Let $D$ be a bounded planar region with a piecewise smooth, simple closed boundary $C$, oriented counterclockwise, and let $P,Q$ have continuous first partial derivatives on an open neighbourhood of its closure. Green's theorem states thatFor a region with holes, orient every boundary component with $D$ on its left: the outer component is counterclockwise and the inner components are clockwise.

Choose $P=-y/2$ and $Q=x/2$. Their indicated partial derivatives differ by one, so the area isThis is signed area as a boundary integral; reversing the direction of $C$ changes the sign of the integral.

The gradient and constant Hessian matrix areAt ${\mathbf{x}}_0=(1,0{)}^T$, $\nabla F=(10,-8{)}^T$ and $H_F^{-1}\nabla F=(1,-1{)}^T$. Hence one multivariable Newton iteration givesAt this point the gradient vanishes, so every subsequent iterate is $(0,1)$. The one-step result follows because the gradient of a quadratic polynomial is affine and its linearization is exact. Since the leading principal minors of $H_F$ are $6>0$ and $8>0$, it is a positive-definite matrix. Alternatively,directly proves that $(0,1)$ is the unique global minimum, with value $-1$.

For a twice continuously differentiable function $F$, its gradient and Hessian matrix areA critical point solves $\nabla F=0$. The Jacobian matrix of this vector equation is $H_F$, so the multivariable Newton iteration isEquivalently ${\mathbf{x}}_{k+1}={\mathbf{x}}_k-H_F({\mathbf{x}}_k{)}^{-1}\nabla F({\mathbf{x}}_k)$ when the Hessian matrix is invertible. Solving the linear system is the direct computational form. A critical point found in this way need not be a local minimum; its type requires a separate check.

Back-substitution in the Euclidean algorithm gives the Bezout identityMultiplying by five produces a particular solution $(x,y)=(-115,60)$. Since $259/7=37$ and $497/7=71$ are coprime integers, all solutions of the linear Diophantine equation areReplacing $t$ by $t+2$ gives the smaller equivalent starting solutionIndeed $259\cdot 27-497\cdot 14=35$. Conversely, subtracting this solution from any other gives $37(x-27)=-71(y+14)$. Coprimality forces $71$ to divide $x-27$, and then $y+14$ is the corresponding multiple of $-37$. This proves completeness of the family.

The Euclidean algorithm givesThe last nonzero remainder is the greatest common divisor, so $\boxed{\gcd (259,497)=7}$.

For a fixed attachment ratio $\delta >0$, making the disc much heavier than the rod givesIf the disc is also small relative to the attachment distance, $\alpha /\delta =a/d\to 0$, its moment of inertia about its own center is negligible compared with $M{d}^2$. ThusThis is a simple pendulum with a point mass at the disc's center and effective length $d$; it is length $\ell$ only if $d=\ell$. Even before the small-angle approximation, its equation tends to $\ddot{\theta }=-(g/d)\sin \theta$.

If $\delta$ is allowed to tend to zero together with the other parameters, the bare conditions $\alpha \ll 1$ and $\mu \gg 1$ do not alone justify that reduction. The needed dominance conditions are $\mu \delta \gg 1$, $\mu {\delta }^2\gg 1$, and $\alpha /\delta \ll 1$, so that the rod contributes negligible gravitational torque and moment of inertia, and the disc behaves as a point mass. They follow automatically in the combined limit with fixed $\delta >0$.

The small-angle approximation $\sin \theta =\theta +O({\theta }^3)$ reduces ii to $\ddot{\theta }+{\omega }^2\theta =0$, whereThe coefficient is positive, so the solution is simple harmonic motion, $\theta (t)=C\cos \omega t+D\sin \omega t$. Here $\omega$ is the angular frequency; the ordinary frequency and period are

The rod and disc contribute restoring torques $-mg(\ell /2)\sin \theta$ and $-Mgd\sin \theta$, respectively. Using i in the rotational equation of motion givesOnly the disc's center of mass enters its gravitational torque; its finite radius enters its moment of inertia. The formula is also regular at $M=m$, although the source happens to exclude that value.

The disc is rigidly attached, so its rotation axis is perpendicular to its face. Its moment of inertia about its center is $M{a}^2/2$: with surface mass density $M/(\pi a^2)$, an annulus of radius $s$ contributes $s^2[M/(\pi a^2)]2\pi sds$, whose integral is $M{a}^2/2$. The parallel axis theorem adds $M{d}^2$ at the pivot. Adding the rod's moment of inertia gives

With $L=1/\sqrt{n^2-1}$, one has $\nu =n$. Since $n\ge 3>\sqrt{3}$, ii proves boundedness. Write the positive rational number $n=p/q$ in lowest terms. The radius and its angular derivative repeat after $\Delta \theta =2\pi /n$. After $p$ such radial oscillations,so both the position and the radial and transverse velocities return to their initial values. The elapsed time iswhich is finite and positive because $r$ is bounded above and below by positive constants. Thus the motion is periodic. The orbit is a closed, generally self-intersecting rosette orbit, with $p$ radial oscillations during $q$ complete revolutions; for integral $n$ it has $n$ evenly spaced lobes around the origin. It never reaches the origin and is not a circular orbit.

