ex. 5.} $y = \sqrt{x^2 - 1}$.

Using VI, with $u = x^2 - 1$,
\[
\frac{dy}{dx} = \frac{d}{dx}(x^2-1)^{\frac12} = \frac12 (x^2-1)^{-\frac12} \frac{d}{dx}(x^2-1)
\]
\[
= \frac12 (x^2-1)^{-\frac12}(2x) = \frac{x}{\sqrt{x^2-1}}.
\]

\textbf{Ex. 6.} $x^2 + xy - y^2 = 1$.

We can consider $y$ a function of $x$ determined by the equation. Then
\[
d(x^2) + d(xy) - d(y^2) = d(1) = 0,
\]
that is,
\[
2x\,dx + x\,dy + y\,dx - 2y\,dy = 0,
\]
\[
(2x + y)\,dx + (x - 2y)\,dy = 0.
\]
Consequently,
\[
\frac{dy}{dx} = \frac{2x + y}{2y - x}.
\]

\textbf{Ex. 7.} $x = t + \frac{1}{t}, \quad y = t - \frac{1}{t}$.

In this case
\[
dx = dt - \frac{dt}{t^2}, \qquad dy = dt + \frac{dt}{t^2}.
\]
Consequently,
\[
\frac{dy}{dx} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} = \frac{t^2 + 1}{t^2 - 1}.
\]

\textbf{Ex. 8.} Find an approximate value of $y = \left(\frac{1-x}{1+x}\right)^{\frac13}$ when $x = 0.2$.
Using the formula just proved,

\[
 d \cos u = d \sin \left(\frac{\pi}{2}-u\right) = \cos \left(\frac{\pi}{2}-u\right) d\left(\frac{\pi}{2}-u\right) = -\sin u\,du.
\]

\subsection{38. Proof of IX, X, XI, and XII.}
Differentiating both sides of the equation
\[
\tan u = \frac{\sin u}{\cos u}
\]
and using the formulas just proved for the differentials of $\sin u$ and $\cos u$,
\[
 d\tan u = \frac{\cos u\, d\sin u - \sin u\, d\cos u}{\cos^2 u}
 = \frac{\cos^2 u\,du + \sin^2 u\,du}{\cos^2 u}
 = \sec^2 u\,du.
\]

By differentiating both sides of the equations
\[
\cot u = \frac{\cos u}{\sin u}, \qquad \sec u = \frac{1}{\cos u}, \qquad \csc u = \frac{1}{\sin u},
\]
and using the formulas for the differentials of $\sin u$ and $\cos u$, we obtain the differentials of $\cot u$, $\sec u$ and $\csc u$.

\textit{Example 1.} $y = \sin^2(x^2 + 3)$.

Since
\[
\sin^2(x^2 + 3) = [\sin(x^2 + 3)]^2,
\]
we use the formula for $u^2$ and so get
\[
 dy = 2\sin(x^2 + 3)\, d\sin(x^2 + 3)
 = 2\sin(x^2 + 3)\cos(x^2 + 3)\, d(x^2 + 3)
 = 4x\sin(x^2 + 3)\cos(x^2 + 3)\, dx.
\]

\textit{Ex. 2.} $y = \sec 2x\tan 2x$.

\[
\frac{dy}{dx} = \sec 2x \frac{d}{dx}(\tan 2x) + \tan 2x \frac{d}{dx}(\sec 2x)
\]
\[
= \sec 2x\sec^2 2x (2) + \tan 2x\sec 2x\tan 2x (2)
\]
\[
= 2\sec 2x\,(\sec^2 2x + \tan^2 2x).
\]

\subsection*{EXERCISES}
In the following exercises show that the derivatives and differentials have the values given:

1. $y = 2\sin 3x + 3\cos 2x$, \qquad $\dfrac{dy}{dx} = 6(\cos 3x - \sin 2x)$.
with the axes.) By differentiation we see at once that these components are the derivatives of the function
\[
\frac{\sigma}{\sqrt{(\xi_1-x_1)^2+(\xi_2-x_2)^2+(\xi_3-x_3)^2}}
\]
with respect to the co-ordinates $x_1, x_2, x_3$ respectively. The force vector apart from a constant factor is therefore the gradient of the function
\[
\frac{1}{r}=\frac{1}{\sqrt{(\xi_1-x_1)^2+(\xi_2-x_2)^2+(\xi_3-x_3)^2}}.
\]
If a field of force is obtained from a scalar function by forming the gradient, this scalar function is often called the \emph{potential function} of the field. We shall consider this concept from a more general point of view in the study of work and energy (Chapter V, p. 350, and Chapter VI, pp. 415, 168--81).

