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Assume $\lambda >0$, as required for normalization of the exponential distribution. Its cumulative distribution function is the integral of its probability density function from $-\infty$ to $x$:For the expected value, integration by parts givesSimilarly,The endpoint terms vanish because exponential decay dominates each fixed power of $x$. ThereforeThe mean has the same units as $X$, while the variance has their square; $\lambda$ has reciprocal units.

The printed point is not on the printed surface, so the requested tangent plane does not exist. At $(2,-2,9)$, the left side of the defining equation is $x^{2}y=-8$, whereas the right side is $4z{e}^{x+y}-11=25$. This is a direct counterexample to the premise that the stated point belongs to the surface.

To make the obstruction precise, set $G(x,y,z)=x^{2}y-4z{e}^{x+y}+11$. The specified surface is the level set $G=0$, but $G(2,-2,9)=-33$. Although $\nabla G(2,-2,9)=(-44,-32,-4)$, the resulting plane $11x+8y+z=15$ is tangent to the different level set $G=-33$ through that point. It is not an answer for the printed surface. A tangent plane requires the point to lie on the surface.

If the intended correction preserves $x=2,y=-2$, the surface equation instead forces $z=3/4$. At the corrected point $p_{\ast }=(2,-2,3/4)$,The corresponding conditional tangent plane isThis conditional calculation identifies a possible repair; it does not silently replace the source's coordinate $9$.

Convert the fixed position to $x=0.20\mathrm{m}$. The element's displacement is then $y(t)=0.27\sin (2.4-500t)\mathrm{mm}$. Taking time derivatives at fixed position gives its transverse velocity and acceleration:Equivalently the prefactors in SI units are $-0.135\mathrm{m}{\mathrm{s}}^{-1}$ and $-67.5\mathrm{m}{\mathrm{s}}^{-2}$. The phase is measured in radians. At $t=4\mathrm{s}$,These are the transverse motions of the material element, not the propagation speed of the wave along the string.

The gradient is $\nabla f=(2x-4x^3,-2y-4y^3)$, so the real stationary points are $(0,0)$ and $(\pm 1/\sqrt{2},0)$. The Hessian matrix is diagonal, with entries $2-12x^2$ and $-2-12y^2$. At the origin it has opposite-sign eigenvalues, giving a saddle point. At each of the other two stationary points it is negative definite, giving strict maxima of value $1/4$. These are also the global maxima, sinceThere are no local minima, and the function is unbounded below. The zero contour has two lobes touching at the saddle; positive contours are closed curves around the maxima.

Under the stated change of variables,Inside the resulting circular cap the constraint $X>-1/2$ is automatic. Comparing its circle and horizontal boundary givesHere $a$ is the positive half-width used in the source figure, independent of the pulse-height notation in the preceding question. The Jacobian determinant is $\partial (X,Y)/\partial (x,y)=4xy>0$ in the chosen quadrant, so the integral is one-quarter of the circular segment area:

Write the initial triangular pulse as the function $h$, extended by zero to negative arguments:At $t=0$, the initial displacement and zero velocity give $f+g=h$ and $-c{f}^{\prime }+c{g}^{\prime }=0$ on the positive half-line. Fix the harmless opposite additive constants by taking both travelling waves to vanish at positive infinity. ThenThe second formula follows from the fixed-end boundary condition $f(-ct)+g(ct)=0$. Thus $f$ is one-half of the odd extension $H$ of the initial displacement, andThis is the odd-extension solution for a semi-infinite fixed string: the reflected displacement reverses sign. Evaluating the travelling profiles givesThe first profile is a positive triangle supported on $[L,3L]$, with peak $a/2$ at $2L$. The second has a negative triangle on $[0,2L]$ with minimum $-a/2$ at $L$, and a positive triangle on $[2L,4L]$ with maximum $a/2$ at $3L$.

Multiply the secant function integrand by $(\sec \theta +\tan \theta )/(\sec \theta +\tan \theta )$:The numerator is the derivative of the denominator. The formula holds on any real interval avoiding zeros of $\cos \theta$; on such an interval $\sec \theta +\tan \theta$ cannot vanish because $(\sec \theta +\tan \theta )(\sec \theta -\tan \theta )=1$.

