18 02A Topic 22 Parametric equations continued Author Jeremy Orloff Read TB 17 4 Circles and ellipses x t a cos t circle x2 y 2 a2 y t a sin t 2 2 x t a cos t ellipse xa2 yb2 1 y t b sin t x t a cos t y t a sin t is called parametric form x2 y 2 a2 is called symmetric form Example 1 For the circle r t b cos t i sin t j find the curvature ds v r0 t b sin t i cos t j v r0 t b dt Using formula 4 in topic 21 a b cos t i sin t j a v b2 b3 1 b I e curvature of a circle 1 ra v 3 dius bigger circle smaller curvature d d dt 1 Alternate method t 2 ds ds dt b y T r t Example 2 Circle with different parameterization r t bhcos t2 sin t2 i Since curvature is geometric it should be independent of parameterization We ll use formula 4 from topic 21 to verify this in this example v h 2bt sin t2 2bt cos t2 i a h 2b sin t2 4bt2 cos t2 2b cos t2 4bt2 sin t2 i 8b2 t3 1 A little algebra a v 8b2 t3 k and v 2bt 8b 3 t3 b 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 j dT dT dt cos t i sin t j N 1 b and N cos t i sin t j ds ds dt b R 1 b C r t RN bhcos t i sin ti bh cos t sin ti 0 center of circle These circle examples 1 radius explain the the name radius of curvature for 1 and the name center of curvature for C continued 1 t 2 x 18 02A topic 22 2 Lines parametric form t param X A t v x a1 v 1 t y a2 v 2 t symmetric form x a1 y a2 v1 v2 7 v A 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 k 0 1 We can also write this in column vector form as 2 t 2 1 1 Example 5 For the line in example 4 find x t y t z t v T a N R C x t t y t 2 2t z t 1 t v t hx0 y 0 z 0 i h1 2 1i constant velocity T t v v 16 h1 2 1i a t dv 0 dt 0 easy using any of the formulas for in topic 21 R N C are undefined Example 6 Cycloid will use this often Roll a wheel circle of radius a along the x axis and follow the trajectory of a point on the wheel this is a cycloid Brachistochrone Bernouilli Tautochrone Huygens P P O P P C Q To find parametric equations we use vectors r OQ QC CP OQ ha 0i amount rolled a radius QC h0 ai CP h a sin a cos i r ha a sin a a cos i ah sin 1 cos i a sin i 1 cos j x a sin y a 1 cos Note symmetric form of equations is hard to write down Note physical units like speed and acceleration will depend on t continued 18 02A topic 22 3 Example 7 For the cycloid in example 6 find and the arclength of one arch dr ah1 cos sin i 2ahsin2 2 sin 2 cos 2i d q dr ds 2a sin2 2 2a sin 2 d d 2ahsin2 2 sin 2 cos 2i Note T hsin 2 cos 2i a unit vector 2a sin 2 At the cusp ds d 0 i e physically you must stop to make a sudden 180 degree turn ds For one arch 0 2 d 2a sin 2 r R ds s A d AP 0 d 4a cos 2 0 P 4a cos 2 OA 4a Wren s theorem To find the curvature r a sin i a 1 cos j v a 1 cos i a sin j and a a sin i a cos j a v a2 1 cos cos sin sin k a2 cos 1 k a2 sin2 2 k a2 1 cos Using v 2a sin 2 a v a3 23 2 3 2 1 4a sin1 2 v 3 1 cos 3 2 a2 1 cos Note is undefined when 2n i e at the cusps Example 8 Where symmetric form loses information Find the symmetric form for x a cos2 t y a sin2 t Easily we get x y a with x y non negative The symmetric form shows a line while the parametric equations show which part of the line is actually traced out Intersection of two planes If N 1 and N 2 are the normals to the planes then the line of intersection is perpen dicular to both normals I e it has direction vector N 1 N 2 Note the direction vector does not have to be a unit vector The direction of the direction vector is the direction for the line continued 18 02A topic 22 4 Example 9 Find the intersection of the planes x y z 1 and y 2 Normals N 1 h1 1 1i and N 2 h0 1 0i Direction vector v N 1 N 2 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 curvature and 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 z v 1 a2 a v a sin t i a cos t j k Formula 4 a v v 3 1 a2 1 a2 3 2 1 1 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 v 4 N v a v h sin t cos t ai h a sin t a cos t 1i h cos t a2 cos t sin t a2 sin t 0i 1 a2 h cos t sin t 0i Now formula 5 implies 1 a2 h cos t sin t 0i v 4 N Since v 1 a2 this gives N h …
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