\section{The Divergence and Curl of a Vector Field}
By differentiation we have assigned to every function or scalar a vector field, the gradient. Similarly, by differentiation we can assign to every vector field a certain scalar, known as the divergence of the vector field. Given a specific co-ordinate system, the $x$-system, we define the divergence of the vector $u$ as the function
\[
\mathrm{div}\,u = \frac{\partial u_1}{\partial x_1} + \frac{\partial u_2}{\partial x_2} + \frac{\partial u_3}{\partial x_3},
\]
to the sum of the partial derivatives of the three components with respect to the corresponding co-ordinates. Suppose now that we change the co-ordinate system to the $\xi$-system. If the divergence is really to be a scalar function associated with the vector field and independent of the particular co-ordinate system, we must have
\[
\mathrm{div}\,u = \frac{\partial \omega_1}{\partial \xi_1} + \frac{\partial \omega_2}{\partial \xi_2} + \frac{\partial \omega_3}{\partial \xi_3},
\]
where $\omega_1, \omega_2, \omega_3$ are the components of $u$ in the $\xi$-system. In fact, the truth of the equation
\[
\frac{\partial u_1}{\partial x_1} + \frac{\partial u_2}{\partial x_2} + \frac{\partial u_3}{\partial x_3} = \frac{\partial \omega_1}{\partial \xi_1} + \frac{\partial \omega_2}{\partial \xi_2} + \frac{\partial \omega_3}{\partial \xi_3}
\]
can be verified immediately by applying the chain rule and the transformation formulae of p. 84.
is no longer an ``affine'' transformation, but a transformation to general curvilinear co-ordinates.

We again assume that when $(x,y)$ ranges over a region $R$ of the $xy$-plane the corresponding point $(\xi,\eta)$ ranges over a region $B$ of the $\xi\eta$-plane, and also that for each point of $B$ the corresponding $(x,y)$ in $R$ can be uniquely determined, in other words, that the transformation is one-to-one. The inverse transformation we again denote by $x=g(\xi,\eta)$, $y=h(\xi,\eta)$.

By the co-ordinates of a point $P$ in a region $R$ we can mean any number-pair which serves to specify the position of the point $P$ in $R$ uniquely. Rectangular co-ordinates are the simplest case of co-ordinates which extend over the whole plane. Another typical case is the system of polar co-ordinates in the $xy$-plane, introduced by the equations

$$
\xi = r = \sqrt{x^2 + y^2},
$$

$$
\eta = \theta = \arctan(y/x) \qquad (0 \leq \theta < 2\pi).
$$

When we are given a system of functions $\xi = \phi(x,y)$, $\eta = \psi(x,y)$ as above, we can in general assign to each point $P(x,y)$ the corresponding values $(\xi,\eta)$ as new co-ordinates. For each pair of values $(\xi,\eta)$ belonging to the region $B$ uniquely determines the pair $(x,y)$, and thus uniquely determines the position of the point $P$ in $R$, this entitles us to call $\xi,\eta$ the co-ordinates of the point $P$. The ``co-ordinate lines'' $\xi = \text{const.}$ and $\eta = \text{const.}$ are then represented in the $xy$-plane by two families of curves, which are defined implicitly by the equations $\phi(x,y)=\text{const.}$ and $\psi(x,y)=\text{const.}$ respectively. These co-ordinate curves cover the region $R$ with a co-ordinate net (usually curved), for which reason the co-ordinates $(\xi,\eta)$ are also called curvilinear co-ordinates in $R$.

