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Unformatted text preview: S E C T I O N 14.3 Arc Length and Speed (ET Section 13.3) 487 (b) The velocity vectors point in the same direction but may have different lengths. (c) The velocity vectors may point in opposite directions. SOLUTION (a) The length of the velocity vector is the speed of the particle. Therefore, if the speeds of the cars are different the velocities are not identical. The statement is false. (b) The velocity vector is tangent to the curve. Since the cars travel in the same direction, their velocity vectors point in the same direction. The statement is true. (c) Since the cars travel in the same direction, the velocity vectors point in the same direction. The statement is false. 3. A mosquito flies along a parabola with speed v( t ) = t 2 . Let L ( t ) be the total distance traveled at time t . (a) How fast is L ( t ) changing at t = 2? (b) Is L ( t ) equal to the mosquitos distance from the origin? SOLUTION (a) By the Arc Length Formula, we have: L ( t ) = t t w r w ( t ) w dt = t t v( t ) dt Therefore, L w ( t ) = v( t ) To find the rate of change of L ( t ) at t = 2 we compute the derivative of L ( t ) at t = 2, that is, L w ( 2 ) = v( 2 ) = 2 2 = 4 (b) L ( t ) is the distance along the path traveled by the mosquito. This distance is usually different from the mosquitos distance from the origin, which is the length of r ( t ) . r( t ) Distance L ( t ) Distance from the origin t t 4. What is the length of the path traced by r ( t ) for 4 t 10 if r ( t ) is an arc length parametrization? SOLUTION Since r ( t ) is an arc length parametrization, the length of the path for 4 t 10 is equal to the length of the time interval 4 t 10, which is 6. Exercises In Exercises 16, compute the length of the curve over the given interval. 1. r ( t ) = w 3 t , 4 t 3 , 6 t + 1 w , t 3 SOLUTION We have x ( t ) = 3 t , y ( t ) = 4 t 3, z ( t ) = 6 t + 1 hence x w ( t ) = 3 , y w ( t ) = 4 , z w ( t ) = 6 . We use the Arc Length Formula to obtain: L = 3 w r w ( t ) w dt = 3 x w ( t ) 2 + y w ( t ) 2 + z w ( t ) 2 dt = 3 3 2 + 4 2 + 6 2 dt = 3 61 2. r ( t ) = 2 t i 3 t k , 11 t 15 SOLUTION We have, x ( t ) = 2 t , y ( t ) = 0, z ( t ) =  3 t . Hence x w ( t ) = 2 , y w ( t ) = , z w ( t ) =  3 Using the Arc Length Formula we get: L = 15 11 w r w ( t ) w dt = 15 11 x w ( t ) 2 + y w ( t ) 2 + z w ( t ) 2 dt = 15 11 2 2 + 2 + ( 3 ) 2 dt = 13 ( 15 11 ) = 4 13 . 488 C H A P T E R 14 CALCULUS OF VECTORVALUED FUNCTIONS (ET CHAPTER 13) 3. r ( t ) = 2 t , ln t , t 2 , 1 t 4 SOLUTION The derivative of r ( t ) is r w ( t ) = 2 , 1 t , 2 t . We use the Arc Length Formula to obtain: L = 4 1 w r w ( t ) w dt = 4 1 2 2 + 1 t 2 + ( 2 t ) 2 dt = 4 1 4 t 2 + 4 + 1 t 2 dt = 4 1 2 t + 1 t 2 dt = 4 1 2 t + 1 t dt = t 2 + ln t 4 1 = ( 16 + ln 4 ) ( 1 + ln 1 ) = 15 + ln 4 4. r ( t ) = 2 t 2 + 1 , 2 t...
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This note was uploaded on 07/15/2009 for the course MATH 210 taught by Professor Hubscher during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 Hubscher
 Arc Length, Vectors

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