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Let $r=|{\mathbf{r}}_1-{\mathbf{r}}_2|$ and $\hat{\mathbf{r}}=({\mathbf{r}}_1-{\mathbf{r}}_2)/r$. A differentiable central potential energy gives the internal forcesThere are no external forces. Differentiating the total kinetic energy and using Newton's second law therefore givesSince $\dot{r}=\hat{\mathbf{r}}\cdot ({\mathbf{v}}_1-{\mathbf{v}}_2)$, the last expression is $-U^{\prime }(r)\dot{r}=-dU/dt$. ThusThis conserves total mechanical energy, including the kinetic energy of centre-of-mass motion, not merely the relative kinetic energy.

Write ${\mathbf{r}}_i=\mathbf{X}+{\mathbf{r}}_i^{\prime }$ and ${\mathbf{v}}_i=\mathbf{V}+{\mathbf{v}}_i^{\prime }$, where $\mathbf{V}=\dot{\mathbf{X}}$. The definition of the centre of mass and its derivative giveExpand the total angular momentum:The two middle terms vanish. Thus the angular momentum decomposition about the centre of mass isThe first term is orbital angular momentum of the centre; the second is angular momentum about the centre of mass.

About a fixed inertial origin, the total angular momentum is $\mathbf{L}={\sum }_i{m}_i{\mathbf{r}}_i\times {\dot{\mathbf{r}}}_i$. Differentiation givesbecause ${\dot{\mathbf{r}}}_i\times {\dot{\mathbf{r}}}_i=0$. Separate the external and internal contributions. The combined torque of an internal pair isThe first equality uses Newton's third law, and the second uses the given central-force condition. All internal torques therefore cancel, leavingAn equal-and-opposite pair which is not central need not have zero combined torque; this explains why both internal-force hypotheses matter.

For constant particle masses define the centre of mass by $M\mathbf{X}={\sum }_i{m}_i{\mathbf{r}}_i$, with $M={\sum }_i{m}_i$. Summing Newton's second law over all particles givesEach pair of internal forces cancels by Newton's third law. HenceOnly the action-reaction condition is needed for this translation law; centrality of the internal forces is needed for the angular momentum result in part (ii).

Let $q(t)=\mathbf{r}(t)\cdot \mathbf{B}$. Since the field is constant, dotting the equation of motion with $\mathbf{B}$ eliminates the magnetic term:This is a harmonic oscillator with frequency $\omega =\sqrt{k/m}$. Initially $q(0)={\mathbf{r}}_0\cdot \mathbf{B}$ and $\dot{q}(0)=0$, because the particle starts at rest. ThereforeThe magnetic force modifies the transverse motion but not this projection. The formula also holds trivially when $\mathbf{B}=0$.

Write $\mathbf{v}=\dot{\mathbf{r}}$. The original PDF contains the cross product $\dot{\mathbf{r}}\times \mathbf{r}$; the TeX aid drops its velocity dot. Since $\mathbf{v}\times \mathbf{v}=0$, differentiation and the equation of motion giveThe radial restoring force contributes no torque. The vector triple-product identity yieldsMeanwhileThe derivatives cancel, provingThis is the magnetic axial angular-momentum invariant. Its first term is the negative of the ordinary angular momentum projected on $\mathbf{B}$; retaining that sign is essential.

Dot Newton's second law with $\dot{\mathbf{r}}$. The magnetic cross-product term is perpendicular to this velocity, soThusorThe two halves are the kinetic energy and the isotropic spring's potential energy. The expression in the question is twice their sum. The Lorentz force changes the direction of motion but supplies no power, because it is perpendicular to velocity; this explains why the magnetic field does not appear in the conserved mechanical energy. The sign corresponds to charge $-e$.

Assume $P,Q$ are continuously differentiable on a neighbourhood of a bounded region with a positively oriented piecewise smooth simple boundary. To prove Green theorem, first take a region between the graphs $y=f(x)$ and $y=g(x)$, $a\le x\le b$, with $f\le g$. Counterclockwise orientation traverses the lower graph to the right and the upper graph to the left; vertical end pieces contribute nothing to $Pdx$. Thus the fundamental theorem of calculus givesLikewise, for a region between graphs $x=u(y)$ and $x=v(y)$, the right boundary is traversed upwards and the left one downwards. Horizontal end pieces contribute nothing to $Qdy$, soFor a polygonal region, triangulate its interior. Each triangle is of both graph types with piecewise linear bounds, so adding the formulae gives the result: the double integrals add and the integrals on internal edges cancel because the adjacent triangles traverse them in opposite directions. For a regular piecewise smooth simple boundary, choose inscribed simple polygonal approximations to its piecewise smooth parametrization. Their line integrals converge to the original line integral, since the parametrizations converge uniformly and their piecewise constant tangent vectors converge in the integral norm on each smooth arc. Their regions converge in area: the symmetric difference lies in a shrinking neighbourhood of the original boundary, which has planar measure zero. Boundedness of the continuous integrand then gives convergence of the area integrals. Passing to the limit proves the formula for the stated boundary class. Combining the two identities givesPositive orientation is essential for the sign.

Choose $P=-y/2$ and $Q=x/2$, for which $Q_x-P_y=1$. The boundary formula for planar area isFor the ellipse, take $a,b>0$ so the given increasing-angle parametrization is counterclockwise. Then $dx=-a\sin \theta d\theta$ and $dy=b\cos \theta d\theta$, soIf signed nonzero parameters are allowed, the unsigned area is $\pi |ab|$, with orientation corrected accordingly.

