Curves & Motion

Study Sheet

Curves & Motion

Vector-valued functions, speed, arc length, curvature

Vector-Valued Functions

Tip
Differentiate coordinate by coordinate

A path r(t)=(x(t),y(t),z(t))\mathbf r(t) = (x(t), y(t), z(t)) has velocity r′(t)\mathbf r'(t) (tangent to the path) and speed ∣r′(t)∣|\mathbf r'(t)|. Arc length integrates the speed: L=∫ab∣r′(t)∣ dtL = \int_a^b |\mathbf r'(t)|\,dt. The helix (cos⁡t,sin⁡t,t)(\cos t, \sin t, t) has CONSTANT speed 2\sqrt2 — its length is just 2\sqrt2 times the time elapsed.

Concept
Curvature measures how hard you turn

For a graph y=f(x)y = f(x): κ=∣y′′∣(1+y′2)3/2\kappa = \dfrac{|y''|}{(1 + y'^2)^{3/2}}. A circle of radius RR has constant curvature 1R\tfrac1R — so 1κ\tfrac1\kappa is the radius of the circle that best hugs the curve. The parabola y=x2y = x^2 turns hardest at its vertex, where κ=2\kappa = 2.

Example
Projectile motion, decomposed

For r(t)=(3t,  4t−5t2)\mathbf r(t) = (3t,\; 4t - 5t^2): velocity (3,4−10t)(3, 4-10t), acceleration (0,−10)(0, -10). The top of the arc is where velocity is horizontal — perpendicular to gravity — at t=25t = \tfrac25. Physics questions become dot-product questions.

Side note
Significance: kinematics is vector calculus

GPS arc lengths, roller-coaster g-forces (curvature times speed squared), and orbital mechanics are all r(t)\mathbf r(t) computations. When engineers design a highway transition curve they are literally prescribing a curvature function — the "clothoid" ramps curvature linearly so the steering wheel turns at constant speed.

Try it
Try it: speed vs. velocity

A particle follows r(t)=(3cos⁡t,3sin⁡t)\mathbf r(t) = (3\cos t, 3\sin t). Work: velocity (−3sin⁡t,3cos⁡t)(-3\sin t, 3\cos t) — always changing direction; speed 9sin⁡2t+9cos⁡2t=3\sqrt{9\sin^2t + 9\cos^2t} = 3 — never changing at all. Uniform circular motion in one line: constant speed, perpetually turning velocity, acceleration pointing inward with magnitude v2r=3\tfrac{v^2}{r} = 3.

Proofs & Why It Matters

Tip
Proof: arc length = ∫|r′(t)| dt

Chop [a,b][a, b] into small steps Δt\Delta t. Each step moves the point by r(t+Δt)−r(t)≈r′(t) Δt\mathbf r(t + \Delta t) - \mathbf r(t) \approx \mathbf r'(t)\,\Delta t, a straight segment of length ∣r′(t)∣Δt|\mathbf r'(t)|\Delta t (Pythagoras applied to the coordinate increments).

Reminder — The Pythagorean theorem:  a2+b2=c2a^2+b^2=c^2

Summing the segments gives a Riemann sum ∑∣r′(ti)∣ Δt\sum |\mathbf r'(t_i)|\,\Delta t, and the limit as Δt→0\Delta t \to 0 is by definition ∫ab∣r′(t)∣ dt\int_a^b |\mathbf r'(t)|\,dt. ■\blacksquare

Tip
Proof: κ = |y″|/(1 + y′²)^3/2 for graphs

Curvature is the turning rate of the tangent angle per unit of ARC length: κ=∣dϕds∣\kappa = \left|\tfrac{d\phi}{ds}\right| where tan⁡ϕ=y′\tan\phi = y'. Differentiate in xx: sec⁡2ϕ dϕdx=y′′\sec^2\phi\,\tfrac{d\phi}{dx} = y'', so dϕdx=y′′1+y′2\tfrac{d\phi}{dx} = \tfrac{y''}{1 + y'^2}. Meanwhile arc length grows at dsdx=1+y′2\tfrac{ds}{dx} = \sqrt{1 + y'^2}.

