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14.4%20Ex%2060

# 14.4%20Ex%2060 - S E C T I O N 14.4 Curvature(ET Section...

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S E C T I O N 14.4 Curvature (ET Section 13.4) 531 Hint: Observe that T ( t ) = cos θ ( t ), sin θ ( t ) . y x r ( t ) ( t ) T ( t ) = ⟨ cos ( t ), sin ( t ) FIGURE 19 The curvature is the rate of change of θ ( t ) . SOLUTION Since T ( t ) is a unit vector that makes an angle θ ( t ) with the positive x -axis, we have T ( t ) = cos θ ( t ), sin θ ( t ) . Differentiating this vector using the Chain Rule gives: T ( t ) = − θ ( t ) sin θ ( t ), θ ( t ) cos θ ( t ) = θ ( t ) sin θ ( t ), cos θ ( t ) We compute the norm of the vector T ( t ) : T ( t ) = θ ( t ) sin θ ( t ), cos θ ( t ) = | θ ( t ) | ( sin θ ( t )) 2 + ( cos θ ( t )) 2 = | θ ( t ) | · 1 = | θ ( t ) | When r ( s ) is a parametrization by arc length we have: κ ( s ) = d T ds = d T dt dt d θ d θ ds = θ ( t ) 1 | θ ( t ) | d θ ds = d θ ds as desired. 60. A particle moves along the path y = x 3 with unit speed. How fast is the tangent turning (i.e., how fast is the angle of inclination changing) when the particle passes through the point ( 2 , 8 ) ? SOLUTION The particle has unit speed hence the parametrization is in arc length parametrization. Therefore, the change in the angle of inclination is, by Exercise 59: d θ ds = κ ( s ) In particular at the point ( 2 , 8 ) we have: d θ ds ( 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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