A conservative vector field is the gradient of a single-valued differentiable scalar potential. Its line integral is path independent and equals the potential difference. If $\psi \mathbf{A}=\nabla F$ with a twice continuously differentiable potential, equality of mixed partial derivatives givesThe product rule then gives ${\psi }_y{A}_1+\psi A_{1,y}={\psi }_x{A}_2+\psi A_{2,x}$. Rearranging proves the integrating-factor equation for a planar vector field:This argument does not require division by $\psi$, so it remains valid even at its zeros. A nonvanishing integrating factor is normally sought when one wants to recover useful potential curves for the original field.

TakeThe real eigenvalues are one and two, with eigenlines spanned by $(1,0{)}^T$ and $(1,1{)}^T$ respectively. These independent vectors form a real eigenbasis, so $E$ is a diagonalizable matrix over $\mathbb{R}$. Their inner product is one, and any two nonzero vectors on those eigenlines have nonzero inner product. Rescaling can normalize their lengths but cannot make them orthogonal. Hence no orthonormal eigenbasis exists.

UseIt has independent complex eigenvectors $(1,-i{)}^T$ and $(1,i{)}^T$, with eigenvalues $i$ and $-i$. Taking these as columns gives an invertible complex matrix $P$ with $P^{-1}DP=\mathrm{diag}(i,-i)$, so $D$ is a diagonalizable matrix over $\mathbb{C}$. It cannot be diagonalized over $\mathbb{R}$, since a real diagonal matrix would have real eigenvalues, whereas similarity preserves the characteristic polynomial.

Take the Jordan blockIts only eigenvalue is one. An eigenvector satisfies $(C-I)v=0$, so its second coordinate vanishes; the eigenspace is one-dimensional. A diagonalizable matrix of size two requires a basis of two independent eigenvectors, which this matrix lacks even over $\mathbb{C}$.

TakeIts characteristic polynomial is ${\lambda }^2+1$, whose roots are $i$ and $-i$. Thus this real matrix has no real eigenvalues.

Can be false. Use $G=S_3$ and the normal subgroup $H=A_3$ from the root solution. The subgroup $H$ is cyclic of order three, hence an abelian group, and the quotient group has the two cosets $A_3$ and $(12)A_3$, so is cyclic of order two and abelian. Yet $(12)(23)=(123)$ whereas $(23)(12)=(132)$, proving that $S_3$ is not abelian. An abelian kernel and an abelian quotient do not force all elements of the original group to commute.

Always true. Elements of a subgroup $H$ commute when they are elements of an abelian group $G$, so $H$ is abelian. In the quotient group, $(gH)(kH)=gkH=kgH=(kH)(gH)$. Therefore $G/H$ is abelian too.

Can be false. Take the Klein four-group $G=C_2\times C_2$ and $H=C_2\times \{e\}$. It is a normal subgroup because $G$ is an abelian group. Both $H$ and $G/H$ have order two and are cyclic groups. But every nonidentity element of $G$ has order two, so none generates its four elements. Thus $G$ is not cyclic.

Always true. Suppose $G=⟨g⟩$. If $H$ is nontrivial, let $d$ be the least positive integer with $g^d\in H$. For any $g^k\in H$, divide $k=qd+r$ with $0\le r<d$. Since $g^r=g^k(g^d{)}^{-q}\in H$, minimality forces $r=0$. Hence $H=⟨g^d⟩$ is a cyclic group. The trivial subgroup is cyclic as well. Every coset in $G/H$ is $g^{k}H=(gH{)}^k$, so the quotient group is generated by $gH$. This proves the subgroups and quotients of a cyclic group assertion.

Write the surface as the level set $F(x,y,z)=x^2{y}^3-y+1-z=0$. The gradient is normal to a regular level surface because differentiating $F$ along any surface curve gives $\nabla F\cdot \mathbf{v}=0$ for its tangent vector. At the specified point,Thus a normal vector and a unit normal are $(2,2,-1)$ and $(2,2,-1)/3$, respectively. The tangent-plane displacement is perpendicular to that normal, givingReversing the normal's sign gives the same plane.

For a regular differentiable curve $\mathbf{x}(t)$, its speed is $ds/dt=|{\mathbf{x}}^{\prime }(t)|>0$, where $s$ is arc length. The oriented unit tangent vector is $\hat{\mathbf{T}}={\mathbf{x}}^{\prime }/|{\mathbf{x}}^{\prime }|$. The curvature of a space curve is $\kappa =|d\hat{\mathbf{T}}/ds|$. Equivalently, for a twice differentiable regular curve,For the helix, differentiation gives ${\mathbf{x}}^{\prime }=(-a\sin t,a\cos t,b)$ and speed $v=\sqrt{a^2+b^2}$. ConsequentlyIndeed $d\hat{\mathbf{T}}/dt=(-a\cos t,-a\sin t,0)/v$ has magnitude $|a|/v$, and division by $ds/dt=v$ gives the result. This requires $(a,b)\ne (0,0)$; the excluded constant curve has no well-defined unit tangent vector. For $a=0,b\ne 0$ the curve is a straight line with zero curvature of a space curve.

The exterior radial Laplace equation has $\psi =a+b/r$. Decay at infinity forces $a=0$ and the value at $r=2$ forces $b=2$. ThereforeTogether the three solutions satisfy their separate regional equations and specified boundary values. Their radial derivatives jump at the interfaces, so they should not be asserted to solve one globally smooth Poisson equation with only the listed volume sources and no interface terms.