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The cross product is $\mathbf{a}\times \mathbf{b}=(1,-7,-3)$. Taking its dot product with $\mathbf{c}$ gives the scalar triple product

The two dot products are $\mathbf{a}\cdot \mathbf{b}=2-1-6=-5$ and $\mathbf{b}\cdot \mathbf{c}=-1-3=-4$. Therefore

Differentiate before substituting the instantaneous side length. The related rates equation isAt the stated dimensions and rate,The diagonal grows more slowly than the changing side, since $b/d<1$.

Let $b_0=b(0)$ in centimetres and take $t$ in seconds. The constant rate gives $b(t)=b_0+4t$. By the Pythagorean theorem, the diagonal length isfor times when the side length is nonnegative. No initial value of $b$ is specified, so it must remain as a parameter. Choosing the time origin when $b=0$ gives the special convention $d(t)=\sqrt{100+16t^2}\mathrm{cm}$.

Differentiate the outer sine, then the inner cosine. The chain rule gives

For $a>0$, write $a^x=e^{x\ln a}$ and apply the chain rule:This also covers $a=1$, for which both sides are zero.

Use completing the square by adding and subtracting $3^2$:

Write the partial fraction decomposition as $A/(x-3)+B/(x+7)$. After multiplying through, $7x+29=A(x+7)+B(x-3)$. Setting $x=3$ gives $A=5$, and setting $x=-7$ gives $B=2$, so

For $m=-1$, set $u=\ln x$, $du=dx/x$. This integration by substitution givesThis case is treated separately because the denominator $m+1$ in part (a) vanishes.

For real powers work on $x>0$. Apply integration by parts with $u=\ln x$ and $dv=x^{m}dx$, so $v=x^{m+1}/(m+1)$ when $m\ne -1$. Then

Substitution into the circle gives $x^2{\sec }^{2}x=25$, or $x=5|\cos x|$ since $x\ge 0$. Necessarily $x\le 5$, and each admissible $x$ fixes a unique $y=x\tan x$.

On $[0,\pi /2)$, $x-5\cos x$ is strictly increasing from a negative value to a positive value, giving one root. On $(\pi /2,3\pi /2)$, the equation is $x+5\cos x=0$. This function is strictly convex, positive at the two interval ends and negative at $\pi$, giving exactly two roots.

There are none on the remaining possible interval $3\pi /2<x\le 5$. Put $\delta =x-3\pi /2>0$; then $\cos x=\sin \delta \le \delta$, soHence the number of simultaneous real solutions is $3$, as marked in the preceding graph.

The tangent has poles at $\pi /2$ and $3\pi /2$, so these are vertical asymptotes of $x\tan x$. The zeros on the given interval are $0,\pi ,2\pi$. Near zero, $x\tan x=x^2+O(x^4)$, so the graph starts with a horizontal tangent line.

Each of its three branches is increasing. Indeedand the numerator starts at zero with derivative $2{\cos }^{2}x\ge 0$, so it is positive for $x>0$. The first branch rises from zero to $+\infty$, the middle branch rises from $-\infty$ through $(\pi ,0)$ to $+\infty$, and the last rises from $-\infty$ to $(2\pi ,0)$.

The initial condition gives $e^{y(0)}=2=1+C$, so $C=1$. ConsequentlyIts logarithm has positive argument for every real $x$.

Multiply the separable differential equation by $e^y$: $e^y{y}^{\prime }=e^x$. The left side is $(e^y{)}^{\prime }$, hence $e^y=e^x+C$. ThereforeThe real general solution is taken on any connected interval where the logarithm's argument is positive.

