Solving the equation for r:
[tex]\begin{gathered} V=400(9.025\cdot10^{-5}-r^2) \\ r^2=9.025\cdot10^{-5}-\frac{V}{400} \\ r=\sqrt[]{9.025\cdot10^{-5}-\frac{V}{400}} \end{gathered}[/tex]
With the first equations, we can establish some limits for V:
With the lowest value for r (r=0):
[tex]\begin{gathered} V=400(9.025\cdot10^{-5}-0^2) \\ V=400(9.025\cdot10^{-5}) \\ V=3.61\cdot10^{-2} \end{gathered}[/tex]
With the highest value for r (r=9.5x10^-3)
[tex]\begin{gathered} V=400(9.025\cdot10^{-5}-(9.5\cdot10^{-3})^2) \\ V=400(9.025\cdot10^{-5}-9.025\cdot10^{-5}) \\ V=400(0) \\ V=0 \end{gathered}[/tex]
According to the radius range, velocity can be between 0 and 3.61x10^-2
It is also necessary to check the domain of the function considering it is a square root. The argument of an square root cannot be less than 0. Then:
[tex]\begin{gathered} 9.025\cdot10^{-5}-\frac{V}{400}\ge0 \\ 9.025\cdot10^{-5}\ge\frac{V}{400} \\ V\leq400(9.025\cdot10^{-5}) \\ V\leq3.61\cdot10^{-2} \end{gathered}[/tex]
This is the same limit for velocity obtained before. Then, we can say for velocity that:
[tex]0\leq V\leq3.61\cdot10^{-2}[/tex]
