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Let the diameter of the spherical ball be \(d\). Given, \(d = \frac{5}{2}(3x+1)\). Therefore, the radius \(r\) is:
\(r = \frac{d}{2} = \frac{5}{4}(3x+1)\)
The volume \(V\) of the spherical ball is given by:
\(V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \left(\frac{5}{4}(3x+1)\right)^3 = \frac{4}{3}\pi \left(\frac{125}{64}(3x+1)^3\right) = \frac{125\pi}{48}(3x+1)^3\)
Now, we need to find the rate of change of volume with respect to \(x\), which is \(\frac{dV}{dx}\):
\(\frac{dV}{dx} = \frac{125\pi}{48} \cdot 3(3x+1)^2 \cdot 3 = \frac{125\pi}{16}(3x+1)^2 \cdot 3 = \frac{375\pi}{16}(3x+1)^2\)
We need to find the rate of change when \(x=1\):
\(\frac{dV}{dx}\Big|_{x=1} = \frac{375\pi}{16}(3(1)+1)^2 = \frac{375\pi}{16}(4)^2 = \frac{375\pi}{16} \cdot 16 = 375\pi\)
Correct Answer: \(375\pi\)