Let $D=y(3y-2)$. Implicit differentiation of the two constraints gives the linear systemAdding them, then substituting back, gives at regular pointsThe relevant Jacobian determinant is $2(1+a)yz(3y-2)$. Thus the special parameter value is $\boxed{a=-1}$, assuming the other factors are nonzero. At that parameter, adding the original constraints gives $(x+1{)}^2=0$: the intersection curve lies wholly in the plane $x=-1$ and cannot be parametrized by $x$ on an interval. Calling its derivatives “infinite” describes a vertical projection, not a finite-valued differentiable pair $y(x),z(x)$ at $a=-1$; the displayed ratios then have simultaneous zero numerators on the intersection and are undefined.

There are also point-dependent failures when $y=0$, $y=2/3$ or $z=0$. These can produce vertical tangents or require a different local parameter, and simultaneous numerator zeros need separate analysis. The implicit function theorem guarantees finite $y^{\prime },z^{\prime }$ only where the stated Jacobian determinant is nonzero. This distinction is the geometry of a singular projection of an implicit curve.

For $x>0$, $\sqrt{1+x^2}$ increases from $1$, while $2/\sqrt{1+x^2}$ decreases from $2$. They meet at $(1,\sqrt{2})$, after which their order reverses. In particular, their positive separation on $0<x<1$ is $(1-x^2)/\sqrt{1+x^2}$.

Use the hyperbolic substitution $x=\sinh t$, $dx=\cosh tdt$. With $T=\mathrm{arsinh}1$, the definite integral for the area becomesSince $\sinh T=1$, $\cosh T=\sqrt{2}$ and $T=\log (1+\sqrt{2})$, the answer is

Rationalize the difference of square roots before taking the limit. For positive $x$, divide the resulting numerator and denominator by $x^2$:Each factor now has a finite nonzero denominator limit, giving

The velocity vector of the helix is $r^{\prime }(t)=(-\sin t,\cos t,1)$, whose Euclidean norm is $\sqrt{2}$. Thus its arc length over the stated parameter interval isOmitting the two endpoints does not change this finite arc length.

For a plane wave, ${\partial }_t$ acts as $-i\omega$ and ${\partial }_x$ as $ik$. Substitution givesso, for $\omega \ne {\omega }_d$, the dispersion relation isAssume ${\omega }_n>{\omega }_d>0$ and real angular frequency. The ratio is negative exactly on ${\omega }_d<\omega <{\omega }_n$, giving imaginary wavenumbers and an evanescent wave. Thus the propagating frequency gap is $({\omega }_d,{\omega }_n)$. At ${\omega }_d$ there is no finite-$k$ solution, while ${\omega }_n$ has $k=0$; neither endpoint is an ordinary finite-wavelength propagating mode.

Choose the branch $k=(\omega /c)\sqrt{(\omega -{\omega }_n)/(\omega -{\omega }_d)}$, so the phase velocity is positive in the allowed bands. Direct differentiation gives the phase velocity and group velocityThe opposite spatial branch reverses both signs. On the chosen branch, the requested limits areThe band-edge limits are taken from their propagating sides. More precisely, with $ϵ=\omega -{\omega }_n\downarrow 0$ and $\delta ={\omega }_d-\omega \downarrow 0$,

For the final case, set $K=k_{0}c>0$. Substitution of $T(t)e^{i{k}_{0}x}$ at ${\omega }_d=0$ gives $T^{(3)}+(3iK/2)T^{\prime \prime }+K^2{T}^{\prime }=0$. Its characteristic polynomial factors as $r(r+2iK)(r-iK/2)$, henceThe initial data give $A+B+C=1$, $-2B+C/2=0$ and $-4B-C/4=-5$. Thus $B=1$, $C=4$, $A=-4$, andThe time-independent contribution is essential: static modes lost by cancelling frequency would be missed if one divided the ${\omega }_d=0$ dispersion relation by $\omega$ and retained only its two nonzero-frequency roots.