Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generated
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Differentiation of sums, scalar multiples, inner products, cross products and squared norms of differentiable curves in R3

Statement

Let I⊆R be an interval with at least two points (Intervals of R: the nine order-convex forms, nondegeneracy, and length), let t0∈I, let x,y:I→R3 be differentiable at t0 (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral) and let λ:I→R be differentiable at t0 (The derivative f′(c)=lim⁡x→cf(x)−f(c)x−c of f:A→R at a point c∈A that is a limit point of A, and differentiability on a set). Write x′(t0),y′(t0)∈R3 for the derivatives, ⟨⋅,⋅⟩ for the Euclidean inner product and ∥⋅∥2 for the Euclidean norm (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn), and × for the cross product (The cross product in R3). Then:

  1. x+y and λx are differentiable at t0, with (x+y)′(t0)=x′(t0)+y′(t0) and (λx)′(t0)=λ′(t0)x(t0)+λ(t0)x′(t0);
  2. t↦⟨x(t),y(t)⟩ is differentiable at t0, with derivative ⟨x′(t0),y(t0)⟩+⟨x(t0),y′(t0)⟩;
  3. t↦x(t)×y(t) is differentiable at t0, with derivative x′(t0)×y(t0)+x(t0)×y′(t0);
  4. t↦∥x(t)∥22 is differentiable at t0, with derivative 2⟨x′(t0),x(t0)⟩.

Facts & Assumptions

Given: An interval I⊆R with at least two points, a point t0∈I, curves x,y:I→R3 and a function λ:I→R, with x,y,λ differentiable at t0.

[L1]

A function f:I→Rm is differentiable at t0 if and only if every component fk is differentiable at t0 as a real function on I, and then f′(t0)k=fk′(t0); moreover sums and scalar multiples of differentiable curves are differentiable with (f+g)′(t0)=f′(t0)+g′(t0) and (μf)′(t0)=μf′(t0) for μ∈R (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral).

[L3]

The Euclidean inner product and norm are ⟨u,w⟩=∑k<3ukwk and ∥u∥2=⟨u,u⟩; the inner product is bilinear and symmetric (The Euclidean inner product ⟨x,y⟩=∑k<nxkyk on Rn, Laws of finite sums and finite products).

[L4]

For all u,v∈R3 the cross product is bilinear-alternating: (u,v)↦u×v is bilinear, u×u=0, and u×v=−(v×u); in coordinates (u×v)0=u1v2−u2v1, (u×v)1=u2v0−u0v2, (u×v)2=u0v1−u1v0 (The cross product in R3, The cross product is bilinear, alternating, and orthogonal to both factors).

Proof

technique · direct
1.1givenL1L2

By [L1] the components xk,yk:I→R are differentiable at t0 with xk′(t0)=x′(t0)k and yk′(t0)=y′(t0)k, and by [L2] each of the functions xk+yk, λxk, xkyk is differentiable at t0 with the sum, scalar-multiple and product rules; this proves clause 1 componentwise, since (x+y)k=xk+yk and (λx)k=λxk.

2.1step 1.1L2L3

The scalar function t↦⟨x(t),y(t)⟩=∑k<3xk(t)yk(t) is a finite sum of products of functions differentiable at t0, hence by [L2] differentiable at t0 with derivative ∑k<3(xk′(t0)yk(t0)+xk(t0)yk′(t0)); by [L3] this is ⟨x′(t0),y(t0)⟩+⟨x(t0),y′(t0)⟩, which is clause 2.

2.2step 1.1L1L2L4

Each component of t↦x(t)×y(t) is, by [L4], one of the differences x1y2−x2y1, x2y0−x0y2, x0y1−x1y0 of products of functions differentiable at t0, so [L2] makes it differentiable at t0 with derivative (x1′y2+x1y2′)−(x2′y1+x2y1′), and by [L4] and bilinearity this is the corresponding component of x′(t0)×y(t0)+x(t0)×y′(t0); reassembling the three components with [L1] gives clause 3.

3.1step 2.1L3algebra∎

Applying clause 2 at y:=x gives differentiability of t↦⟨x(t),x(t)⟩=∥x(t)∥22 with derivative ⟨x′(t0),x(t0)⟩+⟨x(t0),x′(t0)⟩=2⟨x′(t0),x(t0)⟩, the two terms being equal by symmetry of the inner product [L3]; this is clause 4.

Depends on

Dependency tree · 2 levels

Sources