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Math Help - Maximum and Minimub values need help!

  1. #1
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    Maximum and Minimub values need help!

    Need help with this question, dunno what to do. Please help me

    The position function of an object that moves in a straight line is s(t)=1+2t - \frac {8}{t^2+1}, 0<t<2. Calculate the maximum and minimum velocities of the object over the given time interval.
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    Quote Originally Posted by haddad287 View Post
    Need help with this question, dunno what to do. Please help me

    The position function of an object that moves in a straight line is s(t)=1+2t - \frac {8}{t^2+1}, 0<t<2. Calculate the maximum and minimum velocities of the object over the given time interval.
    Velocity is ds/dt. So differentiate. Max and min values in the time interval will occur either at stationary points (set ds/dt=0, solve for t and sub to find s at these points). Also find s(0) and s(2) ie the end points. Then compare the values and pick the biggest and smallest.
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  3. #3
    Junior Member asemh's Avatar
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    I think you first have to derive it, set it to zero, find the x's and plug it into the main function to find the minimum and maximum values
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    Is this the correct differiantion of it?

    s'(t)=-16(t^2+1)^-2
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    Quote Originally Posted by haddad287 View Post
    Is this the correct differiantion of it?

    s'(t)=-16(t^2+1)^-2
    2 + 16t(t^2+1)^-2
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  6. #6
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    where did u get the 2 from?

    Here is what I did
    =1+8(t^2+1)^-1
    =1+8(-1)(t^2+1)^-2(2t)
    =1+8(-2t)(t^2+1)^-2
    =1-16t(t^2+1)^-2
    =16t(t^2+1)^-2
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  7. #7
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    Quote Originally Posted by haddad287 View Post
    where did u get the 2 from?

    Here is what I did
    =1+8(t^2+1)^-1
    =1+8(-1)(t^2+1)^-2(2t)
    =1+8(-2t)(t^2+1)^-2
    =1-16t(t^2+1)^-2
    =16t(t^2+1)^-2
    Your original question had a 2t term.
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  8. #8
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    ok, but I took the (2t) and multiplied with the (-1) which gave me (-2t). That is this step here

    =1+8(-1)(t^2+1)^-2(2t)
    =1+8(-2t)(t^2+1)^-2
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    Quote Originally Posted by haddad287 View Post
    ok, but I took the (2t) and multiplied with the (-1) which gave me (-2t). That is this step here

    =1+8(-1)(t^2+1)^-2(2t)
    =1+8(-2t)(t^2+1)^-2
    Go back to your original question. It has 3 terms.
    Differentiate the 1 to get 0.
    Differentiate the 2t to get 2. (This is where the 2 comes from)
    Differentiate the last term to get.....see previous post
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  10. #10
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    ohh ok thanks. ill finish rest of question and see how i do >.<
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  11. #11
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    Quote Originally Posted by haddad287 View Post
    ohh ok thanks. ill finish rest of question and see how i do >.<
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  12. #12
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    stuck again! >.<

    0 = 2 + 16t(t^2+1)^-2

    help? how do I get t by itself?
    Last edited by haddad287; March 27th 2010 at 04:11 PM.
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  13. #13
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    I tried! I got down to the lowest part before solving for t, I dont think tis possiable.

    I did this

    0=2+16t(t^2+1)^-2

    -2=\frac {16t}{(t^2+1)^2}

    -2(t^2+1)^2=16t
    -(t^2+1)^2=8t
    -(t^4+2t^2+1)=8t
    -t^4+2t^2-1=8t
    -t^4-2t^2+8t+1=0

    well???
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  14. #14
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    v(t) = 2+\frac{16t}{(t^2+1)^2}

    note that v(t) > 0 for all values of t in the given interval 0 < t < 2.

    to find the max and min velocity ...

    v'(t) = \frac{16(1-3t^2)}{(t^2+1)^3}

    v'(t) = 0 at t = \sqrt{\frac{1}{3}}

    v'(t) > 0 for t < \sqrt{\frac{1}{3}}

    and

    v'(t) < 0 for t > \sqrt{\frac{1}{3}}


    therefore, v(t) is a maximum at t = \sqrt{\frac{1}{3}}

    v\left(\sqrt{\frac{1}{3}}\right) is the maximum velocity in the given interval.

    the minimum velocity would occur at one of the two endpoint times ( t = 0 or t = 2) if they are included in the interval.

    recheck the given interval to see if it is really 0 \le t \le 2.
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