Question

Find the curvature of r(t) = (-7 sin(t), cos(2t), –3t) at t = ž. (Use symbolic notation and fractions where needed.) k () =

Answer 1

Here olt) = (-7 sin(t), cos(24), -34) ; t= Then g'(t) = d r(t) 2. dt = d (-7 sint, Cos(24), -3t). = {-7 lost t), -sin(24) (2011), -310) = 't) -3> < -7 Cas(4), - 2 sin(at), -3) .: O'Y 5x) = (-7001(%), - 2 sin(2.), =) g'%) = (-740), -2 sin(ti) , -3) az ola) = { 0, 0; +3 ) ii "(t) = Also, a 20 (4) <-7 cosct), -2 sin121), -3) dt -71-sint), -2005(2+(2011), o). g"(t) = ( 7 Sint), -4 105(24), o) .: 77%) < < 7 siml), -4605/2.7);0). (+((), - { des(), …) =(7,-461), 0) lui =) 7" 1) =<7, 4, o T'TA) x >" ) k. 0 - 3 7 4 0 i 10 +12.) - j (0+21) + K (0-0) =) r" (1/4) X&"(TA) = 121 – 213 + OK.

hen given by. Cerrature at t = This K(T) = 1 x'(TM) * g."(54) 18'1T) 19 (12)²+ (217² +0² (5027 Os +(-3)) il Jsos 51444441 (59) = (3) 3 165 (3)3 ros 9 K(T/₂) = Tros is the desired answer. 9
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