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Thread: 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,
    $\displaystyle z=\sqrt{x^2+y^2}$
    when $\displaystyle \frac{dx}{dt} = -5sin (t) $ , $\displaystyle \frac{dy}{dt} = 12 cos (t)$ , $\displaystyle x(0) = 5 $ , $\displaystyle y(0)=0$

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

    Can someone help me explain the significance of this answer (when $\displaystyle \frac{d^2 z}{dt^2} = \frac{-110}{12}$ at $\displaystyle t=\frac{\pi}{2}$) ????
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  2. #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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