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14.4%20Prel%201-4,%20Ex%201%20-%209

14.4%20Prel%201-4,%20Ex%201%20-%209 - 500 C H A P T E R 14...

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500 C H A P T E R 14 CALCULUS OF VECTOR-VALUED FUNCTIONS (ET CHAPTER 13) 14.4 Curvature (ET Section 13.4) Preliminary Questions 1. What is the unit tangent vector of a line with direction vector v = 2 , 1 , 2 ? SOLUTION A line with direction vector v has the parametrization: r ( t ) = −−→ O P 0 + t v hence, since −−→ O P 0 and v are constant vectors, we have: r ( t ) = v Therefore, since v = 3, the unit tangent vector is: T ( t ) = r ( t ) r ( t ) = v v = 2 / 3 , 1 / 3 , 2 / 3 2. What is the curvature of a circle of radius 4? SOLUTION The curvature of a circle of radius R is 1 R , hence the curvature of a circle of radius 4 is 1 4 . 3. Which has larger curvature, a circle of radius 2 or a circle of radius 4? SOLUTION The curvature of a circle of radius 2 is 1 2 , and it is larger than the curvature of a circle of radius 4, which is 1 4 . 4. What is the curvature of r ( t ) = 2 + 3 t , 7 t , 5 t ? SOLUTION r ( t ) parametrizes the line 2 , 0 , 5 + t 3 , 7 , 1 , and a line has zero curvature. 5. What is the curvature at a point where T ( s ) = 1 , 2 , 3 in an arc length parametrization r ( s ) ? SOLUTION The curvature is given by the formula: κ ( t ) = T ( t ) r ( t ) In an arc length parametrization, r ( t ) = 1 for all t , hence the curvature is κ ( t ) = T ( t ) . Using the given information we obtain the following curvature: κ = 1 , 2 , 3 = 1 2 + 2 2 + 3 2 = 14 6. What is the radius of curvature of a circle of radius 4? SOLUTION The definition of the osculating circle implies that the osculating circles at the points of a circle, is the circle itself. Therefore, the radius of curvature is the radius of the circle, that is, 4.
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