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A parametric curve interface supplies its parameter domain, evaluation, and derivatives where they exist. Robust geometric algorithms also need restricted subcurves and certified range or derivative bounds on parameter intervals. Tangent, speed and curvature are derived from point and derivative enquiries when the relevant derivatives are regular. Finite point samples alone cannot certify the absence of every plane intersection of an arbitrary smooth curve.

Two adjacent Bézier curves of degrees $n,m$ and interval lengths $h,k$ have a $C^1$ join precisely when their endpoint positions and their derivatives in the common parameter agree. Parallel endpoint polygon edges alone give tangent-direction agreement; their scaled vectors must be equal for parametric first-derivative continuity.

Core numerical enquiries return a surface point and its derivatives through second order, together with the parameter domain, patch boundaries and regularity information. At a regular point these determine its tangent plane, normal vector, first and second fundamental forms and curvatures. Subdivision and certified enclosing volumes provide spatial search and intersection support. Singular charts or discontinuity boundaries require one-sided or alternative-chart information; a zero cross product does not define a unique normal vector.

A partial derivative measures the local response of a function to one input while its other inputs stay fixed. For a function $f$ on an open subset of ${\mathbb{R}}^2$, its partial derivatives at $(a,b)$ are the limitswhen the respective limits exist. Each is an ordinary derivative of a one-variable slice: $x\mapsto f(x,b)$ or $y\mapsto f(a,y)$. This is why the other variable is treated as a constant during partial differentiation.^[ref]

For an original example, take the polynomial functionHolding $y$ fixed gives $f_x=2x+y$, while holding $x$ fixed gives $f_y=x+4y$. At $(1,1)$, the height is $f=4$ and the two partial derivatives are $f_x=3$ and $f_y=5$. The values answer different experiments on the same mathematical model: change $x$ alone, or change $y$ alone.

The two exact slices through that point areDividing either height change by $t\ne 0$ gives $3+t$ or $5+2t$; taking the limit $t\to 0$ recovers the two slopes. At $t=0.1$, the finite height changes are $0.31$ and $0.52$, although their linear approximations are $0.3$ and $0.5$. A derivative measures the limiting response, rather than an exact finite change.

For a function of $n$ variables, ${\partial }_{j}f(a)$ varies only its $j$th coordinate: it is the same difference quotient along $e_j$, the $j$th standard basis vector. This gives a practical sensitivity analysis: one can compare how a mathematical model responds to changes in its individual inputs, provided the chosen system of units and the quantities held fixed are stated. Comparing the numerical sizes of two partial derivatives in different units does not by itself identify a physically preferable input.

Existence of the partial derivatives at one point is weaker than differentiability. The example under partial derivatives do not imply continuity shows how both coordinate tests can miss a bad diagonal approach. When the first partial derivatives are continuous in a neighborhood, they do give differentiability there, allowing the total differential to combine all small input changes.

Suppose $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and satisfies $f(a)=f(b)$, with $a<b$. Apply the extreme value theorem to both $f$ and $-f$ to obtain a maximum and a minimum. If $f$ is constant, every interior derivative is zero. Otherwise some value differs from the common endpoint value, so either a maximum or a minimum occurs at an interior point $c$. At an interior maximum, difference quotients are nonpositive from the right and nonnegative from the left; since the derivative exists, it must be zero. The minimum case is identical with the inequalities reversed. This proves Rolle's theorem: $\boxed{f^{\prime }(c)=0}$ for some $c\in (a,b)$.

For the mean value theorem, subtract the endpoint chord:Then $g(a)=g(b)=0$, so Rolle's theorem supplies $c\in (a,b)$ with $g^{\prime }(c)=0$. HenceIf $f^{\prime }(t)>0$ everywhere, applying the mean value theorem to any $x<y$ gives $f(y)-f(x)=f^{\prime }(c)(y-x)>0$. Thus $f$ is strictly increasing. The converse is false: $\boxed{f(t)=t^3}$ is strictly increasing but $f^{\prime }(0)=0$.

