Multivariable-Calculus · Unit 6 · Video 3 · Interactive Practice

Arc Length and the Unit Tangent Vector: v = T ds/dt

IKey Formulas

FormulaNameWhat it says
dsdt=∣v⃗∣=∣dr⃗dt∣\dfrac{ds}{dt} = |\vec v| = \left|\dfrac{d\vec r}{dt}\right|SpeedArc length changes at the rate of the speed
length=∫ab∣v⃗∣ dt\text{length} = \displaystyle\int_a^b |\vec v|\,dtArc lengthIntegrate the speed over the time interval
T^=v⃗∣v⃗∣\hat T = \dfrac{\vec v}{|\vec v|}Unit tangentDirection of motion, length 11; needs ∣v⃗∣≠0|\vec v| \ne 0
v⃗=T^ dsdt\vec v = \hat T\,\dfrac{ds}{dt}Velocity factoredDirection times speed (chain rule: dr⃗dt=dr⃗dsdsdt\frac{d\vec r}{dt} = \frac{d\vec r}{ds}\frac{ds}{dt})

Key Insight: One arch of the cycloid has length ∫02π2−2cos⁡t dt=∫02π2sin⁡t2 dt=[−4cos⁡t2]02π=4+4=8.\displaystyle\int_0^{2\pi}\sqrt{2 - 2\cos t}\,dt = \int_0^{2\pi} 2\sin\tfrac{t}{2}\,dt = \Big[-4\cos\tfrac{t}{2}\Big]_0^{2\pi} = 4 + 4 = 8.

IIVisualization 1 — Arc Length Is Integrated Speed

The arc length s(t)s(t) is the signed area under the speed graph ∣v⃗∣=2sin⁡(t/2)|\vec v| = 2\sin(t/2), starting from the reference time.

IIIVisualization 2 — Velocity Is Direction Times Speed

Every velocity splits into a unit direction T^\hat T and a length ds/dtds/dt: v⃗=T^ dsdt\vec v = \hat T\,\dfrac{ds}{dt}.

IVVisualization 3 — Why dr⃗/dsd\vec r/ds Is the Unit Tangent

For finite Δt\Delta t the chord Δr⃗\Delta\vec r is shorter than Δs\Delta s and tilted off T^\hat T; both errors vanish as Δt→0\Delta t \to 0.

💡 Dividing Δr⃗≈T^ Δs\Delta\vec r \approx \hat T\,\Delta s by Δt\Delta t and letting Δt→0\Delta t \to 0 turns the approximation into the identity v⃗=dr⃗dt=T^ dsdt\vec v = \dfrac{d\vec r}{dt} = \hat T\,\dfrac{ds}{dt}.

VQuiz Questions

Problem 1 · Length of a Helix Turn

Given: the helix r⃗(t)=⟨3cos⁡t, 3sin⁡t, 4t⟩\vec r(t) = \langle 3\cos t,\ 3\sin t,\ 4t\rangle — find the length of one turn, 0≤t≤2π0 \le t \le 2\pi.

✅ Correct! The speed is the constant 55, so the length is 5⋅2π=10π5 \cdot 2\pi = 10\pi.
❌ Not quite. 6π6\pi is the circumference of the circle the helix projects to — it ignores the climb 4t4t in the zz-direction.
❌ Not quite. 8π8\pi is only the height gained, 4⋅2π4 \cdot 2\pi. The speed combines all three components of v⃗\vec v.
❌ Close, but… 9+16=259 + 16 = 25 is ∣v⃗∣2|\vec v|^2. The speed is its square root, 55.
❌ Not quite. Compute v⃗=⟨−3sin⁡t, 3cos⁡t, 4⟩\vec v = \langle -3\sin t,\ 3\cos t,\ 4\rangle, then its length, then integrate.
Show solution

Differentiate: v⃗=dr⃗dt=⟨−3sin⁡t, 3cos⁡t, 4⟩\vec v = \dfrac{d\vec r}{dt} = \langle -3\sin t,\ 3\cos t,\ 4\rangle.

∣v⃗∣=9sin⁡2t+9cos⁡2t+16=25=5.|\vec v| = \sqrt{9\sin^2 t + 9\cos^2 t + 16} = \sqrt{25} = 5.

Integrate the speed:

length=∫02π5 dt=10π.\text{length} = \int_0^{2\pi} 5\,dt = 10\pi.

Problem 2 · The Unit Tangent

Given: r⃗(t)=⟨3t, t2⟩\vec r(t) = \langle 3t,\ t^2\rangle — find the unit tangent vector T^\hat T at t=2t = 2.

✅ Correct! v⃗=⟨3,4⟩\vec v = \langle 3, 4\rangle has length 55, and T^=v⃗/5\hat T = \vec v/5 has length 11.
❌ Close, but… ⟨3,4⟩\langle 3, 4\rangle is the velocity itself, with length 55. T^\hat T must have length 11 — divide by ∣v⃗∣|\vec v|.
❌ Not quite. Dividing by 3+4=73 + 4 = 7 does not give length 11: (3/7)2+(4/7)2=5/7\sqrt{(3/7)^2 + (4/7)^2} = 5/7. Divide by the length 32+42=5\sqrt{3^2 + 4^2} = 5.
❌ Not quite. You divided by ∣v⃗∣2=25|\vec v|^2 = 25; that vector has length 15\tfrac{1}{5}. Divide by ∣v⃗∣=5|\vec v| = 5.
Show solution

v⃗=dr⃗dt=⟨3, 2t⟩\vec v = \dfrac{d\vec r}{dt} = \langle 3,\ 2t\rangle, so at t=2t = 2: v⃗=⟨3, 4⟩\vec v = \langle 3,\ 4\rangle.

