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Three explicit functions separate the hypotheses from the conclusion. These are original mathematical illustrations, not measured data.

The polynomial function $h(x)=x^3-x=x(x-1)(x+1)$ is a continuous function on $[-3/2,3/2]$, with $h(-3/2)=-15/8$ and $h(3/2)=15/8$. The intermediate value theorem guarantees at least one zero of a function in the interior, while the polynomial factorization displays three: $-1,0,1$. Thus strict opposite signs do not establish uniqueness.

The everywhere-defined functionhas $j(-1)=-1$ and $j(1)=1$, yet no zero of a function. Its jump discontinuity at zero is exactly where the continuity hypothesis fails. A drawn vertical connector would be misleading: the function graph contains neither the intermediate heights nor a zero of a function at that jump. Even continuity only on the interior is insufficient: setting $q(x)=-1$ for $0\le x<1$ and $q(1)=1$ gives opposite endpoint signs on $[0,1]$ with no zero of a function, because the left one-sided limit at $1$ is not $q(1)$.

The polynomial function $t(x)=x^2$ on $[-1,1]$ is a continuous function, and $t(-1)=t(1)=1$, yet $t(0)=0$. Its root of a polynomial at zero has multiplicity of a root two. The function graph touches its horizontal tangent line there and stays positive on both sides: existence of a zero of a function does not require a sign change. More generally, near a real root of a polynomial $r$, a polynomial factorization $p(x)=(x-r{)}^{2m}u(x)$ with $m\ge 1$ and $u(r)\ne 0$ has the same sign on both sides sufficiently close to $r$, since the continuous function $u$ retains its sign. The intermediate value theorem does not rule out values outside the range between endpoint outputs.^[ref]

For the first-order ordinary differential equation $y^{\prime }=f(x,y)$ defined on an open subset $D$ of the plane, an isocline of slope $k$ is the level setHere $k$ is a constant. At every point of this isocline, the direction field prescribes a tangent vector proportional to $(1,k)$ for a solution of a differential equation. It does not generally prescribe the tangent line of the isocline itself. The level set can have several components or fail to be a smooth curve, and points outside $D$ are excluded.

For example, take $y^{\prime }=x-y$ on the whole plane. Its isoclines are the parallel straight linesEach line has slope $1$, while the direction field along it has slope $k$. Thus the isocline $y=x$ carries horizontal segments even though the line itself rises. A function graph of a solution of a differential equation follows the segments, rather than following this line. Only the isocline with $k=1$, namely $y=x-1$, is also a solution of a differential equation.^[ref]

The first-order linear differential equation has the exact familyThe initial condition $y(x_0)=y_0$ selects $A=(y_0-x_0+1)e^{x_0}$ and hence a unique solution of a differential equation for every real $x$. The picture shows $y(0)=-1,0,1,2$. For $A=0$ the solution of a differential equation stays on the $k=1$ isocline; for other $A$ it moves through different isoclines. These formulas check the pictured trajectories by direct differentiation, independently of how the direction field was sampled.

The zero-slope isocline $f(x,y)=0$ is called a nullcline. Below the nullcline $y=x$ in this example, $y^{\prime }>0$; above it, $y^{\prime }<0$. For $A>0$, the solution of a differential equation meets it at $x=\log A$, where $y^{\prime }=0$ and $y^{\prime \prime }=1$, a strict local minimum. The nullcline is therefore a useful guide to turning points, although it is not itself a solution of a differential equation. In the equivalent planar autonomous differential equation $\dot{x}=1$, $\dot{y}=x-y$, it is the nullcline for the second component; there is no equilibrium point, because the first component never vanishes.

This makes isoclines useful even when no explicit solution of a differential equation is available: organize a direction field by common slopes, read where function graphs rise or fall, and identify possible turning points on nullclines.

A linear approximation replaces a differentiable function near a reference input by its value plus the first-order change supplied by its derivative. In one dimension, this is the tangent line $f(a)+f^{\prime }(a)(x-a)$. The chain rule combines such local approximations through function composition; it does not assert their equality to the original functions over a finite interval.

