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Find speed of particle

  1. Dec 9, 2008 #1
    1. Find the speed (As a function of t) of a particle whose position at time t seconds is c(t)=(sint+t,cost+t)



    2. speed = [x'(t)^2 + y'(t)^2]^1/2



    3.
    x'(t) = cost + 1
    y'(t)= -sint + 1

    speed = [(cos t + 1)^2 +(-sint + 1)^2]^1/2
    =[cos^2 t+ 2cost + 1 +sin^2 t - 2sint + 1]^1/2
    =[cos^2 t + sin^2 t + 2 cost - 2sint +2]^1/2
    =[1 + 2(cost - sint + 1)]^1/2



    -Thanks
     
  2. jcsd
  3. Dec 9, 2008 #2

    Dick

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    Or (2*(cos(t)-sin(t))+3)^(1/2). That looks ok to me.
     
  4. Dec 9, 2008 #3

    HallsofIvy

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    No, Dick. It would be (2cos(t)- 2sin(t))+3)^(1/2). The "3" would not be multiplied by 2.
     
  5. Dec 9, 2008 #4

    Dick

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    No, Halls. There's an extra set of parentheses around the cos(t)-sin(t). When you cut and pasted you only deleted one of them.
     
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