At the stated point, the radial distance is $7\pi /2$, so $4\pi -t=7\pi /2$ and $t=\pi /2$. Differentiate the parametric curve:At $t=\pi /2$, these are $\dot{x}=-1$ and $\dot{y}=-7\pi /2$. Since $\dot{x}\ne 0$, the tangent line has finite slope

The distance from the origin is $r=4\pi -t$. The direction angle, measured counterclockwise from the positive horizontal axis, is $\theta =\pi /2-t$ modulo $2\pi$. Thus the path is an Archimedean spiral, turning clockwise through two complete revolutions as its radius decreases linearly from $4\pi$ to zero.

It starts at $(0,4\pi )$, reaches $(7\pi /2,0)$ at $t=\pi /2$, crosses $(0,-3\pi )$ at $t=\pi$, returns to the positive vertical axis at $(0,2\pi )$ when $t=2\pi$, and ends at the origin when $t=4\pi$. These crossings fix both the scale and direction of the sketch.

For $x>0$, logarithmic differentiation givesSince $f(x)>0$, the derivative vanishes precisely when $\ln x=1$. HenceThe derivative is positive before $e$ and negative after $e$, so this is a global maximum on the given positive domain.

Differentiating the hyperbolic cosine with the chain rule gives $g^{\prime }(x)=3{\cosh }^{2}x\sinh x$. For real $x$, $\cosh x>0$, so the only zero of the derivative is $\sinh x=0$, namely $x=0$. ThusSince $\cosh x\ge 1$, this stationary point is the global minimum.

Let $u=x^{-5}$, which tends to zero as $|x|\to \infty$. Factor the denominator and use the cosecant expansion near zero:Substitution gives the first two nonzero terms and the requested omitted order:More precisely the next term is $7/(360x^{18})$, followed by $O(x^{-28})$. The outside factor $x^2$ is essential when converting the remainder order from powers of $u$ to powers of $x$.

Choose the upward orientation of the paraboloid cap. Its boundary is the circle $x^2+y^2=4$ at $z=1$, oriented counterclockwise as viewed from above. By Stokes theorem,On the boundary, $dz=0$ and $A\cdot dr=ydx-3xydy$. Parametrize by $x=2\cos \theta$, $y=2\sin \theta$, $0\le \theta \le 2\pi$. ThenThe second term integrates to zero over a full period, while the first contributes $-4\pi$. Equivalently, replace the cap by the upward disk at $z=1$: there $(\nabla \times A{)}_z=-3y-1$, whose integral is $-4\pi$ by symmetry.

An orientation is not stated in the original question. The displayed result uses the upward normal, also the outward normal on the curved top of the enclosed cap; the opposite orientation gives $4\pi$.

Take the sphere's outward normal. Since $F=k(x,y,z)$ with constant $k$, its divergence is $3k$. The divergence theorem givesThe translated centre does not enter the answer, because the divergence is constant. Equivalently the constant field produced by translating coordinates has zero net flux through a closed sphere. The inward orientation would negate the result.

Choose the plane normal to be $+\hat{y}$. Only the second field component contributes, and on $y=0$ it is $e^{-(x^2+z^2)}$. Thus the surface integral factors into two Gaussian integrals:The original question does not specify an orientation; choosing $-\hat{y}$ instead gives $-\pi$.

The printed first component $\sin y/y$ is undefined on the entire integration plane. Its removable continuous extension there is $1$, since ${\lim }_{y\to 0}\sin y/y=1$. The stated flux uses this natural extension; in fact the normal component alone is already well-defined and integrable on the plane, so the first component does not affect either oriented value.

The divergence isFor any closed piecewise smooth surface bounding a volume, use the divergence theorem with the outward orientation:Reversing the orientation still gives zero. The field is smooth everywhere, so there is no hidden singular source inside the surface.

View $a=a(b,c)$ with $b=b(d,e)$ and $c=c(d,e)$. Hold $e$ fixed and apply the multivariable chain rule:Dividing the changes along the fixed-$e$ variation by $dd$ gives exactly the two terms in the question. Thus the simpler expression isThe subscripts specify which coordinate is held fixed and must be retained.

Away from zeros of $\sin x$, an integrating factor for a differential one-form is $\mu (x)=1/\sin x$. Multiplication converts the form toTherefore the implicit solution family is $xy+\ln |\sin x|=C$. On each graph interval between consecutive multiples of $\pi$, this becomesNo such open interval contains zero. At $x=k\pi$, the undivided graph equation would read $\cos (k\pi )=0$, which is impossible for a finite derivative; hence no differentiable graph extends across one of these points.

If solutions are instead understood as arbitrary parametrized integral curves of the differential form, the vertical lines $x=k\pi$ are additional solutions: along them $dx=0$ and $x\sin xdy=0$. They were lost on division by $\sin x$ and are not graphs $y(x)$ over an interval.

If the form is $dF$ for a scalar function with continuous second partial derivatives, then $p=F_x$ and $q=F_y$. Equality of the mixed partial derivatives therefore requiresThis is necessary for an exact differential. For continuously differentiable coefficients on a simply connected open domain it is also sufficient; on a domain with holes, the equality alone need not give a single-valued global potential.

Set $w=z^3$. Then $w^2+w+1=0$ gives $w=e^{2\pi i/3}$ or $e^{-2\pi i/3}$. Taking all three cube roots of each yields six distinct solutions.

Equivalently, multiply the original polynomial by $z^3-1$ to obtain $z^9-1$. The original roots are exactly the ninth roots of unity that are not cube roots of unity: a cube root would give $1+1+1=3$, not zero. ThusAll six have modulus one, and their distinct arguments show that no roots have been repeated or omitted from the degree-six polynomial.