We shall once again point out how closely those two interpretations of our system of equations are interrelated. The curves in the $\xi\eta$-plane which in the mapping correspond to straight lines parallel to the axes in the $xy$-plane can be directly regarded as the co-ordinate curves for the curvilinear co-ordinates $x=g(\xi,\eta)$, $y=h(\xi,\eta)$ in the $\xi\eta$-plane; conversely, the co-ordinate curves of the curvilinear co-ordinate system $\xi=\phi(x,y)$, $\eta=\psi(x,y)$ in the $xy$-plane in the mapping are the images of the straight lines parallel to the axes in the $\xi\eta$-plane. Even in the interpretation of $(\xi,\eta)$ as curvilinear co-ordinates in the $xy$-plane
complete symmetry. Lagrange's rule is usually expressed as follows:

\textit{To find the extreme values of the function $f(x,y)$ subject to the subsidiary condition $\phi(x,y)=0$, we add to $f(x,y)$ the product of $\phi(x,y)$ and an unknown factor $\lambda$ independent of $x$ and $y$, and write down the known necessary conditions}

\[
f_x + \lambda \phi_x = 0, \qquad f_y + \lambda \phi_y = 0,
\]

for an extreme value of

\[
\Gamma = f + \lambda \phi
\]

In conjunction with the subsidiary condition $\phi = 0$ these serve to determine the co-ordinates of the extreme value and the constant of proportionality $\lambda$.

Before proceeding to prove the rule of undetermined multipliers rigorously we shall illustrate its use by means of a simple example. We wish to find the extreme values of the function

\[
u = xy
\]

on the circle with unit radius and centre the origin, that is, with the subsidiary condition

\[
x^2 + y^2 - 1 = 0.
\]

According to our rule, by differentiating $xy + \lambda(x^2 + y^2 - 1)$ with respect to $x$ and to $y$ we find that at the stationary points the two equations

\[
y + 2\lambda x = 0,
\]

\[
x + 2\lambda y = 0
\]

have to be satisfied. In addition we have the subsidiary condition

\[
x^2 + y^2 - 1 = 0.
\]

On solving we obtain the four points

\[
\xi = \tfrac{1}{2}\sqrt{2}, \qquad \eta = \tfrac{1}{2}\sqrt{2},
\]
\[
\xi = -\tfrac{1}{2}\sqrt{2}, \qquad \eta = -\tfrac{1}{2}\sqrt{2},
\]
\[
\xi = \tfrac{1}{2}\sqrt{2}, \qquad \eta = -\tfrac{1}{2}\sqrt{2},
\]
\[
\xi = -\tfrac{1}{2}\sqrt{2}, \qquad \eta = \tfrac{1}{2}\sqrt{2}.
\]

The first two of these give a maximum value $u = \tfrac{1}{2}$, the second two a minimum value $u = -\tfrac{1}{2}$, of the function $u = xy$. That the first two do really give the greatest value and the second two the least value of the function $u$ can be seen as follows: on the circumference the function must assume a greatest and a least value, and since the circumference has no boundary point, these points of greatest and least value must be stationary points for the function.
Hence by the definition of $\lambda$ and $\mu$
\[
f_x + \lambda \phi_x + \mu \psi_x = 0.
\]

Similarly, if we multiply the equations
\[
\phi_y + \phi_z \frac{\partial z}{\partial y} + \phi_t \frac{\partial t}{\partial y} = 0
\]
and
\[
\psi_y + \psi_z \frac{\partial z}{\partial y} + \psi_t \frac{\partial t}{\partial y} = 0
\]
by $\lambda$ and $\mu$ respectively and add them to the equation
\[
f_y + f_z \frac{\partial z}{\partial y} + f_t \frac{\partial t}{\partial y} = 0,
\]
we obtain the further equation
\[
f_y + \lambda \phi_y + \mu \psi_y = 0.
\]

We thus arrive at the following result.

