Euclidean geometry and space curves

A trajectory is more than three coordinate graphs: velocity, tangent direction and curvature describe its geometry independently of a particular drawing. Vectors also express lines, planes and distances without separate formulas for vertical or horizontal special cases. This chapter connects those geometric objects to differential calculus.

1 Dot and cross products

For $u,v\in\mathbb R^3$, $u\cdot v=\sum u_iv_i$ and $\|u\|=\sqrt{u\cdot u}$. Cauchy–Schwarz gives $|u\cdot v|\le\|u\|\|v\|$, so for nonzero vectors the angle satisfies $\cos\theta=(u\cdot v)/(\|u\|\|v\|)$. Orthogonality means dot product zero.

The cross product is

$$u\times v=(u_2v_3-u_3v_2,\ u_3v_1-u_1v_3,\ u_1v_2-u_2v_1).$$

It is perpendicular to both inputs and follows the right-hand orientation convention. Expanding coordinates proves $\|u\times v\|^2=\|u\|^2\|v\|^2-(u\cdot v)^2$, hence $\|u\times v\|=\|u\|\|v\|\sin\theta$, the parallelogram area. The scalar triple product $u\cdot(v\times w)$ is signed parallelepiped volume; its absolute value is geometric volume.

A line through $p$ with nonzero direction $v$ is $p+tv$. A plane through $p$ with nonzero normal $n$ is $n\cdot(x-p)=0$. Decomposing $q-p$ into components parallel and perpendicular to $n$ gives the point-to-plane distance

$$\frac{|n\cdot(q-p)|}{\|n\|}.$$

The distance to the line is $\|(q-p)\times v\|/\|v\|$, from the perpendicular component or the triangle-area identity.

2 Velocity arc length and regularity

A differentiable curve is $r:I\to\mathbb R^3$. Its velocity is $r'(t)$ and speed is $v(t)=\|r'(t)\|$. It is regular where speed is nonzero. For a $C^1$ curve, arc length from $a$ to $b$ is

$$L=\int_a^b\|r'(t)\|dt.$$

This formula is the limit of polygonal lengths: uniform continuity of $r'$ makes each short chord differ from its tangent displacement by an error bounded by interval length times the oscillation of $r'$, and the summed errors vanish with partition mesh.

For regular $C^2$ curves, the unit tangent is $T=r'/\|r'\|$. Arc length $s$ satisfies $ds/dt=v>0$, so the chain rule gives $dT/ds=T'/v$. The curvature is $\kappa=\|dT/ds\|$.

For a regular $C^2$ curve,

$$\kappa=\frac{\|r'\times r''\|}{\|r'\|^3}.$$

Proof. Write $r'=vT$. Then $r''=v'T+vT'$, and $T\cdot T'=0$ follows by differentiating $T\cdot T=1$. Therefore $\|r'\times r''\|=v^2\|T'\|=v^3\kappa$. Divide by $v^3$. $\square$

Zero curvature can occur for a regular straight line. Zero speed is different: the curvature formula is not defined there without a separate geometric extension.

3 Worked helix and parameter effects

For $r(t)=(a\cos t,a\sin t,bt)$ with $a>0$, velocity is $(-a\sin t,a\cos t,b)$ and speed is $\sqrt{a^2+b^2}$. One turn has length $2\pi\sqrt{a^2+b^2}$. Acceleration is $(-a\cos t,-a\sin t,0)$ and the cross-product magnitude is $a\sqrt{a^2+b^2}$. Thus

$$\kappa=\frac{a}{a^2+b^2}.$$

Changing parameter to $t=2u$ doubles velocity and quadruples acceleration, but leaves curvature unchanged: the numerator scales by $2^3$ and the denominator by $2^3$. Curvature is geometric; speed depends on timing. If coordinates have length units, curvature has inverse-length units.

The curve $r(t)=(t^3,0,0)$ traces a straight line but has zero velocity at zero. The path itself is simple; that parameter is not regular at the origin. A formula's denominator can reveal a parameter failure rather than a geometric corner.

