13.3 Arc Length Review: curve in space: rt f t i g t j h t k Tangent vector: r'(t0 ) f ' t i g ' t j h ' t k Tangent line at t t0: s r( t 0) s r '( t 0 ) Velocity: v(t ) r '( t ) speed = | v ( t ) | Acceleration a(t ) v '( t ) r ''( t ) 1 Projectile motion: r(t ) gt2 j v t r initial velocity vr , initial position 2 00 00 g 32ft gravitational constant sec2 Initial speed |v0 | , initial height h, launching angle : 1 2 r(t ) v00cos t , h v sin t gt 2 Recall the length of a curve in the plane: b curve as a graph y f x: length from t a to t b 1 [f '(x)]2 dx a b curve in parametric form : x f t and y g t : length [f '(t)]22 [ g '(t)] dt a A curve in 3-space : r t f t, g t , h t b Length or Arc Length [f '(t)]2 [ g '(t)] 2 [ h '(t)] 2 dt a b Length |r '(t ) | dt a Example 1: A line from a to b : r()()tt a b a , 0t 1 1 1 length rt dt b a dt b a = distance from a to b 0 0 Example 2: 2 2 2 Compute the length (circumference) of circle: ()()x x00 y y r parametric description: x x00 rcos( t ), y y rsin( t ) 0 t 2 2 2 2 length = rt dt r2sin 2 ( t ) r 2 cos 2 ( t ) dt rdt2 r 0 0 0 Problem : A drunken bee travels along the path r(tt ) cos(2 ),sin(2 t), t for 10 seconds. It then travels at constant speed in a straight line for 10 more seconds. How far did the bee travel? velocity: vr(t ) '( t ) 2sin(2 t ),2cos(2 t),1 Speed = | v(t ) | 9 3 10 distance traveled along the helix: |r'(t ) | dt 30 0 At time t0 10, it travels along the tangent line tangent line: u(s ) r( t00 ) s v ( t ) 20 20 length = |u'(s) | ds |v(t ) | ds 0 3 (20 10) 30 10 10 Total distance traveled: 60 ft Arc length function describes distance from a beginning point ta : t ds s( t ) r u du Notice: |r '(t ) | a dt rt gives position on the curve as a function of time its more natural to look at the odometer or the mile markers to tell where you are after you have traveled a certain distance s we would like to have ts as a function of so that we could reparametrize in terms of s. Using the arc length function s s t we can find t as a function of s t t s giving us rr t s or just s Example: Parametrize the following curve by arc length: rtt 3sin(t22 ) i 3cos( ) k r't 6 t cos( t22 ) i 6 t sin( t ) k 2 |r 't | 36 t2 cos 2 ( t 2 ) 36 t 2 sin( t 2 ) 36tt 6 t s s s( t ) 6 udu 3ut22 |ut 3 Solve for t : t= u0 3 0 ss Plug into r(t ) : r( t ( s )) r ( s/ 3) 3sin( )ik 3cos( ) 33 ss Arc length parametrization: r(s ) 3sin( ) i 3cos( ) k 33bb Notice: |r '(s ) | 1 length from s a to s b: r s ds 1 ds b a aa tt is an arc length parameter if |r '( ) | 1 ! 