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Math Help - Significance of dz/dt=0 for z=sqrt(x^2 + y^2)

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    Significance of dz/dt=0 for z=sqrt(x^2 + y^2)

    For the curve in the (x,y) plane,
    z=\sqrt{x^2+y^2}
    when \frac{dx}{dt} = -5sin (t)      , \frac{dy}{dt} = 12 cos (t) , x(0) = 5  , y(0)=0

    I calculated \frac{dz}{dt} = \frac{12 y cos (t) - 5 x sin (t)}{\sqrt{x^2+y^2}}=0 at the point when t=\frac{\pi}{2} , x=0, y= 12.

    Can someone help me explain the significance of this answer (when \frac{d^2 z}{dt^2} = \frac{-110}{12} at t=\frac{\pi}{2}) ????
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    Re: Significance of dz/dt=0 for z=sqrt(x^2 + y^2)

    If a second derivative is negative at a critical point, then you have a local maximum.
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