∣v⃗∣=32+42=5,T^=v⃗∣v⃗∣=⟨35, 45⟩.|\vec v| = \sqrt{3^2 + 4^2} = 5, \qquad \hat T = \frac{\vec v}{|\vec v|} = \left\langle \tfrac{3}{5},\ \tfrac{4}{5}\right\rangle.

Check: (35)2+(45)2=9+1625=1\left(\tfrac{3}{5}\right)^2 + \left(\tfrac{4}{5}\right)^2 = \tfrac{9 + 16}{25} = 1, and v⃗=5 T^\vec v = 5\,\hat T — direction times speed.

Problem 3 · Part of a Cycloid Arch

Given: the cycloid r⃗(t)=⟨t−sin⁡t, 1−cos⁡t⟩\vec r(t) = \langle t - \sin t,\ 1 - \cos t\rangle with speed ∣v⃗∣=2−2cos⁡t|\vec v| = \sqrt{2 - 2\cos t} — find the arc length from t=0t = 0 to t=2π3t = \tfrac{2\pi}{3}, and the speed dsdt\dfrac{ds}{dt} at t=2π3t = \tfrac{2\pi}{3}.

Arc length from t=0t = 0 to t=2π/3t = 2\pi/3?

Speed ds/dtds/dt at t=2π/3t = 2\pi/3?

✅ Correct! s=4−4cos⁡π3=2s = 4 - 4\cos\tfrac{\pi}{3} = 2 and dsdt=2sin⁡π3=3\tfrac{ds}{dt} = 2\sin\tfrac{\pi}{3} = \sqrt{3}.
❌ Not quite. 83\tfrac{8}{3} assumes a third of the time covers a third of the length. The speed is not constant — the point moves slowly near the cusp, so the first third of the time covers only 22 of the 88.
❌ Close, but… −4cos⁡π3=−2-4\cos\tfrac{\pi}{3} = -2 is only the upper limit. Subtract the value at t=0t = 0: −2−(−4)=2-2 - (-4) = 2.
❌ Check the arc length. Use 2−2cos⁡t=4sin⁡2t22 - 2\cos t = 4\sin^2\tfrac{t}{2}, so the integrand is 2sin⁡t22\sin\tfrac{t}{2} with antiderivative −4cos⁡t2-4\cos\tfrac{t}{2}.
❌ Close, but… 2−2cos⁡2π3=32 - 2\cos\tfrac{2\pi}{3} = 3 is ∣v⃗∣2|\vec v|^2; the speed is its square root.
❌ Check the speed. dsdt=∣v⃗∣=2sin⁡t2\tfrac{ds}{dt} = |\vec v| = 2\sin\tfrac{t}{2}; evaluate at t=2π3t = \tfrac{2\pi}{3}.
Show solution

Arc length. By the half-angle identity 2−2cos⁡t=4sin⁡2t22 - 2\cos t = 4\sin^2\tfrac{t}{2}, and sin⁡t2≥0\sin\tfrac{t}{2} \ge 0 on [0,2π][0, 2\pi], so ∣v⃗∣=2sin⁡t2|\vec v| = 2\sin\tfrac{t}{2}.

s=∫02π/32sin⁡t2 dt=[−4cos⁡t2]02π/3=−4cos⁡π3+4cos⁡0=−2+4=2.s = \int_0^{2\pi/3} 2\sin\tfrac{t}{2}\,dt = \Big[-4\cos\tfrac{t}{2}\Big]_0^{2\pi/3} = -4\cos\tfrac{\pi}{3} + 4\cos 0 = -2 + 4 = 2.

Speed.

dsdt=2sin⁡π3=2⋅32=3.\frac{ds}{dt} = 2\sin\tfrac{\pi}{3} = 2 \cdot \tfrac{\sqrt{3}}{2} = \sqrt{3}.

A third of the time covers only a quarter of the arch's length 88, because the point starts from rest at the cusp.

Problem 4 · Counting From a Reference Point

Given: the helix r⃗(t)=⟨3cos⁡t, 3sin⁡t, 4t⟩\vec r(t) = \langle 3\cos t,\ 3\sin t,\ 4t\rangle, with arc length ss measured from the point where t=2t = 2 and increasing in the direction of motion — find ss at t=0t = 0.

✅ Correct! s(t)=5(t−2)s(t) = 5(t - 2), so the point at t=0t = 0 lies 1010 units before the reference point: s=−10s = -10.
❌ Close, but check the sign. The distance is 1010, but t=0t = 0 comes before the reference point, and ss increases with tt — so ss is negative there.
❌ Not quite. s=0s = 0 at the reference point, t=2t = 2, not at t=0t = 0. The starting time of the parametrization is not automatically the reference point.
❌ Not quite. −8-8 uses only the vertical rate 44. The speed is ∣v⃗∣=9+16=5|\vec v| = \sqrt{9 + 16} = 5.
Show solution

From Problem 1 the speed is constant: dsdt=∣v⃗∣=5\dfrac{ds}{dt} = |\vec v| = 5. With s(2)=0s(2) = 0:

s(t)=∫2t5 du=5(t−2).s(t) = \int_2^{t} 5\,du = 5(t - 2). s(0)=5(0−2)=−10.s(0) = 5(0 - 2) = -10.

Arc length is negative before the reference point and positive after it; its size, 1010, is the distance along the helix between t=0t = 0 and t=2t = 2.

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