14.4%20Ex%2060

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

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SECTION 14.4 Curvature (ET Section 13.4) 531 Hint: Observe that T ( t ) = h cos θ ( t ), sin ( t ) i . y x r ( t ) ( t ) T ( t ) = ⟨ cos ( t ), sin ( t ) FIGURE 19 The curvature is the rate of change of ( t ) . SOLUTION Since T ( t ) is a unit vector that makes an angle ( t ) with the positive x -axis, we have T ( t ) = h cos ( t ), sin ( t ) i . Differentiating this vector using the Chain Rule gives: T 0 ( t ) = - 0 ( t ) sin ( t ), 0 ( t ) cos ( t ) ® = 0 ( t ) h− sin ( t ), cos ( t ) i We compute the norm of the vector T 0 ( t ) : k T 0 ( t ) k=k 0 ( t ) h− sin ( t ), cos ( t ) ik=| 0 ( t ) | q ( sin ( t )) 2 + ( cos ( t )) 2 =| 0 ( t ) 1 =| 0 ( t ) | When r ( s ) is a parametrization by arc length we have: κ ( s ) = ° ° ° ° d T ds ° ° ° ° = ° ° ° ° d T dt ° ° ° ° ¯ ¯ ¯ ¯ dt d d ds ¯ ¯ ¯ ¯ = ¯ ¯ 0 ( t ) ¯ ¯ 1 | 0 ( t ) | ¯ ¯ ¯ ¯ d ds ¯ ¯ ¯ ¯ = ¯ ¯ ¯ ¯ d ds
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This homework help was uploaded on 04/22/2008 for the course MATH 32A taught by Professor Gangliu during the Winter '08 term at UCLA.

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