The triangular roof has height $(w/2)\tan \theta$ and two slopes of length $(w/2)\sec \theta$. Hence the area constraint and cost areThe wall height is fixed, so its cost is constant. Assume the physical case $A,h,\beta >0$. Minimize $C$ subject to the area constraint by Lagrange multipliers:Eliminating the positive common factors givesThus the optimum has $0<\theta <\pi /4$ andSet $D=A-wh>0$, so the area constraint gives $\tan \theta =4D/w^2$. Substituting into the preceding equation and multiplying by $w^3$ yields $w^4-16D^2=16hwD$. ThereforeFor a global check, the square of the total roof length isIts derivative has the sign of $P(w)=w^4+16Ahw-16A^2$. Since $P^{\prime }(w)>0$, $P(0)<0$ and $P(A/h)>0$, it has one root in this interval, with $Q$ decreasing before that root and increasing afterwards. This proves the unique global minimum, rather than merely a stationary design. The minimum roof length at fixed wall height and cross-sectional area is independent of $\alpha$ and of the positive factor $\beta$. If $\beta =0$, every feasible shape has equal cost; the displayed conditions then describe a possible minimizer but are not necessary.

Take a face on the flat base and let $x,y$ be its positive half-widths, with height $z>0$. The upper vertices meet the spherical boundary at $x^2+y^2+z^2=a^2$, and the rectangular parallelepiped has volume $V=4xyz$. A Lagrange multiplier givesMultiplication by $x,y,z$ respectively shows $x^2=y^2=z^2$. Positivity and the constraint give $x=y=z=a/\sqrt{3}$. ThereforeBoundary choices with a zero dimension give zero volume, so this is the maximum in this configuration.

The maximum-volume cuboid in a hemisphere also has a bound covering arbitrary orientations. Let $u_i$ be any box's half-edge lengths, $v_i$ its centre coordinates in its orthonormal edge basis, and $n_i$ the vertical unit-vector coordinates. Remaining above the base plane requires $\sum n_i{v}_i\ge \sum |n_i|u_i$, so $|v_j|\ge u_j$ for at least one index. Its farthest vertex obeyswhere containment supplies the first inequality. The arithmetic-geometric mean inequality applied to the three terms $u_i^2$ for $i\ne j$ and $4u_j^2$ yields $2u_1{u}_2{u}_3\le a^3/(3\sqrt{3})$. Thus its full volume $8u_1{u}_2{u}_3$ cannot exceed the value attained above; tilting cannot improve it. Boundary contact is allowed for the maximum; strict containment would give the same supremum.

The two derivatives are $2x$ and $2(x-1)$. At the intersection their slopes are $1$ and $-1$, so their direction vectors $(1,1)$ and $(1,-1)$ have zero dot product. Thus the tangent lines are perpendicular and their angle is

At an intersection, $x^2=(x-1{)}^2=x^2-2x+1$, so $x=1/2$. Substitution into either parabola givesThe resulting linear equation for $x$ has only one solution, so this is the unique intersection.

For $p\le 0$, the terms $k^{-p}$ do not tend to zero, so the series diverges. For $p>0$, group the positive terms into the dyadic blocks $2^j\le k<2^{j+1}$. Each block contains $2^j$ terms, and monotonicity givesWhen $p>1$, the upper bounds form a convergent geometric series with ratio $2^{1-p}<1$, proving convergence. When $0<p\le 1$, the lower bounds do not tend to zero, so infinitely many blocks each make a nonvanishing contribution and the sum diverges. Thus the p-series criterion isThis dyadic comparison is the mechanism of the Cauchy condensation test. In particular $p=1$ is the divergent harmonic series.

Apply the ratio test to the infinite series whose partial sums are $S_N$. Assuming the terms are eventually nonzero and the limit exists, letIf $\ell <1$, the series has absolute convergence, by eventual comparison with a decaying geometric series. If $\ell >1$, including an infinite limit, the terms fail to tend to zero and the series diverges. If $\ell =1$, the test is inconclusive: both $\sum 1/k$ and $\sum 1/k^2$ have that ratio limit, but only the second converges. A finite partial sum itself requires no convergence test; the issue is the limit as $N\to \infty$.