The oscillatory numerator $\cos (\nu \theta )+\nu \sin (\nu \theta )$ has amplitude $\sqrt{1+{\nu }^2}$. Therefore the reciprocal radius stays bounded away from zero exactly whenIn this case the radial bounds areFor completeness, the potential energy per unit mass is $\Phi =-1/(ar)+1/(2r^2)$, and conservation of mechanical energy givesThis effective potential diverges positively at zero, has one minimum at $r=a(1+L^2)$, and tends to zero from below at infinity. Since the initial radial velocity is negative, the particle first reaches an inner turning point. If $L^2\ge 1/2$, the subsequent outward motion has no finite outer turning point and escapes. At equality the escape speed tends to zero, but the radius is still unbounded; infinity is approached only as time tends to infinity. HenceThe minimum of $U$ permits a circular orbit for other initial data, but the nonzero initial radial velocity here excludes that exceptional solution.

The initial transverse velocity gives $h=a(L/a)=L$. From $\dot{r}=-h{u}^{\prime }$, the initial conditions are $u(0)=1/a$ and $u^{\prime }(0)=1/a$. DefineThe Binet equation becomes the constant-coefficient linear ordinary differential equationIts particular solution is $({\nu }^2-1)/(a{\nu }^2)$. Applying the two initial conditions to the complementary solution givesThe physical branch is the connected range containing $\theta =0$ on which the right-hand side is positive. A zero corresponds to escape to infinite radius, rather than continuation to negative $r$.

The source calls $F$ an inward force magnitude, but here $F(r)=(r-a)/(a{r}^3)$ changes sign at $r=a$. The stated initial condition has inward radial velocity, so the particle immediately enters $r<a$, where the force is outward. The consistent interpretation is that $F$ is the signed inward radial component of the force per unit mass.

The rotation axis is $L=\{(x,1,0):x\in \mathbb{R}\}$. The distance of the generating function graph from this axis is $\sqrt{x}$, so a parametrization isThe surface of revolution therefore has equation $(y-1{)}^2+z^2=x$: it is a circular elliptic paraboloid opening along the positive $x$ direction, with its vertex and its rim at $x=1$ excluded. Translating the axis does not change the surface area formula, soThe cancellation makes the improper integral finite at the vertex. Neither the missing vertex nor the missing boundary circle changes the surface area, and no cap is included.

A parametrization of the surface of revolution isIts tangent vectors are orthogonal, with norms $\sqrt{1+f^{\prime }(x{)}^2}$ and $|f(x)|$. ThusThe absolute value is necessary if the function graph passes below the axis. The formula gives the curved surface area; it adds no end discs. Zeros of $f$ can be treated by splitting the integral into regular portions and taking limits.

Parametrize the function graph by $\mathbf{R}(x,y)=(x,y,f(x,y))$. Its two tangent vectors are ${\mathbf{R}}_x=(1,0,f_x)$ and ${\mathbf{R}}_y=(0,1,f_y)$. A small parameter rectangle maps, to first order, to a tangent parallelogram with area $|{\mathbf{R}}_x\times {\mathbf{R}}_y|dxdy$. Sincethe surface area of a graph isFor example, $f\in C^1$ ensures a regular surface chart; more general domains can be handled by the corresponding convergent improper integral.

For the cardioid, $r^{\prime }=-a\sin \theta$. The expression under the square root simplifies toSince $0\le \theta \le {\theta }_2\le \pi$, the cosine is nonnegative. ThereforeThe upper half of the cardioid starts at $(2a,0)$, passes through $(0,a)$, bends to negative $x$, and ends at the cusp $(0,0)$ when ${\theta }_2=\pi$. Its full upper-half arc length is $4a$; reflecting gives $8a$ for the full cardioid.

The fundamental theorem of calculus givesThis decreases from $2a$ to zero, so the arc length increases fastest at ${\theta }_2=0$, interpreted as the right-hand rate at the initial endpoint. If only positive endpoints are allowed, the rate has supremum $2a$ as ${\theta }_2\downarrow 0$, rather than an attained interior maximum.

In polar coordinates, $x=r(\theta )\cos \theta$ and $y=r(\theta )\sin \theta$. The product rule givesSquaring and adding cancels the mixed terms and gives $(x^{\prime }{)}^2+(y^{\prime }{)}^2=(r^{\prime }{)}^2+r^2$. Hence the arc length is

For a piecewise continuously differentiable parametric curve, the arc length is the integral of speed:The Pythagorean theorem gives the length of an infinitesimal displacement as $\sqrt{d{x}^2+d{y}^2}$. Integration counts traversal length, including repeated portions if the parametrization retraces them. Piecewise continuous derivatives, or suitable absolute continuity, justify the formula; mere differentiability as stated in the source is not by itself the usual sufficient regularity hypothesis.

The vector triple product identity givesEliminate $V$ with iii to obtain the reciprocal basis involution:This formula actually holds without the unit-length assumptions. Under those assumptions it also says $|{\mathbf{b}}^{\prime }\times {\mathbf{c}}^{\prime }|=1/|V|$; normalizing this cross product alone would lose the sign of $V$ and could return $-\mathbf{a}$.

Since $V\ne 0$, iii gives ${\mathbf{a}}^{\prime }\cdot ({\mathbf{b}}^{\prime }\times {\mathbf{c}}^{\prime })=1/V\ne 0$. The criterion in a therefore proves that ${\mathbf{a}}^{\prime },{\mathbf{b}}^{\prime },{\mathbf{c}}^{\prime }$ are non-coplanar. In particular they form a basis of ${\mathbb{R}}^3$.