Let $F$ be twice continuously differentiable on an interval around a zero of a function $t_{\ast }$, with $|F^{\prime }(t)|\ge m>0$ and $|F^{\prime \prime }(t)|\le M$ there. Choose a closed interval $[t_{\ast }-\rho ,t_{\ast }+\rho ]$ inside that interval with $M\rho /(2m)<1$. Starting inside it makes the Newton root-finding iteration well defined, keeps it inside, and gives the displayed quadratic convergence bound for $e_r=t_r-t_{\ast }$. The nonzero derivative and sufficiently close start are essential hypotheses. The scalar estimate follows from the Taylor remainder argument in MIT's root-finding notes.^[ref]

For a short proof outline, apply Taylor's theorem between $t_r$ and $t_{\ast }$; subtract the defining tangent line equation. The linear error cancels, leaving $e_{r+1}=F^{\prime \prime }({\xi }_r)e_r^2/(2F^{\prime }(t_r))$. The displayed bound shrinks the error and preserves the chosen interval. If $F^{\prime \prime }(t_{\ast })\ne 0$, the error ratio tends to $|F^{\prime \prime }(t_{\ast })/(2F^{\prime }(t_{\ast }))|$; if it vanishes, convergence can be faster.

The Mach angle of a supersonic flow with Mach number $M>1$ is $\mu =\arcsin (M^{-1})$. Infinitesimal pressure disturbances lie on the corresponding Mach cone, and a weak oblique shock approaches this angle as its strength tends to zero.

The Mach angle is the half-angle between a Mach cone generator and its backward motion axis for a steady point source in a uniform three-dimensional medium with isotropy and constant sound speed. It obeysThe relation concerns weak acoustic waves in linear acoustics. It is not the angle of a finite oblique shock produced by finite turning of the fluid flow; that angle also depends on the deflection and the gas properties.^[ref]

For a short proof, choose an old emission in the moving-point acoustic wavefronts construction and set $s=-\tau >0$. Its circle has center $(-Us,0)$ and radius $cs$. A tangent line from the present source at the origin meets the circle at a right angle to its radius. The resulting right triangle has hypotenuse $Us$ and opposite side $cs$, giving the displayed sine ratio. The emission age cancels: every wavefront has the same tangent directions, so a continuum of expanding spheres produces one simple Mach cone.

In an original example, stipulate uniform gas with sound speed $c=340\mathrm{m}{\mathrm{s}}^{-1}$ and a source moving through it at $U=680\mathrm{m}{\mathrm{s}}^{-1}$. Then $M=2$ and $\mu =\arcsin (1/2)=30^{\circ }$. An emission one second earlier is now a sphere centered $680\mathrm{m}$ behind the source with radius $340\mathrm{m}$. The selected sound speed defines this example; it is not a constant for air at every temperature. The Mach angle decreases as $M$ increases: its derivative is $-1/(M\sqrt{M^2-1})$ in radians, and its limits are $90^{\circ }$ as $M\downarrow 1$ and $0$ as $M\to \infty$.

The mean value theorem says that an average rate of change is attained as an instantaneous rate somewhere between the endpoints. Let $a<b$ and let the real-valued function $f:[a,b]\to \mathbb{R}$ be continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. Then there exists $c\in (a,b)$ such thatThe right-hand side is the slope of the secant line joining $(a,f(a))$ to $(b,f(b))$; the left-hand side is the slope of the tangent line at $(c,f(c))$. The derivative need not be continuous, and the function need not be differentiable at the endpoints. ^[ref]

For the polynomial $f(x)=x^2$ on $[1,3]$, the secant line has slope $(9-1)/(3-1)=4$. Since the derivative is $f^{\prime }(x)=2x$, the required interior point is $c=2$. The tangent line there is $y=4+4(x-2)=4x-4$, whereas the secant line is $y=1+4(x-1)=4x-3$. Their equal slopes make them parallel, but they are distinct straight lines: the theorem equates rates, not heights.

The practical strength is that the unknown point $c$ can disappear from an estimate. Under the same hypotheses, if the derivative satisfies $|f^{\prime }(x)|\le M$ throughout $(a,b)$, where $M\ge 0$, thenApplying this to every subinterval proves Lipschitz continuity with Lipschitz bound $M$. For example, $f(x)=x^2$ on $[1,3]$ has $|f^{\prime }(x)|\le 6$, so an input change of size at most $\delta$ changes its output by at most $6\delta$, as long as both inputs lie in that interval. A local derivative bound therefore controls finite changes without computing an integral. The sign of the derivative similarly gives the derivative criterion for monotonicity.

