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If the speed distribution, in m/s, of a flow is given by $v = 2x^3 + 2y^2 - 3z$, then the acceleration of the fluid at the point with coordinates $[2, 1, 5]$, in meters, will be greater than $20, \text{m/s}^2 $.

How can I prove that?

I tried to apply the definition of material derivative by expressing velocity in terms of magnitude and a unit vector, but I was unsuccessful. I suspect that some form of maximization must be applied, but I don't know which one. Because, depending on the direction of velocity, acceleration changes but remains restricted to a range of values.

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    $\begingroup$ $v = 2x^3 + 2y^2 - 3z$ is a single number, so it isn’t the velocity field. Did you mean that it’s the speed field? $\endgroup$
    – Ghoster
    Commented Dec 11 at 13:40
  • $\begingroup$ Yes. This question is from a civil service examination, so in my opinion it was designed to be answered quickly. Therefore, I don't think you need the velocity field to find the maximum and minimum possible acceleration. $\endgroup$
    – lucas bublitz
    Commented Dec 11 at 13:50
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    $\begingroup$ Then please edit your post to correct your incorrect description of $v$. An unclear question cannot be clarified by a comment. $\endgroup$
    – Ghoster
    Commented Dec 11 at 13:53
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    $\begingroup$ I'm not sure that the claim is true. Suppose the velocity of the fluid is everywhere parallel to the y axis. Then its speed at $(2,1,5)$ is $3$ m/s, and its acceleration is $4y \frac {dy}{dt} = 12$ m/s/s. $\endgroup$
    – gandalf61
    Commented Dec 11 at 18:29
  • $\begingroup$ In direct relation to what gandalf61 points out: it would be better if you cite the precise problem, including if possible the name of the source, page number, etc. $\endgroup$
    – Amit
    Commented Dec 11 at 18:33

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