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Math Help - Derivatives

  1. #1
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    Derivatives


    1st one: differentiate y=(x^3/2)+(48/x)

    2nd one: The fixed point A has coordinates (8, -6,5) and the variable point P has coordinates (t,t2t).
    a. Show that AP^2=6t^2-24t+125
    b. Hence find the value of t for which the distance AP is least.
    c. Determine this least distance.
    thank u
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  2. #2
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    Quote Originally Posted by thereddemon View Post

    1st one: differentiate y=(x^3/2)+(48/x)

    2nd one: The fixed point A has coordinates (8, -6,5) and the variable point P has coordinates (t,t2t).
    a. Show that AP^2=6t^2-24t+125
    b. Hence find the value of t for which the distance AP is least.
    c. Determine this least distance.
    thank u
    y = \frac{x^3}{2} + 48 x^{-1}
    y' = \frac{3x^2}{2} - 48 x^{-2}

    For the second

    AP^2 = (t - 8)^2+(t+6)^2+(2t-5)^2 expand this.

    For the second part, take the derivative and set this to zero. I'm sure you can do the rest.
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  3. #3
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    wow fast response! thank u so much! u rock
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