Taking $M=0$ shows that a function with zero derivative throughout a real interval is a constant function, provided it is continuous at any included endpoints. The interval condition is essential: on the domain of a function $(-2,-1)\cup (1,2)$, the function equal to zero on the left interval and one on the right interval has zero derivative everywhere. It is a locally constant function, yet not a constant function on the whole domain of a function. The mean value theorem cannot cross the gap.

A concise proof subtracts the secant line: set $m=(f(b)-f(a))/(b-a)$ and $h(x)=f(x)-f(a)-m(x-a)$. Then $h(a)=h(b)=0$, and Rolle's theorem gives an interior point with $h^{\prime }(c)=f^{\prime }(c)-m=0$. The geometric idea is to remove the average trend and find a horizontal tangent line in what remains.

The Newton root-finding iteration seeks a zero of a function $F$ by repeatedly solving its tangent line approximation. At a current guess $t_r$ with derivative $F^{\prime }(t_r)\ne 0$, setGeometrically, follow the tangent line to the horizontal axis, then return to the function graph for the next tangent line. Close to a suitable zero of a function, the error becomes dramatically smaller.^[ref]

For a concrete square root calculation, take the polynomial $F(t)=t^2-2$ and start at $t_0=1$. The Newton root-finding iteration becomes $t_{r+1}=(t_r+2/t_r)/2$, givingThe errors use the unrounded iterates. After the first step, each iterate overestimates $\sqrt{2}$, and the exact identityexhibits quadratic convergence: once close, the number of correct decimal digits roughly doubles per step. This speed makes the Newton root-finding iteration useful when accurate derivatives are available.

Global convergence requires additional care. For the polynomial $F(x)=x^3-2x+2$, starting at $x_0=0$ gives $F(0)=2$, $F^{\prime }(0)=-2$, hence $x_1=1$. Then $F(1)=1$, $F^{\prime }(1)=1$ gives $x_2=0$. The sequence repeats $0,1,0,1,\dots$ forever, although a real polynomial root lies in $(-2,-1)$ by the intermediate value theorem. Neither point in the cycle is a zero of a function, and neither has zero derivative. A maintained sign-changing interval and fallback bisection method can protect a root solver from unsuitable Newton steps.

Newton's law of cooling models an object's temperature $y$ using a positive constant rate $k$ and ambient temperature $T_a$. The displayed ordinary differential equation assumes one temperature represents the object, the environment remains at $T_a$, and the proportional rate stays fixed over the modeled range. It also describes warming when $y<T_a$. These are modeling assumptions: ^[ref] explain the proportional law through a tangent line approximation and caution about large temperature differences.

Dimensional analysis requires $k$ to have units of inverse time. The product $k(t-t_0)$ in the exponential function solution is therefore dimensionless. The equilibrium point $y=T_a$ attracts both warmer and colder starting states: the excess temperature undergoes exponential decay. See initial value problem for computed cooling times, direction field for the family of trajectories, and separable differential equation for a derivation.

At a regular point of a differentiable plane curve, the normal line passes through that point perpendicular to its tangent line. If the tangent slope is a finite nonzero number $m$, the normal slope is $-1/m$, since the direction vectors $(1,m)$ and $(1,-1/m)$ have zero dot product. A horizontal tangent has a vertical normal, and a vertical tangent has a horizontal normal. For a regular level set $F(x,y)=c$, the nonzero gradient $\nabla F$ gives a normal direction because differentiating $F$ along the curve gives $\nabla F\cdot {\mathbf{r}}^{\prime }=0$.

At a point of a twice continuously differentiable regular curve with positive curvature $\kappa$, its osculating circle has radius $1/\kappa$ and center $c=r+N_p/\kappa$, where $N_p$ is the principal normal vector. It matches the point, tangent line and curvature. For a space curve, it lies in the plane spanned by the unit tangent vector and principal normal vector. The parabola example under curvature of a plane curve gives center $(0,1/2)$ and radius $1/2$ at the origin.

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.

Let $r_A=|\mathbf{P}-\mathbf{A}|$, $r_B=|\mathbf{P}-\mathbf{B}|$, and define the outward-from-focus unit vectors ${\mathbf{u}}_A=(\mathbf{P}-\mathbf{A})/r_A$ and ${\mathbf{u}}_B=(\mathbf{P}-\mathbf{B})/r_B$. The ellipse is the nondegenerate level curve $r_A+r_B=s$, with $s>|\mathbf{A}-\mathbf{B}|$. The previous gradient result givesIf $\mathbf{t}$ is a unit tangent at $P$, orthogonality to the level-curve gradient impliesTherefore $|\mathbf{t}\cdot {\mathbf{u}}_A|=|\mathbf{t}\cdot {\mathbf{u}}_B|$. These absolute dot products are the cosines of the acute angles made by the two focal lines with the unoriented tangent line. Thus those angles are equal.

