Find the length l of the arc traveled from time t 2

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Find the length L of the arc traveled from time t = 2 to time t = 4.
Problem 14(b) - Spring 2007 The velocity vector of a particle moving in space equals v ( t ) = 2 t i - 2 t j + t k at any time t 0. Find the length L of the arc traveled from time t = 2 to time t = 4. Solution: Using the arclength formula, L = Z 4 2 | v ( t ) | dt = Z 4 2 p (2 t ) 2 + ( - 2 t ) 2 + t 2 dt = Z 4 2 9 t 2 dt = Z 4 2 3 t dt = 3 2 t 2 4 2 = 3 2 (16 - 4) = 18 .
Problem 15(b) - Spring 2008 Suppose a particle moving in space has velocity v ( t ) = h sin t , cos 2 t , e t i and initial position r (0) = h 1 , 2 , 0 i . Find the position vector function r ( t ).
Problem 15(b) - Spring 2008 Suppose a particle moving in space has velocity v ( t ) = h sin t , cos 2 t , e t i and initial position r (0) = h 1 , 2 , 0 i . Find the position vector function r ( t ). Solution: We find r ( t ) by integrating r 0 ( t ) = v ( t ): r ( t ) = Z t 0 v ( t ) dt + r (0) = h- cos t , 1 2 sin 2 t , e t i t 0 + h 1 , 2 , 0 i = h- cos t , 1 2 sin 2 t , e t i - h- cos 0 , 1 2 sin 0 , e 0 i + h 1 , 2 , 0 i = h- cos t , 1 2 sin 2 t , e t i - h- 1 , 0 , 1 i + h 1 , 2 , 0 i = h- cos t , 1 2 sin 2 t , e t i + h 2 , 2 , - 1 i . So, r ( t ) = h 2 - cos t , 2 + 1 2 sin 2 t , - 1 + e t i .
Problem 19(a) - Fall 2007 Suppose a particle moving in space has the velocity v ( t ) = h 3 t 2 , 2 sin(2 t ) , e t i . Find the acceleration of the particle. Write down a formula for the speed of the particle (you do not need to simplify the expression algebraically).
Problem 19(a) - Fall 2007 Suppose a particle moving in space has the velocity v ( t ) = h 3 t 2 , 2 sin(2 t ) , e t i . Find the acceleration of the particle. Write down a formula for the speed of the particle (you do not need to simplify the expression algebraically). Solution: Recall the acceleration vector a ( t ) = v 0 ( t ). Hence, a ( t ) = h 6 t , 4 cos(2 t ) , e t i . Recall that the speed ( t ) is the length of the velocity vector. Hence, speed ( t ) = q 9 t 4 + 4 sin 2 (2 t ) + e 2 t .
Problem 19(b) - Fall 2007 Suppose a particle moving in space has the velocity v ( t ) = h 3 t 2 , 2 sin(2 t ) , e t i . If initially the particle has the position r (0) = h 0 , - 1 , 2 i , what is the position at time t ?
Problem 19(b) - Fall 2007 Suppose a particle moving in space has the velocity v ( t ) = h 3 t 2 , 2 sin(2 t ) , e t i . If initially the particle has the position r (0) = h 0 , - 1 , 2 i , what is the position at time t ? Solution: To find the position r ( t ), we first integrate the velocity v ( t ) and second use the initial position value r (0) = h 0 , - 1 , 2 i to solve for the constants of integration . r ( t ) = Z h 3 t 2 , 2 sin 2 t , e t i dt = h t 3 + x 0 , - cos(2 t ) + y 0 , e t + z 0 i . Plugging in the position at t = 0, we get: h 0 3 + x 0 , - cos(0) + y 0 , e 0 + z 0 i = h x 0 , - 1 + y 0 , 1 + z 0 i = h 0 , - 1 , 2 i . Thus, x 0 = 0, y 0 = 0 and z 0 = 1. Hence, r ( t ) = h t 3 , - cos 2 t , e t + 1 i .
Problem 24(b) - Fall 2006 Suppose a particle moving in space has velocity v ( t ) = h sin t , cos 2 t , e t i and initial position r (0) = h 1 , 2 , 0 i . Find the position vector function r ( t ).
Problem 24(b) - Fall 2006 Suppose a particle moving in space has velocity v ( t ) = h sin t , cos 2 t , e t i and initial position r (0) = h 1 , 2 , 0 i . Find the position vector function r ( t ). Solution: The position vector function r ( t ) is the integral of its derivative r 0 ( t ) = v ( t ): r ( t ) = Z v ( t ) dt = Z h sin t , cos 2 t , e t i dt = h- cos( t ) + x 0 , 1 2 sin(2 t ) + y 0 , e t + z 0 i .

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