S E C T I O N14.4Curvature(ET Section 13.4)531Hint:Observe thatT(t)=cosθ(t),sinθ(t).yxr(t)(t)T(t)= 〈cos (t), sin (t)〉FIGURE 19The curvature is the rate of change ofθ(t).SOLUTIONSinceT(t)is a unit vector that makes an angleθ(t)with the positivex-axis, we haveT(t)=cosθ(t),sinθ(t) .Differentiating this vector using the Chain Rule gives:T(t)= −θ(t)sinθ(t),θ(t)cosθ(t)=θ(t)−sinθ(t),cosθ(t)We compute the norm of the vectorT(t):T(t)=θ(t)−sinθ(t),cosθ(t)= |θ(t)|(−sinθ(t))2+(cosθ(t))2= |θ(t)| ·1= |θ(t)|Whenr(s)is a parametrization by arc length we have:κ(s)=dTds=dTdtdtdθdθds=θ(t)1|θ(t)|dθds=dθdsas desired.60.A particle moves along the pathy=x3with unit speed. How fast is the tangent turning (i.e., how fast is the angleof inclination changing) when the particle passes through the point(2,8)?SOLUTIONThe particle has unit speed hence the parametrization is in arc length parametrization. Therefore, the changein the angle of inclination is, by Exercise 59:dθds=κ(s)In particular at the point(2,8)we have:dθds(2,8)=κ(s)(2,8)(1)We, thus, must find the curvature at the given point. We use the formula for the curvature of a graph in the plane:κ(x)=|y(x)|(1+y(x)2)
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