Divide: κ=∣dϕ/dxds/dx∣=∣y′′∣(1+y′2)3/2\kappa = \left|\tfrac{d\phi/dx}{ds/dx}\right| = \dfrac{|y''|}{(1 + y'^2)^{3/2}}. ■\blacksquare

Going Deeper: Explanations & Worked Problems

Concept
Velocity, speed, and why arc length is an integral

A moving point r(t)\mathbf r(t) has two different "how fast" questions hiding in it. The VELOCITY r′(t)\mathbf r'(t) is a vector: it points along the instantaneous direction of travel (tangent to the path) and its components are just the one-variable derivatives of each coordinate.

The SPEED ∣r′(t)∣|\mathbf r'(t)| is a number — the length of that vector. Distance traveled is speed accumulated over time, which is why arc length is L=∫ab∣r′(t)∣ dtL = \int_a^b|\mathbf r'(t)|\,dt and not anything simpler: over a tiny interval Δt\Delta t the point moves approximately r′(t)Δt\mathbf r'(t)\Delta t, a straight step of length ∣r′(t)∣Δt|\mathbf r'(t)|\Delta t, and adding infinitely many tiny straight steps IS the integral. Note what this is not: it is not ∣r(b)−r(a)∣|\mathbf r(b) - \mathbf r(a)|, the straight-line displacement — a runner on a circular track covers a full lap of distance with zero displacement.

Example
Worked: the helix length, with the speed computed honestly

Find the length of r(t)=(cos⁡t,sin⁡t,t)\mathbf r(t) = (\cos t, \sin t, t) for 0≤t≤2π0 \le t \le 2\pi.

Step 1 — differentiate coordinate-wise: r′(t)=(−sin⁡t,cos⁡t,1)\mathbf r'(t) = (-\sin t, \cos t, 1).

Step 2 — speed: ∣r′(t)∣=sin⁡2t+cos⁡2t+1=1+1=2|\mathbf r'(t)| = \sqrt{\sin^2t + \cos^2t + 1} = \sqrt{1 + 1} = \sqrt2, using the Pythagorean identity to collapse the first two squares.

Step 3 — notice the speed is CONSTANT: the integral needs no antiderivative work, L=∫02π2 dt=22 πL = \int_0^{2\pi}\sqrt2\,dt = 2\sqrt2\,\pi.

Step 4 — sanity check by unrolling: the helix wraps around a cylinder of circumference 2π2\pi while climbing 2π2\pi; unrolled flat it is the hypotenuse of a right triangle with both legs 2π2\pi, length (2π)2+(2π)2=22 π\sqrt{(2\pi)^2 + (2\pi)^2} = 2\sqrt2\,\pi ✓ — the same answer with no calculus at all.

Example
Worked: curvature of y = x² away from the vertex

Find κ\kappa for y=x2y = x^2 at x=1x = 1.

Step 1 — derivatives: y′=2x=2y' = 2x = 2 and y′′=2y'' = 2 at x=1x = 1.

Step 2 — plug into the formula: κ=∣y′′∣(1+y′2)3/2=2(1+4)3/2=255\kappa = \dfrac{|y''|}{(1 + y'^2)^{3/2}} = \dfrac{2}{(1 + 4)^{3/2}} = \dfrac{2}{5\sqrt5}.

Step 3 — compare with the vertex, where y′=0y' = 0 gives κ=2\kappa = 2: at x=1x = 1 the curvature is 255≈0.18\tfrac{2}{5\sqrt5} \approx 0.18, roughly a tenth of the vertex value.

Step 4 — interpret via the osculating circle: the best-fitting circle at x=1x=1 has radius 1κ=552≈5.6\tfrac1\kappa = \tfrac{5\sqrt5}{2} \approx 5.6, versus radius 12\tfrac12 at the vertex — the parabola straightens out fast as you leave the bottom, and the formula quantifies exactly how fast.