Question

Q2 (6 points) If C is a smooth plane curve with parametrization r r(t),t E [a, b], then the curvature K(t) of C at the point r(t) is defined to be the magnitude of the rate of change -ll dT of the unit tangent vector with respect to the arc length. That is, = ds () [2p] Show that K(t) = ||F (C) xr" (t)|| r (3 (ii) [4p] If C is a simple closed curve in R2, then the total curvature of C is defined to be x2 K ds. Find the total curvature of the curve 1 b2 a2 C Drag and drop your images or click to browse...

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Answer 1

Q2 (6 points) If C is a smooth plane curve with parametrization r rt), t e [a, b), then the curvature k(t) of C at the point r(t) is defined to be the magnitude of the rate of change -l d'T of the unit tangent vector with respect to the arc length. That is, K ds ()Xr() r(O () 12p] Show that K(t) Ans.: Curvature Let C be a smooth curve with position vector r (s) where s is the arc length parameter The curvature K of C is given by, K= ds Here T is the unit tangent vector. The letter used to denote the curvature is the greek letter kappa denoted K Using the chain rule, is given by dt dT dT dT ds dt ds dt ds dT T' dt

T into dT ds dT Substitute of C is given by, the curvature K T' The unit tangent vector to the curve r(), T, is given by '(t) T = r r'()-T() ds Butr'(t dt ds Thus, r'(t)T dt ds By taking the derivarive of r'()-T with respect to t, this gives, dt d's r'(t) dt ds dt d's ds product of r'()aT andr')T+Tis gis given by The cross dt dt dt

ds T T+ ds r'()xr"(t)T dt dt dt ds d ds TxT' dt TxT+ dt dt (Since the cross product of a vector by itself is always the zero vector.) But TxT 0 = ds ITTsin dt (t)xr"() TxT dt Since TIT', sin e 2 _ Irexr'(l=IP(e/f irilain플-rofi메 F()xr() Substitute T-'(t)xr" (t)| .ntox 3 (t) T'(t this gives, F _FOxr*] K J))