18.02A Topic 22: Parametric equations continued.Author: Jeremy OrloffRead: TB: 17.4Circles and ellipsesx(t) = a cos ty(t) = a sin t⇒ circle x2+ y2= a2x(t) = a cos ty(t) = b sin t⇒ ellipsex2a2+y2b2= 1x(t) = a cos t, y(t) = a sin t is called parametric formx2+ y2= a2is called symmetric form.Example 1: For the circle r(t) = b(cos t i + sin t j) find the curvaturev = r0(t) = b(−sin t i + cos t j), ⇒ |v| =dsdt= |r0(t)| = bUsing formula 4 (in topic 21): a = −b(cos t i + sin t j)⇒ κ =|a × v||v|3= b2/b3= 1/b I.e. curvature of a circle = 1/ra-dius (bigger circle = smaller curvature).Alternate method: φ = t + π/2 ⇒dφds=dφ/dtds/dt=1b.xy??r__T•φ = t + π/2tExample 2: Circle with different parameterization. r(t) = bhcos t2, sin t2iSince curvature is geometric it should be independent of parameterization. We’ll useformula 4 from topic 21 to verify this in this example.v = h−2bt sin t2, 2bt cos t2i, a = h−2b sin t2− 4bt2cos t2, 2b cos t2− 4bt2sin t2i.A little algebra ⇒ a × v = −8b2t3k and |v| = 2bt ⇒ κ =8b2t38b3t3=1b. (same as before!).Example 3: For the circle in example 1, find x(t), y(t), T , N, R (radius of curvature)and C (center of curvature).x(t) = b cos t, y(t) = b sin t.T = v/|v| = −sin t i + cos t jκN =dTds=dT/dtds/dt=−cos t i − sin t jb⇒ κ = 1/b, and N = −(cos t i + sin t j).R = 1/κ = b.C = r(t) + RN = bhcos t i, sin ti + bh−cos t, sin ti = 0 = center of circle.These circle examples (κ = 1/radius) explain the the name ’radius of curvature’ for1/κ and the name ’center of curvature’ for C(continued)118.02A topic 22 2Linesparametric form (t=param.) symmetric form−→X =−→A + t−→v⇔x = a1+ v1ty = a2+ v2t⇔x − a1v1=y − a2v2•−→v77A = (a1, a2)Example 4: Find the line through (0, −2, 1) and (1, 0, 2)−→v =−−→PQ =−→Q −−→P = h1, 2, 1i⇒−→X = ht, −2 + 2t, 1 + ti = t i + (−2 + 2t) j + (1 + t) kWe can also write this in column vector form as0−21+ t121.Example 5: For the line in example 4 find x(t), y(t), z(t), v, T, a, κ, N, R, C.x(t) = t, y(t) = −2 + 2t, z(t) = 1 + t.v(t) = hx0, y0, z0i = h1, 2, 1i (constant velocity). ⇒ T(t) = v/|v| =1√6h1, 2, 1i.a(t) =dvdt= 0.κ = 0 (easy using any of the formulas for κ in topic 21). ⇒ R = ∞, N, C areundefined.Example 6: Cycloid (will use this often)Roll a wheel (circle of radius a) along the x-axis and follow the trajectory of a pointon the wheel –this is a cycloid.Brachistochrone -Bernouilli, Tautochrone –Huygens•P•P??•POO•PCθQTo find parametric equations we use vectors:−→r (θ) =−−→OQ +−−→QC +−−→CP−−→OQ = haθ, 0i = amount rolled (a=radius)−−→QC = h0, ai−−→CP = h−a sin θ, −a cos θi⇒−→r (θ) = haθ − a sin θ, a − a cos θi = ahθ − sin θ, 1 − cos θi = a((θ − sin θ) i + (1 −cos θ) j).⇒ x(θ) = a(θ − sin θ), y(θ) = a(1 − cos θ)Note: symmetric form of equations is hard to write downNote: physical units like speed and acceleration will depend on θ(t).(continued)18.02A topic 22 3Example 7: For the cycloid in example 6 find κ and the arclength of one arch.drdθ= ah1 − cos θ, sin θi = 2ahsin2θ/2, sin θ/2 cos θ/2idrdθ=dsdθ= 2aqsin2θ/2 = 2a|sin θ/2|Note: T =2ahsin2θ/2, sin θ/2 cos θ/2i2a|sin θ/2|= ±hsin θ/2, cos θ/2i (a unit vector!)At the cusp ds/dθ = 0, i.e., physically, you must stop to make a sudden 180 degree turn.For one arch, 0 < θ < 2π,dsdθ= 2a sin θ/2|AP | =Rθ0dsdθdθ= −4a cos θ/2|θ0= 4a cos θ/2⇒ |OA| = 4a (Wren’s theorem)s•P•A : θ = πr(θ)To find the curvature κ.r(θ) = a(θ −sin θ) i + a(1 − cos θ) jv = a(1 − cos θ) i + a sin θ j and a = a sin θ i + a cos θ j.⇒ a × v = a2((1 − cos θ) cos θ − sin θ sin θ) k = a2(cos θ −1) k = −a2sin2(θ/2) kUsing |v| = 2a|sin θ/2|: κ =|a×v||v|3=a2(1−cos θ)a323/2(1−cos θ)3/2=1a23/2√(1−cos θ)=14a|sin θ/2|.Note, κ is undefined when θ = 2nπ, i.e. at the cusps.Example 8: Where symmetric form loses information.Find the symmetric form for x = a cos2t, y = a sin2t.Easily we get: x + y = a, with x, y non-negative.The symmetric form shows a line, while the parametric equations showwhich part of the line is actually traced out.••Intersection of two planes:If−→N1and−→N2are the normals to the planes then the line of intersection is perpen-dicular to both normals. I.e. it has direction vector =−→N1×−→N2Note: the ’direction vector’ does not have to be a unit vector. The direction of thedirection vector is the direction for the line.(continued)18.02A topic 22 4Example 9: Find the intersection of the planes x + y + z = 1 and y = 2.Normals:−→N1= h1, 1, 1i and−→N2= h0, 1, 0i.Direction vector:−→v =−→N1×−→N2= −i + k = h−1, 0, 1i.One point of intersection: (by elimintation) y = 2 ⇒ x + 2 + z = 3 ⇒ P = (1, 2, 0).Answer: The intersection is the line h1, 2, 0i + th−1, 0, 1i.Example 10: For the helix r(t) = cos t i + sin t j + at k find the radius of curvatureand center of curvature for arbitrary t.answer: We will use the formulas (2), (3) and (4) from topic 21,v = −sin t i + cos t j + a k; a = −cos t i − sin t j.⇒ |v| =√1 + a2; a × v = −a sin t i + a cos t j − k.Formula (4) ⇒ κ =|a×v||v|3=√1+a2(1+a2)3/2=11+a2.⇒ radius of convergence = R = 1 + a2.The center of curvature C = r(t) + RN ⇒ we have to find RN.Since we already have a × v we use formula (5) from topic 21,i.e. v × (a × v) = κv4N.v × (a × v) = h−sin t, cos t, ai × h−a sin t, a cos t, −1i= h−cos t − a2cos t, −sin t − a2sin t, 0i= (1 + a2)h−cos t, −sin t, 0iNow, formula 5 implies (1 + a2)h−cos t, −sin t, 0i = κv4NSince v =√1 + a2this givesN = h−cos t, −sin t, 0i and κ = 1/(1 + a2) ⇒ R = 1 + a2.Thus, C = (cos t, sin t, at) + RN = (−a2cos t, −a2sin t,
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