Let $m=4\pi \rho a^3/3$ be the drop's mass, with constant mass density $\rho$. Its volume-growth rule gives $4\pi a^2\dot{a}=c\pi a^{2}v$, hence $\dot{a}=cv/4$ and $\dot{m}=\rho c\pi a^{2}v$. The incoming mist is stationary, so the variable-mass system momentum balance isExpanding the derivative and substituting the accretion rate gives the accreting raindrop with quadratic drag equationsThe extra $c$ contribution is the momentum needed to accelerate collected matter from rest; omitting it would treat the growing drop as a constant-mass particle.

Set $q=v^2/a$ and $B=7c/8+3k/4$. The chain rule givesTo justify the limiting value as well as its stability sign, use $c>0$, as appropriate to volume growth. Divide by $\dot{a}=cv/4$ and integrate the resulting first-order equation:Thus $q$ remains between two positive constants, its initial value and $g/B$. Since $d\sqrt{a}/dt=c\sqrt{q}/8$, the radius is defined for all increasing time and tends to infinity. The correction vanishes, proving the speed-radius attractor of an accreting raindrop:The strict inequality uses $k>0$. The growing radius prevents a finite terminal velocity; instead the acceleration approaches this positive constant.

For an attractive central force, the transverse acceleration vanishes. In polar coordinates,Thus the specific angular momentum $h=r^2\dot{\theta }$ is constant. Assume $h\ne 0$, so angle can parameterize the orbit, and put $u=1/r$. Primes denote angle derivatives. The chain rule givesThe outward radial component of the attractive force is $-f(u)$. Its Newton's second law equation is therefore $m(\ddot{r}-r{\dot{\theta }}^2)=-f(u)$, giving the Binet equationThe sign reflects $f$ being the positive inward force magnitude. Zero angular momentum produces radial motion and is outside this angular parametrization.

Substituting $y=x^p$ in the homogeneous ordinary differential equation gives $p(p-1)-2=(p-2)(p+1)=0$. ThusFor a Green's function decaying at the two ends, use $u(x)=x^2$ on the side of zero and $v(x)=x^{-1}$ on the side of infinity. Their Wronskian is $u{v}^{\prime }-u^{\prime }v=-3$. The function must be continuous at $x=\xi$, or its second derivative would contain a derivative of a Dirac delta. Its derivative jump must be one, to give the specified positive delta source. These two conditions yield the boundary-decaying Green kernel for an inverse-square differential operator:Its derivatives at the join are $-2/3$ on the left and $1/3$ on the right, confirming the jump one. Away from the join each branch solves the homogeneous equation, and both endpoint conditions hold.

Integrate this Green's function against the forcing, whose support is between zero and one. For $0<x\le 1$,For $x\ge 1$, only the right branch contributes. ThereforeOn the first branch, $y^{\prime }=(2x/3)\log x+x/9$ and $y^{\prime \prime }=(2/3)\log x+7/9$, so $y^{\prime \prime }-2y/x^2=1$. The second branch is a multiple of the homogeneous solution $x^{-1}$, so its forcing is zero. At $x=1$, both $y$ and $y^{\prime }$ agree, with values $-1/9$ and $1/9$, so there is no additional delta source. Finally $x^2\log x\to 0$ as $x\downarrow 0$ and $x^{-1}\to 0$ as $x\to \infty$. A homogeneous solution obeying both boundary conditions has $C=D=0$, establishing uniqueness.

Put $\theta =x-t$. Taking the gradient of the velocity potential givesThe motion is not generally incompressible; a velocity potential ensures irrotationality, not zero divergence. Both components have zero time average at each fixed point because their sine and cosine averages vanish.

The fluid acceleration is the material derivative $D\mathbf{u}/Dt={\partial }_t\mathbf{u}+(\mathbf{u}\cdot \nabla )\mathbf{u}$. Direct differentiation givesAveraging over one period at fixed $(x,y)$ therefore yieldsThis need not be the derivative of the mean velocity: the nonlinear advective term remains after averaging.

