18.02A Topic 22: Vector Derivatives: Velocity, Curvature (2 Hours). Read

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18.02A Topic 22: Vector derivatives: velocity, curvature (2 hours). Read: TB: 17.4, 17.5 (x(t), y(t)) • General parametric curve Ò@ ÒÒ Think of it as a point moving in time. ÒrÒ(t) ÒÒ r(t) = x(t) i + y(t) j = hx(t), y(t)i = position vector ÒÒ b b ÒÒ

Examples: (from last time) a v Circle: r(t) = ahcos t, sin ti Line: r(t) = 1 + t 1 = A + tv a v ⇒ r0(t) = ah− sin t, cos ti 2 2 v ⇒ r0(t) = 1 = v v2 o 7o o v ooo •ooo o o A = (a , a ) o o 1 2

Arclength: Geometrically this means choose a starting point P on the curve and refer to points on the curve by their distance along the ÓÓ curve from P . (This is like mileage markers along the highway.) ds ÓÓ ÓÓ dy Notation: s = arclength ÓÓ ÓÓ Geometric approach: (see picture above) •Ó s dx ds dx2 dy 2 ds = p(dx)2 + (dy)2 ⇒ = + . dt dt dt ds Note: = speed (has units distance/time). dt ds Example: Find and arclength for the curve r(t) = ht, t2i between (0, 0) and (1, 1) dt ds Z 1 √ = p1 + (2t)2 ⇒ L = 1 + 4t2 dt. dt 0 Velocity: If something moves it has a velocity = speed and direction. How do we ﬁnd it for r(t)? What does it mean geometrically? dr Answer: Instantaneous velocity = r0(t) = = hx0(t), y0(t)i dt −→ ÓA Reasons: dr ÓÓ ÓÓ dy Over a small time dt the point moves a (vector) dr = dx i + dy j ÓÓ ÓÓ dr dx dy •ÓÓ and its velocity (displacement/time) is = i + j dx dt dt dt ds dr Note: speed = dt = dt . (continued)

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∆r Tangent vector: In the picture above, we see that as ∆t shrinks to 0 the vector ∆t becomes tangent to the curve When the parameter is t we can refer to r0(t) as the ’velocity’. In general, the derivative is given its geometric name: the ’tangent vector’. Physics approach: (get same formulas) ds = speed = |r0(t)| = |hx0(t), y0(t)i| = p(x0)2 + (y0)2. dt Unit tangent vector r(t) The unit vector in the direction of the tangent vector is denoted T = |r0(t)| . It’s called the unit tangent vector. ds 0 Note dt T = r (t). Nomenclature summary: Here are a list of names and formulas. We will motivate and derive them below. r(t) = position. ds s = arclength, speed = v = dt . 0 ds v(t) = r (t) = dt T = tangent vector, velocity. dv d2r a(t) = dt = dt2 = acceleration. T = unit tangent vector, N = unit normal vector. κ = curvature, R = 1/κ = radius of curvature. ϕ = tangent angle. C = Center of curvature = center of best ﬁtting circle (has radius = radius of curvature). Formulas: (explained in the following pages) ds 1. Speed = = |v(t)| = p(x0)2 + (y0)2. dt ds v 2. v = T, T = dt ds/dt d2s ds2 d2s v2 3. a(t) = T + κ N = T + N dt2 dt dt2 R dT |a × v| 4. κ = = . ds |v|3 |x00y0 − x0y00| 4a. For plane curves r(t) = x(t)bi + y(t)bj : κ = . ((x0)2 + (y0)2)3/2 5. v × (a × v) = κv4N. 1 6. C = r + R N = r + N. κ

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Example: Cycloid r(θ) = ahθ − sin θ, 1 − cos θi dr = ah1 − cos θ, sin θi = 2ahsin2 θ/2, sin θ/2 cos θ/2i dθ q dr ds 2 = = 2a sin θ/2 = 2a| sin θ/2| dθ dθ At the cusp ds/dθ = 0, i.e., physically, stop to make a sudden 180 degree turn. ds For one arch, 0 < θ < 2π, dθ = 2a sin θ/2 2ahsin2 θ/2, sin θ/2 cos θ/2i T = = hsin θ/2, cos θ/2i (a unit vector!) 2a sin θ/2 r(θ) R θ ds s •A : θ = π |AP | = 0 dθ dθ θ • = −4a cos θ/2|0 P = 4a cos θ/2 ⇒ |OA| = 4a (Wren’s theorem)

_?? How sharply curved is the trajectory? ?T? Curvature: ?• T That is, how fast does the tangent vector turn in per unit arclenth? o • O ∆sO This is tricky so pay attention. Curvature is the rate T is turning per ∆s T T • • dT unit arclength. That is, κ = . ds (Smaller circle = faster turning = greater curvature.) Note well, curvature is a geometric idea– we measure the rate with respect to arclength. The speed the point moves over the trajectory is irrelevant. T is a unit vector ⇒ T = hcos ϕ, sin ϕi where ϕ is the tangent angle. dT d dϕ y ⇒ = hcos ϕ, sin ϕi = h− sin ϕ, cos ϕi. ds ds ds Both magnitude and direction of dT are useful: _?? ? ds ??N T ?? dT dϕ ?? ϕ Curvature = κ = = . ds ds Direction = N = unit normal = h− sin ϕ, cos ϕi ⊥ T. x Note: the book doesn’t use the absolute value in its deﬁnition of κ, but it’s more stan- dard to include it. y _?? Example: Circle r(t) = b(cos t i + sin t j) ??T ?? ϕ = t + π/2 r0(t) = b(− sin t i + cos t j) ?? •? ds 0 ⇒ |v| = dt = |r (t)| = b and ϕ = t + π/2 r dϕ dϕ dt dϕ/dt 1 t ⇒ ds = dt · ds = ds/dt = b x I.e. curvature of a circle = 1/radius (bigger circle = smaller curvature). |a×v| 2 3 Using formula 4: a = −b(cos t i + sin t j) ⇒ κ = |v|3 = b /b = 1/b.