If the point $(\xi, \eta, \zeta, \tau)$ is an extreme point of $f(x, y, z, t)$ subject to the subsidiary conditions
\[
\phi(x, y, z, t) = 0,
\]
\[
\psi(x, y, z, t) = 0,
\]
and if at that point $\dfrac{\partial(\phi, \psi)}{\partial(z, t)}$ is not zero, then two numbers $\lambda$ and $\mu$ exist such that at the point $(\xi, \eta, \zeta, \tau)$ the equations
\[
f_x + \lambda \phi_x + \mu \psi_x = 0,
\]
\[
f_y + \lambda \phi_y + \mu \psi_y = 0,
\]
\[
f_z + \lambda \phi_z + \mu \psi_z = 0,
\]
\[
f_t + \lambda \phi_t + \mu \psi_t = 0,
\]
and also the subsidiary conditions, are satisfied.

These last conditions are perfectly symmetrical. Every trace of emphasis on the two variables $x$ and $y$ has disappeared from them, and we should equally well have obtained them if, instead of assuming that $\dfrac{\partial(\phi, \psi)}{\partial(z, t)} \neq 0$, we had merely assumed that any one of the Jacobians $\dfrac{\partial(\phi, \psi)}{\partial(x, y)}$, $\dfrac{\partial(\phi, \psi)}{\partial(x, z)}$, $\ldots$, $\dfrac{\partial(\phi, \psi)}{\partial(z, t)}$ did not
Since $\psi_1(x)$ and $\psi_2(x)$ are uniformly continuous, as $\delta \to 0$ the functions $\psi_{1\delta}(x)$ and $\psi_{2\delta}(x)$ tend uniformly to $\psi_1(x)$ and $\psi_2(x)$ respectively, and so

$$
\lim_{\delta \to 0} \int_{\psi_{1\delta}(x)}^{\psi_{2\delta}(x)} f(x,y)\,dy = \int_{\psi_1(x)}^{\psi_2(x)} f(x,y)\,dy
$$

uniformly in $x$. It follows that

$$
\lim_{\delta \to 0} \int_{x_0}^{x_1} dx \int_{\psi_{1\delta}(x)}^{\psi_{2\delta}(x)} f(x,y)\,dy = \int_{x_0}^{x_1} dx \int_{\psi_1(x)}^{\psi_2(x)} f(x,y)\,dy.
$$

On the other hand, as $\delta \to 0$ the region $\bar{R}$ tends to $R$. Hence

$$
\lim_{\delta \to 0} \iint_{\bar{R}} f(x,y)\,dS = \iint_{R} f(x,y)\,dS.
$$

Combining the three equations, we have

$$
\iint_{R} f(x,y)\,dS = \int_{x_0}^{x_1} dx \int_{\psi_1(x)}^{\psi_2(x)} f(x,y)\,dy.
$$

The other statement can be established in a similar way.

A similar argument is available if we abandon the hypothesis

\begin{figure}
\centering
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics{figure_6_1.pdf}
\end{minipage}
\begin{minipage}{0.45\textwidth}
\centering
\includegraphics{figure_6_2.pdf}
\end{minipage}
\caption{Non-convex regions of integration}
\label{figure_6}
\end{figure}

of convexity and consider regions of the form indicated in fig.~\ref{fig:figure_6}. We assume merely that the boundary curve of the region is intersected by every parallel to the $x$-axis and by every parallel to the $y$-axis in a bounded number of points or intervals. By $\int f(x,y)\,dy$ we then mean the sum of the integrals of the function $f(x,y)$ for a fixed $x$, taken over all the intervals which the line $x = \text{const.}$ has in common with the closed region. For non-
each circle $K_m$ is contained in the interior of $R_v$, provided $v$ is sufficiently large; on the other hand, every $R_v$ is bounded and is therefore contained in a circle $K_M$ of sufficiently large radius $M$. Since the integrand $e^{-x^2-y^2}$ is positive everywhere, it follows that

$$
\iint_{K_m} e^{-x^2-y^2} \, dx \, dy \le \iint_{R_v} e^{-x^2-y^2} \, dx \, dy \le \iint_{K_M} e^{-x^2-y^2} \, dx \, dy.
$$

As $m$ and $M$ increase, the integrals over $K_m$ and $K_M$ have the same limit $\pi$, so that the integral over $R_v$ must have the same limit; thus proves that the integral must converge to the limit $\pi$.