Equivalently, the outward unit normal is $\mathbf{n}=({\mathbf{u}}_A+{\mathbf{u}}_B)/|{\mathbf{u}}_A+{\mathbf{u}}_B|$, the bisector of the two unit focal directions. Its denominator is nonzero for a nondegenerate ellipse. Using $|{\mathbf{u}}_A|=|{\mathbf{u}}_B|=1$, the specular reflection formula becomesThis algebra proves the equal-angle statement with the incident and reflected ray directions correctly oriented, rather than confusing the line $BP$ with the outgoing direction $PB$.

The arc length of the helix follows from its parametrization:Thus ${\ell }_h=15\pi \sqrt{5}\mathrm{m}$. Its endpoint is $(0,20,15\pi )$, at height $h=15\pi \mathrm{m}$.

Under the usual intended interpretation that the descending quarter of a circle drops through one radius, its radius is $R=h$. Its arc length is ${\ell }_c=\pi R/2=15{\pi }^2/2\mathrm{m}$, givingThe printed smoothness condition is inconsistent. The terminal tangent vector of the helix is ${\mathbf{r}}^{\prime }(3\pi )=(-10,0,5)$. Every tangent vector to a curve confined to the $yz$ plane has zero $x$ component, so no such regular curve can join this helix with a common tangent line. Moreover, the helix is still rising at that endpoint. Thus no smooth ride satisfying all the literal conditions exists.

Even after allowing a corner, a quarter of a circle reaching the ground does not alone determine its radius: the orientation of its endpoint radii must also be specified. For example, in the $yz$ plane, the circular arctraversed toward decreasing $\theta$, descends monotonically for $0\le {\theta }_0\le \pi /2$ and reaches $z=0$ when $R=h/(\sin {\theta }_0+\cos {\theta }_0)$. Choosing ${\theta }_0=\pi /4$ gives $R=h/\sqrt{2}$, whereas ${\theta }_0=\pi /2$ gives $R=h$. The numerical answer above is conditional on the intended $R=h$ geometry; it does not repair the impossible smooth join.

Let $F$ define the nonsingular plane cubic and let $H_F=\det \mathrm{Hess}F$ be its Hessian curve of a plane cubic equation. The Hessian criterion for a flex applies in the allowed field characteristic. Here is the local calculation, so no characteristic-zero argument is needed.

Move any point $P$ to $[0:1:0]$ and its tangent line to $Z=0$. Smoothness gives a nonzero coefficient $g$ of $Y^{2}Z$, while the coefficients of $Y^3$ and $X{Y}^2$ vanish. ThusThe intersection multiplicity with the tangent line is at least three precisely when $b=0$, which is precisely $H_F(P)=0$ because the field characteristic is not two. The tangent cannot be a component of a nonsingular plane cubic.

If $H_F$ vanishes identically, every point is a flex by this calculation. Otherwise it is a degree-three homogeneous polynomial. Its zero locus meets the cubic by Bézout's theorem; if they share a component, that also supplies a point of intersection. Every such point is smooth and satisfies the criterion, so a flex exists.

Choose a flex as $O=[0:1:0]$, with tangent line $Z=0$. The flex coordinates for a smooth plane cubic giveThe coefficient $a$ is nonzero because $Z=0$ is not a component, and $b\ne 0$ expresses smoothness at $O$. Since the field characteristic is not two, the projective linear changecompletes the square. The equation becomes $Y^{\prime 2}Z=P_3(X,Z)$ after dividing by a nonzero constant, where $P_3$ is a degree-three homogeneous polynomial with nonzero $X^3$ coefficient.

On $Z=1$, the polynomial $P_3(x,1)$ has three distinct roots $r_1,r_2,r_3$ in the algebraically closed field. A repeated root $r$ would make $(r,0)$ a singular point of $y^{\prime 2}=P_3(x,1)$, since both first partial derivatives vanish there. PutA further nonzero scaling of $y^{\prime }$ absorbs the leading coefficient and $(r_2-r_1{)}^3$; the required square root exists in the algebraically closed field. These are projective linear changes of the original coordinates. Renaming the new coordinates gives the Legendre form of an elliptic curveIts projective completion is $Y^{2}Z=X(X-Z)(X-\lambda Z)$, with the chosen flex at infinity.