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Proofs of formulas 3-5: ds Note: we will repeatedly use that v = dt . Formula 3 is an application of the product and chain rules: ds Start with v = T. dt dv d2s ds dT d2s ds dT ds ⇒ a = = T + = T + dt dt2 dt dt dt2 dt ds dt d2s ds2 = T + κN. dt2 dt In physics this is the decomposition of acceleration into tangential and radial compo- nents. Formula 4 now follows from formula 3 since T and N are orthogonal unit vectors: ! d2s ds2 ds ds3 a × v = T + κN × T = κ(N × T). dt2 dt dt dt Since N and T are orthogonal unit vectors N × T is a unit vector ds 3 3 ⇒ |a × v| = ( dt ) κ = v κ. The second part of formula 4 is just the ﬁrst in coordinates: v = x0 bi + y0 bj and a = x00 bi + y00 bj p ⇒ a × v = (x00y0 − x0y00)kb and v = (x0)2 + (y0)2 ⇒ what we want. Formula 5 now follows from what we just did. We found a × v = v3κ(N × T). ⇒ v × (a × v) = v4 κ T × (N × T) = v4κ N. The last equality is easy using your right hand (since T and N are orthogonal unit vectors). Example: Find the curvature of the cycloid r(θ) = a(θ − sin θ)bi + a(1 − cos θ)bj v = a(1 − cos θ)bi + a sin θbj and a = a sin θbi + a cos θbj. p ⇒ a×v = a2((1−cos θ) cos θ−sin θ sin θ) kb = a2(cos θ−1) kb and v = a 2(1 − cos θ) |a×v| a2(1−cos θ) 1 1 ⇒ κ = 3 = = √ = |. |v| a323/2(1−cos θ)3/2 a23/2 (1−cos θ) 4a| sin(θ/2) Example: Circle with diﬀerent parameterization. r(t) = b(cos t2, sin t2) Since curvature is geometric it should be independent of parameterization. We’ll use formula 4 to verify this in this example. v = h−2bt sin t2, 2bt cos t2i, a = h−2b sin t2 − 4bt2 cos t2, 2b cos t2 − 4bt2 sin t2i. 2 3 8b2t3 1 A little algebra ⇒ a × v = −8b t k and |v| = 2bt ⇒ κ = 8b3t3 = b . (same as before!).

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These circle examples (κ = 1/radius) explains the following deﬁnition 1 Radius of curvature = κ The center of curvature and the osculating circle: • 11 The osculating (kissing) circle is the best ﬁtting circle to 1o1 radius of curvature the curve. 11 Radius = radius of curvature. Center along normal direction.

Example: For the helix r(t) = cos tbi + sin tbj + at kb find the radius of curvature and center of curvature for arbitrary t. answer: We will use the formulas (2), (3) and (4), z v = − sin t i + cos t j + a k; a = − cos t i − sin t j. √b b b b b ⇒ |v| = 1 + a2; a × v = −a sin tbi + a cos tbj − kb. √ |a×v| 1+a2 1 Formula (4) ⇒ κ = |v|3 = (1+a2)3/2 = 1+a2 . ⇒ radius of convergence = R = 1 + a2. The center of curvature C = r(t) + RN ⇒ we have to find RN. y Since we already have a × v we could use formula (5). √ ds 2 d2s ds 2 x Instead we note that dt = 1 + a ⇒ dt2 = 0 ⇒ a = κ( dt ) N. But a is already a unit vector ⇒ a = N. ⇒ RN = −(1 + a2)(cos tbi + sin tbj). ⇒ C = (cos t, sin t, t) − ((1 + a2) cos t, (1 + a2) sin t, 0) = (−a2 cos t, −a2 sin t, t). Example: For the parabola r(t) = t i + t2 j find v, a, T, ds/dt, κ, R, N and C for arbitrary t. v = i + 2t j, a = 2j. √ ⇒ ds/dt = |v| = 1 + 4t2, T = v = √ 1 i + √ 2t j. |v| 1+4t2 1+4t2 2 Formula 4: a × v = −2k. ⇒ κ = (1+4t2)3/2 . (Maximum curvature at t = 0 as expected.) (1+4t2)3/2 R = 1/κ = . •jTTTT 2 Tt = 0 Formula 5: v × (a × v) = −4t i + 2 j. ⇒ N = √ 1 (−4t i + 2 j). 2 1+4t2 2 1+4t2 Formula 6: C = r + RN = (t i + t j) + 4 (−4t i + 2 j).