The dyed particle obeys the Lagrangian trajectory equationsExpand $x=ϵx_1+{ϵ}^2{x}_2+O({ϵ}^3)$ and $y=ϵy_1+{ϵ}^2{y}_2+O({ϵ}^3)$, for bounded times. The first-order equations give $x_1=0$, ${\dot{y}}_1=\cos t$, so $y_1=\sin t$. At second order,Integrating with zero initial values verifiesThe velocity averaged along this particle over $0\le t\le 2\pi$ is its displacement divided by $2\pi$. HenceThis Stokes drift is a Lagrangian mean; it does not contradict the zero Eulerian mean velocity at a fixed point. The secular term is a small-parameter expansion, not an assertion of uniform accuracy as $t\to \infty$.

The initial wavefunction has unit norm, since ${\int }_0^a(2/a){\sin }^2(\pi x/a)dx=1$. On its nonzero half, $-{\hslash }^2{\psi }^{\prime \prime }/(2m)=E_{\ast }\psi$ with $E_{\ast }={\hslash }^2{\pi }^2/(2m{a}^2)$. A small domain qualification is important: the derivative jumps at zero, so the state is not in the operator domain of the Dirichlet Hamiltonian operator. Its finite energy expectation value is nevertheless defined by the quadratic form of a positive quantum Hamiltonian. The energy form of a half-well sine state givesThis is also the rigorous weak interpretation of the printed $\int \psi H\psi$. Indeed the distributional second derivative is ${\psi }^{\prime \prime }=-(\pi /a{)}^2\psi +\sqrt{2/a}(\pi /a){\delta }_0$ inside the well. The Dirac delta term pairs to zero because $\psi (0)=0$. One should not instead conclude that this state is an energy eigenstate from its equation on just the positive half.

Its coefficients in the full-well energy eigenstates areFor even $n$, sine orthogonality gives $a_2=-1/\sqrt{2}$ and all other even coefficients zero. For $n=2p+1$, set $u=x/a$ and use $\sin [n\pi (u+1)/2]=(-1{)}^p\cos (n\pi u/2)$. Product-to-sum integration then givesSince these coefficients are $O(n^{-2})$, ${\sum }_n{E}_n|a_n{|}^2$ converges, confirming that the state has finite energy. With $E_n=n^2{E}_1$, the spectral expression isThe kinetic-energy quadratic form equals this spectral sum: expanding the weak derivative in the corresponding orthonormal cosine modes and using Parseval's identity gives $\int |{\psi }^{\prime }{|}^2={\sum }_n{k}_n^2|a_n{|}^2$. Equating it to the directly integrated value $4E_1$ provesIn contrast, ${\sum }_n{E}_n^2|a_n{|}^2$ diverges, consistently showing why $H\psi$ is not an ordinary square-integrable vector even though the energy expectation is finite.

Uniformly continuous. For $x\ge 1$, the derivative satisfiesThe mean value theorem therefore gives a Lipschitz condition with constant two, hence uniform continuity.

A function $f:I\to \mathbb{R}$ is uniformly continuous ifThe crucial point is that $\delta$ works throughout the interval, not just near one chosen point.

Yes, it is bounded on the closed unit interval. Choose $\delta$ for $\epsilon =1$, and choose finitely many grid points $t_j$ such that every $x\in [0,1]$ is within $\delta$ of some $t_j$. Then $|f(x)|\le 1+{\max }_j|f(t_j)|$, a finite bound. Equivalently, uniform continuity implies continuity, and a continuous real function on a compact set is bounded.

Yes, a bounded derivative gives uniform continuity on the half-line. If $|f^{\prime }|\le K$, the mean value theorem gives $|f(x)-f(y)|\le K|x-y|$. This is a Lipschitz condition, so for $K>0$ use $\delta =\epsilon /K$. If $K=0$, the function is constant. As usual for a derivative on this closed half-line, continuity at its endpoint is included.