We obtain a particularly interesting result if for the regions $R_v$ we choose the squares $|x| \le v$, $|y| \le v$. The integral $\iint_{R_v} e^{-x^2-y^2} dx \, dy$ can then be reduced to two simple integrations (cf section 2, p 228):

$$
\iint_{R_v} e^{-x^2-y^2} dx \, dy = \int_{-v}^{v} e^{-x^2} dx \int_{-v}^{v} e^{-y^2} dy = \left(\int_{-v}^{v} e^{-x^2} dx\right)^2 = \left(2 \int_{0}^{v} e^{-x^2} dx\right)^2.
$$

If we now let $v$ tend to $\infty$, we must again obtain the same limit $\pi$. Hence

$$
\left(2 \int_{0}^{\infty} e^{-x^2} dx\right)^2 = \pi
$$

or

$$
\int_{0}^{\infty} e^{-x^2} dx = \frac{1}{2}\sqrt{\pi},
$$

in agreement with Vol I, p 496.

\section{5. Summary and Extensions}

It is useful to consider the concepts of this section again from a single unifying point of view. Our extension of the concept of integral to cases in which the definitions in section 2 (p 224) are not immediately applicable consists in regarding the value of the integral as the limiting value of a sequence of integrals over regions $R_v$, which approximate to the original region of integration $R$ as $v$ increases. For this purpose we regard the region $R$ as open instead of closed, we assign all the points of discontinuity of the function $f$ to the boundary and consider the boundary as not belonging to $R$. We then say that the region is approximated to by a sequence of regions $R_1, R_2, \ldots, R_n, \ldots$ if all the closed regions $R_n$ lie in $R$ and every arbitrarily chosen closed sub-region in the interior of $R$ is also a sub-region of the region $R_n$, provided only that $n$ is sufficiently large. If in particular the sub-regions $R_n$ are so chosen that each one contains the
stood, we may restrict ourselves to the study of the first. We shall therefore only consider affine transformations of the form

\[
x' = ax + by + cz,
\]
\[
y' = dx + ey + fz,
\]
\[
z' = gx + hy + kz
\]

or

\[
x' = ax + by,
\]
\[
y' = cx + dy,
\]

with non-vanishing determinants.

The results of section 3 (p. 25) for linear equations enable us to express the inverse transformation by the formulae

\[
x = a'x' + b'y' + c'z',
\]
\[
y = d'x' + e'y' + f'z',
\]
\[
z = g'x' + h'y' + k'z',
\]

or

\[
x = a'x' + b'y',
\]
\[
y = c'x' + d'y'.
\]

in which $a', b', \ldots$ are certain expressions formed from the coefficients $a, b, \ldots$. Because of the uniqueness of the solution, the original equations also follow from these latter. In particular, from $x = y = z = 0$ it follows that $x' = y' = z' = 0$, and conversely.

The characteristic geometrical properties of affine transformations are stated in the following theorems:

(1) In space the image of a plane is a plane; and in the plane the image of a straight line is a straight line.

For by section 1 (p. 9) we can write the equation of the plane (or the line) in the form

\[
Ax + By + Cz + D = 0
\]

(or

\[
Ax + By + D = 0).
\]

The numbers $A, B, C$ (or $A, B$) are not all zero. The co-ordinates of the image points of the plane (or of the line) satisfy the equation

\[
A(a'x' + b'y' + c'z') + B(d'x' + e'y' + f'z') + C(g'x' + h'y' + k'z') + D = 0
\]

(or

\[
A(a'x' + b'y') + B(c'x' + d'y') + D = 0).
\]

Hence the image points themselves lie on a plane (or a line), for the coefficients

\[
A' = a'A + d'B + g'C,
\]
\[
B' = b'A + e'B + h'C,
\]
\[
C' = c'A + f'B + k'C
\]