This is a separable differential equation. On the solution with $y<1$, integration from the initial condition givesThe right side lies in $[0,1)$, so taking its sine stays on the required branch. The solution isNear the origin, $1-\sqrt{1-x^2}=x^2/2+O(x^4)$, so $y=x^2/2+O(x^4)$ and the tangent line is horizontal. At the other end, $y\to \sin 1$, whileThus the curve has a vertical limiting tangent line as $x\uparrow 1$, even though $x=1$ is excluded from the domain.

For the direction field, the slope is nonnegative everywhere in the square. The segments are horizontal on $x=0$ and on $y=1$, become steeper as $x$ increases at fixed $y<1$, and become flatter as $y$ increases at fixed $x$. Near $x=1$ they are almost vertical unless $y$ is correspondingly close to one. The excluded corner $(1,1)$ has no single limiting slope: the ratio $(1-y^2)/(1-x^2)$ depends on the path of approach.

Let $O=(0:1:0)$ be the point at infinity of the Weierstrass equation of an elliptic curve. It is the identity element for the elliptic curve group law. For an affine rational point $P=(x,y)$, the inverse isIndeed, the two affine intersections with the vertical line have these ordinates, since their sum as roots of the quadratic in $y$ is $-a_{1}x-a_3$. This reflection is valid for a generalized Weierstrass equation of an elliptic curve in every characteristic of a field.

For distinct $P,Q$, draw their line; for $P=Q$, draw the tangent line. Let $R$ be its third intersection with the elliptic curve, counted with intersection multiplicity. Then the sum is the reflection of the third intersection:A vertical line has third intersection $O$, so $P+(-P)=O$. If a tangent has third intersection $O$, then $2P=O$. Define addition with $O$ by $P+O=P$; these conventions include all exceptional cases of the affine construction.

The construction is defined over the coefficient field: after two known intersections, the remaining root is rational over that field. It gives a commutative group operation. To see why it is associative, use the elliptic curve group law from Riemann-Roch: the map $P\mapsto [P-O]$ identifies the elliptic curve with its degree-zero divisor class group. A line has intersection divisor $P+Q+R$ linearly equivalent to $3O$, the divisor of the line at infinity. Thus $[P-O]+[Q-O]+[R-O]=0$. The geometric sum agrees with addition of divisor classes, which is associative.

Here the inverse in the elliptic curve group law is $(x,y)\mapsto (x,-y-1)$. For $P_0+P_1$, the line is $y=2-2x$. Substitution in the Weierstrass equation of an elliptic curve givesThe third intersection is $(3,-4)$, so reflection givesFor the difference, first take $-P_1=(1,-1)$. The line through $P_0$ and $-P_1$ is $y=2-3x$, and now the intersection polynomial is $x(x-1)(x-8)$. Its third point is $(8,-22)$, givingFinally, implicit differentiation gives $(2y+1)y^{\prime }=3x^2-7$. The tangent line at $P_1$ therefore has slope $-4$ and equation $y=4-4x$. Its intersection polynomial factors as $(x-1{)}^2(x-14)$. The double root records the intersection multiplicity of the tangent, and the remaining intersection is $(14,-52)$. Hence

Work near the identity $O$ of a generalized Weierstrass equation of an elliptic curve. On the projective chart where $Y\ne 0$, use the local parameter $t=-x/y$ and the auxiliary coordinate $w=-1/y$. Substitution givesThe coefficient of $w$ in the linearized equation at $(0,0)$ is one. Consequently, successive coefficient comparison determines a unique formal power series $w(t)$, without dividing by any coefficient of the equation. Its initial terms areRecover the affine coordinates as Laurent series $x=t/w(t)$ and $y=-1/w(t)$; in particular $x=t^{-2}+O(t^{-1})$ and $y=-t^{-3}+O(t^{-2})$.

