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Math Help - Finding Highest Velocity from different points

  1. #1
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    Finding Highest Velocity from different points

    Find the highest velocity in each section as defined in the piecewise function
    the bold is the line under "<"

    -291.67t^3+1125t^2 - 1360t+574.67 for 1 < t <1.6

    2500t^3-12750t^2+21600t-12076 for 1.6 < t < 1.8

    -176t^3+1142.4t^2-2402t+17222.6 for 1.8 < t < 2.3


    step 2 is the derivatives

    -875.01t^2+2250t-1360 for 1 < t < 1.6

    7500t^2-25500t+21600 for 1.6 < t <1.8

    -528t^2+2284.8t-2402 for 1.8 < t < 2.3

    step 3 i plugged in the points into the derivative equation

    -875.01(1)^2+2250(1)-1360 = 14.99

    7500(1.6)^2-25500(1.6)+21600 = 0

    -528(1.6)^2+2284.8(1.6)-2402 = -98

    i plugged in the other number and got 59.92

    how do i get the critical points now
    Last edited by unknown2221; November 26th 2008 at 08:18 PM.
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  2. #2
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    To find the critical points you set the derivative=0 or DNE

    I'll do one example and you should know how to do the rest

    -875.01t^2+2250t-1360=0

    Use the quadratic equation to find t.
    t=.97144573, 1.59999535

    To find the maximum velocity on the interval [1, 1.6) (includes 1 but not 1.6)

    You would find
    V(.97144573)
    v(1.59999535)
    V(1)

    And whichever is the biggest is the maximum velocity on that interval.
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  3. #3
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    Quote Originally Posted by Linnus View Post
    To find the critical points you set the derivative=0 or DNE

    I'll do one example and you should know how to do the rest

    -875.01t^2+2250t-1360=0

    Use the quadratic equation to find t.
    t=.97144573, 1.59999535

    To find the maximum velocity on the interval [1, 1.6) (includes 1 but not 1.6)

    You would find
    V(.97144573)
    v(1.59999535)
    V(1)

    And whichever is the biggest is the maximum velocity on that interval.
    can you elaborate more that made no sense to me (serious)
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  4. #4
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    To find the critical points you set the derivative=0 or DNE

    I'll do one example and you should know how to do the rest

    Your derivative is -875.01t^2+2250t-1360 (this is the derivative of the velocity right?- since that is what you gave me)

    So set it equal to 0
    and DNE to find the critical points
    -875.01t^2+2250t-1360=0
    Use the quadratic equation to find t.
    t=.97144573, 1.59999535

    -875.01t^2+2250t-1360=DNE does not have a solution which is why I didn't write anything for it

    Critical points of the velocity function exist at t=.97144573, 1.59999535.

    To find the maximum velocity on the interval [1, 1.6) (includes 1 but not 1.6) This is another way of saying to find the maximum velocity on the interval 1 to 1.6 including 1 but not 1.6

    You would find
    V(.97144573)= plug in .97144573 into the velocity (original function) NOT THE DERIVATIVE
    v(1.59999535)= plug in 1.59999535 into the velocity (original function) NOT THE DERIVATIVE
    V(1)= plug in 1 into the velocity (original function) NOT THE DERIVATIVE

    And whichever has the largest value out of the 3 is the maximum velocity on that interval.

    Is this clear enough?

    This is a calculator active problem right? Because using the quadratic equation to find t is a lot of work...
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  5. #5
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    Quote Originally Posted by Linnus View Post
    To find the critical points you set the derivative=0 or DNE

    I'll do one example and you should know how to do the rest

    Your derivative is -875.01t^2+2250t-1360 (this is the derivative of the velocity right?- since that is what you gave me)

    So set it equal to 0
    and DNE to find the critical points
    -875.01t^2+2250t-1360=0
    Use the quadratic equation to find t.
    t=.97144573, 1.59999535

    -875.01t^2+2250t-1360=DNE does not have a solution which is why I didn't write anything for it

    Critical points of the velocity function exist at t=.97144573, 1.59999535.

    To find the maximum velocity on the interval [1, 1.6) (includes 1 but not 1.6) This is another way of saying to find the maximum velocity on the interval 1 to 1.6 including 1 but not 1.6

    You would find
    V(.97144573)= plug in .97144573 into the velocity (original function) NOT THE DERIVATIVE
    v(1.59999535)= plug in 1.59999535 into the velocity (original function) NOT THE DERIVATIVE
    V(1)= plug in 1 into the velocity (original function) NOT THE DERIVATIVE

    And whichever has the largest value out of the 3 is the maximum velocity on that interval.

    Is this clear enough?

    This is a calculator active problem right? Because using the quadratic equation to find t is a lot of work...
    im just lost when you used the quadratic equation to find t...
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  6. #6
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    Quote Originally Posted by unknown2221 View Post
    im just lost when you used the quadratic equation to find t...
    Uh...I didn't show my work because I used my calculator. That equation isn't pretty at all, and solving for t using the quadratic equation will take a while. If you are allowed to use your calculator, just put it in there. =)
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  7. #7
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    Quote Originally Posted by Linnus View Post
    Uh...I didn't show my work because I used my calculator. That equation isn't pretty at all, and solving for t using the quadratic equation will take a while. If you are allowed to use your calculator, just put it in there. =)
    but i thought you can use integrals to solve this problem from the start...
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  8. #8
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    Quote Originally Posted by unknown2221 View Post
    but i thought you can use integrals to solve this problem from the start...
    Why would you think that?

    These equations do represent the velocity right?
    -291.67t^3+1125t^2 - 1360t+574.67 for 1 < t <1.6
    2500t^3-12750t^2+21600t-12076 for 1.6 < t < 1.8
    -176t^3+1142.4t^2-2402t+17222.6 for 1.8 < t < 2.3

    If so, everything I've done is right.
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