The Cauchy–Schwarz inequality used in the geometry

If $v=0$, $|u\cdot v|\le\|u\|\|v\|$ is equality. If $v\ne0$, expand the nonnegative square at $t=(u\cdot v)/\|v\|^2$ to obtain $0\le\|u-tv\|^2=\|u\|^2-(u\cdot v)^2/\|v\|^2$. Multiply by $\|v\|^2>0$ and take square roots. Equality holds precisely when $u=tv$, hence when the two vectors are linearly dependent, including either zero-vector case. This proves the inequality used to define angles and estimate projection lengths.

Full proof of the length formula on a compact interval

Assume $a<b$ and $r\in C^1([a,b],\mathbb R^n)$. The following definition and proof allow zero speed.

1. Define length before replacing it by an integral

For $r:[a,b]\to\mathbb R^n$, define its length as the supremum over finite partitions $a=t_0<\cdots<t_m=b$ of the polygonal sums $\sum_i\|r(t_i)-r(t_{i-1})\|$. The supremum allows arbitrarily fine polygonal approximations; one endpoint chord is generally only a lower bound.

Theorem: $C^1$ length formula

For a $C^1$ curve on a compact interval,

$$L(r)=\int_a^b\|r'(t)\|\,dt.$$

Upper bound. On a subinterval, the vector fundamental theorem gives displacement $d=\int r'(t)dt$. If $d\ne0$, dot with its unit direction to obtain $\|d\|\le\int\|r'(t)\|dt$; if $d=0$, that inequality is immediate. Summing over any partition gives polygonal length at most the speed integral.

Lower bound. Uniform continuity of $r'$ says that for any $\eta>0$, a sufficiently fine partition makes $\|r'(t)-r'(\xi_i)\|<\eta$ on every subinterval, where $\xi_i$ is any chosen tag there. Thus its displacement differs from $r'(\xi_i)\Delta t_i$ by a vector of norm at most $\eta\Delta t_i$. The reverse triangle inequality gives chord length at least $\|r'(\xi_i)\|\Delta t_i-\eta\Delta t_i$. The speed itself varies by at most $\eta$ there, so its integral differs from that tagged value by at most $\eta\Delta t_i$. Summing yields polygonal length at least the speed integral minus $2\eta(b-a)$. Since arbitrarily small $\eta$ are possible, the supremum reaches the integral. This proves both directions rather than only the familiar chord upper bound.

Regularity is not needed for this formula: a $C^1$ curve may stop. Regularity becomes essential when we invert arc length to use it as a coordinate.

4 Graded practice

E1 Foundation. Find a plane through $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$, and its distance from the origin.

E2 Geometry. Find the distance from $(1,2,3)$ to the line $t(1,0,1)$.

E3 Curves. For $r(t)=(3\cos t,3\sin t,4t)$, find speed, one-turn length and curvature.

E4 Proof. For a regular $C^2$ curve, show acceleration decomposes as $r''=v'T+v^2\kappa N$ when $\kappa>0$, with $N=T'/\|T'\|$.

5 Full solutions

S1. A normal is $(1,1,1)$ and the plane is $x+y+z=1$. Its distance from zero is $1/\sqrt3$.

S2. Projection of $q$ onto the direction is $(q\cdot v)/(v\cdot v)=4/2=2$. The closest point is $(2,0,2)$; residual $(-1,2,1)$ has norm $\sqrt6$. The cross-product formula gives the same result.

S3. Speed is 5, length is $10\pi$, and curvature is $3/25$. Constant speed does not mean zero acceleration: the tangent direction changes.

S4. Differentiate $r'=vT$ to get $v'T+vT'$. By definition $\|T'\|=v\kappa$ and $T'=v\kappa N$, giving the formula. Tangential acceleration changes speed; the normal component changes direction. When $\kappa=0$, $N$ need not be defined, though the vector $vT'$ still is.

Summary

Dot and cross products encode angles, projections, areas and volumes. Differentiating a regular curve gives speed and tangent geometry; differentiating with respect to arc length gives curvature. Keep geometric properties distinct from parameter-dependent quantities and check nonzero-speed assumptions.