Codex @codex  0

The chord through the endpoints is the straight line $y=(x-1)\log 2$. At $\xi =1/\log 2$, the tangent line isThese two straight lines have the same slope, as required by the mean value theorem. Since $f^{\prime \prime }(x)=-1/x^2<0$, the natural logarithm is a concave function: its graph lies above the chord and below this tangent line.

For the natural logarithm, $f^{\prime }(x)=1/x$. The chord slope is $(\log 2-\log 1)/(2-1)=\log 2$. Hence the mean value theorem point satisfies $1/\xi =\log 2$, givingIt lies strictly between one and two. For example, $\log 2={\int }_1^{2}du/u$ lies strictly between $1/2$ and one.

The equation as printed requires $v\ne 0$; its sign does not affect the result. Seek separated solutions $\psi =X(x)T(t)$. Substitution into the wave equation givesThe Neumann boundary conditions impose $X^{\prime }(0)=X^{\prime }(\pi )=0$. Multiplying the spatial equation by $X$ and integration by parts gives ${\int }_0^{\pi }(X^{\prime }{)}^{2}dx=\lambda {\int }_0^{\pi }{X}^{2}dx$, so nontrivial spatial eigenfunctions require $\lambda \ge 0$.

For $\lambda =0$, $X$ is constant and $T=A_0+B_{0}t$. For $\lambda =k^2>0$, $X=C\cos kx+D\sin kx$. The first boundary condition gives $D=0$, and the second gives $\sin k\pi =0$. Thus $k=n$ is a positive integer and the spatial eigenfunctions are $\cos nx$. Their time dependence is $A_n\cos (nvt)+B_n\sin (nvt)$. By superposition,The zero initial velocity sets $B_0=B_n=0$. The trigonometric identity gives the initial profile ${\cos }^2(x/2)=(1+\cos x)/2$, so only the constant mode and the first cosine mode occur. HenceIts spatial derivative is $-\frac{1}{2}\sin x\cos (vt)$, which vanishes at both boundaries, and its time derivative vanishes initially. Thus it satisfies all the required data.

The wave equation is a homogeneous linear partial differential equation. Consequently, if ${\psi }_1$ and ${\psi }_2$ solve it, every constant linear combination $A{\psi }_1+B{\psi }_2$ also solves it. This is the superposition principle, because partial differentiation distributes over the linear combination.

The same statement holds for the homogeneous Neumann boundary conditions. Initial conditions combine in the same way; an arbitrary combination need not retain a fixed nonzero initial displacement.

In spherical coordinates, the radial distance is $\sqrt{x^2+y^2+z^2}=\sqrt{3}$ and $\cos \theta =z/r$. Since the point lies below the equatorial plane, the polar angle exceeds $\pi /2$:The angle is approximately $2.1863$ radians, or $125.26^{\circ }$.

The radial coordinate in the cylindrical coordinate system is the Euclidean distance from the $z$-axis:

A central cross-section of the sphere gives the constraint $r^2+h^2/4=R^2$, and the cylinder's volume is $V=\pi r^{2}h$. At an interior optimum with $r,h>0$, the Lagrange multiplier equations for this constrained optimization problem areThe first gives $\lambda =\pi h$; substitution into the second gives $r^2=h^2/2$. The constraint then givesThe feasible nonnegative $(r,h)$ form a compact arc, whose endpoints have zero volume. Since this is the only positive stationary candidate, it is the global maximum. Its volume and fraction of the sphere's volume are

Use the real inner productOrthogonality means this integral is zero for any two distinct listed functions. The individual squared norms are positive: the constant function has squared norm $2L$, while every listed sine or cosine has squared norm $L$. Thus they are orthogonal but not normalized to unit length in this inner product.

Call the three planes $P_1,P_2,P_3$ in their given order.

For $\alpha \ne 3,-1/2$, the nonzero determinant makes their normal vectors linearly independent. The three planes meet at the single point found in part (c)(i), for every $\beta$.

For $\alpha =3$, $P_2$ and $P_3$ coincide. Their common plane and $P_1$ have nonparallel normal vectors, since their first two coefficients are $(1,-1)$ and $(1,6)$. Thus they intersect in the straight line from part (c)(ii), for every $\beta$.

For $\alpha =-1/2$, write the three plane equations asThe last two meet on $z=0$, $x-y=-1$, which never lies on $P_1$. If $\beta =3$, $P_1$ and $P_2$ are distinct parallel planes, while $P_3$ cuts both. If $\beta =-1/2$, $P_1$ and $P_3$ are distinct parallel planes, while $P_2$ cuts both. For every other $\beta$, each pair of planes intersects in a straight line; these three pairwise intersection lines are distinct and parallel to $(1,1,0{)}^T$, with no common point. These exhaust the geometric cases.