(or

\[
A' = a'A + c'B,
\]
\[
B' = b'A + d'B)
\]

of the co-ordinates $x', y', z'$ (or $x', y'$) cannot all be zero, otherwise the equations

\[
a'A + d'B + g'C = 0,
\]
\[
b'A + e'B + h'C = 0,
\]
\[
c'A + f'B + k'C = 0
\]

(or

\[
a'A + c'B = 0,
\]
\[
b'A + d'B = 0)
\]

would hold, and these we may regard as equations in the unknowns $A, B, C$.
denoting differentiation with respect to $s$) Using the relations $x^2 = 1$, $\dot{x}x = 0$, we obtain the equations $(y-x)\dot{x} = 0$, $(y-x)x = 1$, $(y-x)x = 0$. Hence we have
\[
y-x = \frac{[\dot{x}x]}{[x\dot{x}]}x.
\]

5. Of Ex. 3 and also Ex. 5, p. 19.

7. From the definitions of $\xi_1, \xi_2, \xi_3$ we have $\xi_1 = \dot{x}$, $\dot{x}^2 = 1$, $\xi_2 = x/h$, $\xi_3 = [\xi_1, \xi_2]$, $\pm \sqrt{\xi_3^2} = 1/\tau$. Obviously $\xi_1 = k\xi_2$. To determine $\xi_2, \xi_3$, we calculate their components with respect to a rectangular coordinate system $O\xi_1, O\xi_2, O\xi_3$. From the relations
\[
\xi_1^2 = 1, \quad \xi_2^2 = 1, \quad \xi_1\xi_2 = \xi_2\xi_3 = \xi_3\xi_1 = 0
\]
we obtain by differentiation
\[
\xi_3\xi_1 = -\xi_1\xi_3 = 0, \qquad \xi_3\xi_3 = 0;
\]
hence $\xi_3$ is perpendicular both to $\xi_1$ and to $\xi_2$, and therefore
\[
\xi_3 = \pm \sqrt{(\xi_3^2)}\,\xi_2 = \pm \xi_2/\tau.
\]
We define the sign of $\tau$ so as to give $\xi = -\xi_2/\tau$. This implies that $\tau$ is positive or negative according as the screw defined by the motion of the osculating plane in the direction of increasing $s$ is right-handed or left-handed. To prove the second formula, note that
\[
\xi_2\xi_1 = -\xi_1\xi_2 = -k, \qquad \xi_2\xi_2 = 0, \qquad \xi_2\xi_3 = -\xi_3\xi_2 = 1/\tau.
\]

8. Use Ex. 6 and Ex. 3: (a) $k\xi_2 - l^2\xi_1 + \dfrac{k}{\tau}\xi_3$, (b) $\dfrac{k}{l^2\tau}\,\xi_3 + \dfrac{\xi_3}{\tau}$.

9. $1/|\tau| = \sqrt{\xi_3^2} = 0$, hence $\xi_3$ is a constant vector $\eta$, say $x\eta = \xi_1\eta = \xi_2\eta = 0$, so that $x\eta = \text{const.}$, where $\eta$ is a fixed vector. That is, the curve lies in a fixed plane.

10. (b) If the curve is given by $x = f(t)$, $y = g(t)$, $z = h(t)$, the surface has the parametric equations
\[
x = f(t) + s f'(t),
\]
\[
y = g(t) + s g'(t),
\]
\[
z = h(t) + s h'(t),
\]
then express $\dfrac{\partial^2 z}{\partial x^2}$, $\dfrac{\partial^2 z}{\partial x \partial y}$, $\dfrac{\partial^2 z}{\partial y^2}$ in terms of the derivatives with respect to $t$ and $s$.

Appendix, \S 1, p. 100

1. (a) As $R$ is closed, there is a point $B$ in $R$ whose distance from $A$ is less than that of any other point in $R$. Let $n$ be the normal to $AB$ at $B$. Then no point $C$ in $R$ lies on the same side of $n$ as $A$, for otherwise not only $B$ and $C$, but the whole segment $BC$, would belong to $R$, and on this segment there would be points nearer to $A$ than $B$ is. Hence the parallel to $n$ through $A$ cannot meet $R$.