13.4 Curvature The curvature of a curve r(t ) measures the amount the tangent vector bends. Q curvature at PQ> curvature at P r '(t ) Tangent vector r '(t ) Unit tangent vector T ( r '(s ) ! ) |r '(t ) | Curvature - the magnitude of the rate of change of the unit tangent vector T with respect to the arc length parameter s. dT dTr d d Ts r s ds ds ds ds T r s if s is the arc length parameter Examples: a) The curvature of a straight line: r()t r0 + t v r'()t v what is the arc length parametrization of the line: v v r(s) r + s since then |r' (s) | 1 0 ||v ||v dT 2 r s 0 ( if not arc lenght: r(t ) r+0 t v , then r''( t ) 2 t v ) ds b) The curvature of a circle of radius r: r(t ) r cos( t ), r sin( t ) r '(t ) r sin( t ), r cos( t ) t r 'tr hence s r u du rt or t r/ s 0 ss ss arc length parametrization: r(s) rr cos( ), sin( ) Tr'(s) sin( ),cos( ) rr rr dT 11 s s dT 11 T' cos( ), sin( ) ds r r r r ds r2 r small radius means large curvature! It is usually too difficult to parametrize a curve by arc length. Express curvature in terms of a general parameter t : t ds Recall s( t ) r u du hence: |r '(t ) | a dt dT dT dt dT Tt Tt dt ds ds dt ds rt rt dt xy22 Compute the curvature of an ellipse: 1 Example: ab22 r(t ) a cos( t ),bsin( t ) r '(t ) a sin( t ),bcos( t ) |r'(t ) | a2 sin 2 ( t ) b 2 cos 2 ( t ) asin( t ),bcos( t ) ab22cos( t ), ba sin( t ) T(t ) T'(t ) (a lengthy computation…) a2sin 2 ( t ) b 2 cos 2 ( t ) (a2 sin 2 ( t ) b 2 cos 2 ( t )) 3/2 ab Tt ab | T'(t ) | a2sin 2 ( t ) b 2 cos 2 ( t ) rt (a2 sin 2 ( t ) b 2 cos 2 ( t )) 3/2 Sometimes it is easier to use another formula: rrtt 3 rt Why is this true? r '(t ) ds Recall: Tr and | '(t ) | |r '(t ) | dt Tt d ds d2 s ds ds and rT''(t ) TT ' Hence: r'(tt ) | r '( ) | T T rt 2 dt dt dt dt dt 2 ds d2 s ds ds ds ds TT ' TT ' rtt r T 2 T T' dt dt dt dt dt dt but TTTTTTTT 1 implies ' ' 2 ' 0 i.e., T is perpendicular to T' This implies |TTTT ' | | || ' | sin( ) |T ' | | r'| 2 2 ds ds 3 Put together, we get: |rtt r | | T T' | | r'| = | r'| dt dt Example: Ellipse revisited: r(t ) a cos( t ),bsin( t ) or r(t ) a cos( t ),bsin( t ),0 r '(t ) a sin( t ),bcos( t ),0 r ''(t ) a cos( t ), bsin( t ),0 i j k rrt t asin( t ) b cos( t ) 0 (ab sin22 ( t ) ab cos ( t ))k = ab k acos( t ) b sin( t ) 0 |퐫′ 푡 × 퐫″ 푡 | 푎푏 휅 = = |퐫′ 푡 |3 푎2sin2(푡) + 푏2푐표푠2(푡) 3 2 Example: Curvature of the (elliptical) helix: r(t ) a cos( t ),bsin(t), c t r '(t ) a sin( t ),bcos( t ),c r ''(t ) a cos( t ), bsin(t),0 i j k rrt t asin( t ) b cos( t ) c (0bc sin( t ))i ( ac cos( t ) 0) j ( ab sin22 ( t ) ab cos ( t )) k acos( t ) b sin( t ) 0 bcsin( t )i ac cos( t ) j ab k rrt t b2 c 2sin 2 ( t ) a 2 c 2 cos 2 ( t ) a 2 b 2 |r'(t ) | a2 sin 2 ( t ) b 2 cos 2 ( t ) c 2 |퐫′ 푡 × 퐫″ 푡 | b2 c 2sin 2 ( t ) a 2 c 2 cos 2 ( t ) a 2 b 2 휅 = ′ 3 3/2 |퐫 푡 | a2sin 2 ( t ) b 2 cos 2 ( t ) c 2 2 2 2 2 22 a Circular helix: ba , a c a a a c a 3/2 3/2 22 constant! a22 c a22 c a c rrtt Appplication: Curvature of a plane curve: 3 rt r(t ) x ( t ), y(t),0 r'(t ) x '( t ), y'(t),0 i j k rrt t x' y ' 0 (x ' y '' x '' y ')k xy'' '' 0 |x ' y '' x '' y ' | 3/2 (xy ')22 ( ') Summary: b Length of a curve |r '(t ) | dt a t ds Arc length function s( t ) r u du |r '(t ) | a dt Arc length parametrization rr(ss ) with | '( ) | 1 r '(t ) Unit tangent vector Tr'(s) |r '(t ) | dT Tt rrtt Curvature: r s 3 ds rt rt.