The elliptic curve group law is a regular morphism near $(O,O)$. Applying the local parameter to the sum and passing to completed local rings producesOne can construct the series without any nonunit division. PutThe divided difference is an integral formal power series, including its tangent line specialization at $T_1=T_2$. The line $w=\lambda t+\nu$ meets the cubic in the two chosen points and a third point $R$. Substitution in the equation shows that its cubic and quadratic coefficients in $t$ are $-A,-B$, whereSince $\lambda$ has total degree at least two and $\nu$ at least three, $A$ is a unit. The sum of the roots therefore givesThe last expression reflects the third intersection according to the chord-and-tangent group law. Both denominators have constant term one. This proves integrality of the universal construction and also provides an explicit coefficient algorithm. Expanding it, the resulting formal group of an elliptic curve beginsThe integral coefficient recursion and the unit-denominator addition calculation work universally over $\mathbb{Z}[a_1,a_2,a_3,a_4,a_6]$; thus these coefficients specialize in every characteristic of a field. This is the initial coefficients of the elliptic formal group at infinity construction, rather than a definition requiring division by the elliptic-curve discriminant.

The identity, commutativity and associativity of the elliptic curve group law give the defining identities of a one-dimensional commutative formal group law:Reflection in the generalized Weierstrass equation of an elliptic curve gives its formal inverse explicitly:Successive addition gives the multiplication series $[n{]}_F(T)=nT+O(T^2)$.

In characteristic zero, an especially useful description comes from the invariant differential on an elliptic curve:Integrating with zero constant term constructs the formal logarithmTranslation invariance of the invariant differential on an elliptic curve implies $L(F(X,Y))=L(X)+L(Y)$. Thus the formal logarithm converts the formal group law into the formal additive group in characteristic zero. Expanding this identity to degree three also recovers the displayed coefficients of $F$; in positive characteristic the integral addition construction remains valid although the logarithmic denominators may not exist.

The unit vectors of plane polar coordinates obey ${\dot{\mathbf{e}}}_r=\dot{\theta }{\mathbf{e}}_{\theta }$ and ${\dot{\mathbf{e}}}_{\theta }=-\dot{\theta }{\mathbf{e}}_r$, as follows by differentiating their Cartesian sine/cosine components. Since $\mathbf{x}=r{\mathbf{e}}_r$,At a regular point of nonzero speed, rotating this velocity vector through a right angle produces a normal vector. Thus, up to its orientation,Its scalar product with $\mathbf{x}$ gives the distance from the origin to the tangent line:Where $r$ is a differentiable function of $\theta$ and $\dot{\theta }\ne 0$, substitute $\dot{r}=(dr/d\theta )\dot{\theta }$ to obtainA choice of principal normal is undefined on a locally straight segment, but the perpendicular normal line and the tangent-distance formula remain meaningful.

For a central force, the torque vanishes. Use physical signed angular momentum $h=m{r}^2\dot{\theta }$, with $h\ne 0$. The preceding geometric formula implies $|h|=mpv$, where $v=|\dot{\mathbf{x}}|$. Conservation of energy gives $E=m{v}^2/2+V(r)$. Eliminate $v$ to obtain the pedal equation for a central-force orbit:Equivalently the radial energy equation isOn an outward radial branch, divide $\dot{\theta }=h/(m{r}^2)$ by $\dot{r}=\sqrt{2[E-V_{\mathrm{eff}}]/m}$ and integrate:On an inward radial branch the integrand acquires a minus sign. At a turning point the two branches must be joined; the quoted positive-root integral is a local branch representation, not a single global time-parametrized orbit.

For $V=c/r^2$ with $c>0$, defineA finite moving orbit has $E>0$. In the radial integral, set $s=r_{\min }/r$. Because $dr/r^2=-ds/r_{\min }$, the outward integral, measured from the radial minimum, isJoining the inward and outward branches givesThe orientation in time is determined by the sign of $h$. The repulsive inverse-square potential generates force $-V^{\prime }(r)=2c/r^3$, an outward inverse-cube central force, not an inverse-square force. The orbit approaches infinity along two asymptotic directions, turns once at $r_{\min }$ and escapes again. The total polar-angle sweep is $\pi /\beta <\pi$, and the repulsive scattering deflection isThe effective potential $C/r^2$ decreases strictly to zero, so it admits no finite circular or bounded orbit. The sketch shows this scattering in a repulsive inverse-square potential. In the radial case $h=0$, the formulas dividing by $h$ are inapplicable; the direction is fixed, the turning radius is $\sqrt{c/E}$ and the particle reverses on the same ray.