Question

yo why isn't this deleted yo

Answer 1

The rate of change of the diameter is given by the function
\[D'(x)=0.2\sqrt{x}+0.04x^{0.5}\]
Strange that you would be given the function in this way, since \(\sqrt{x}=x^{0.5}\). Assuming this isn't a typo,
\[D'(x)=0.24\sqrt{x}\]
This is a separable differential equation:
\[\frac{dD}{dx}=0.24\sqrt{x}\\
dD=0.24x^{1/2}~dx\\
\int dD=0.24\int x^{1/2}~dx\\
D(x)=0.24 \frac{x^{3/2}}{\frac{3}{2}}+C\\
D(x)=0.16x^{3/2}+C\]
\(D(x)\) is the diameter of an average leaf at time \(x\). If "the beginning of January" is the initial time (week 0), you're given that \(D(0)=6\):
\[6=0.16(0)^{3/2}+C~~\Rightarrow~~C=6\]
Now, find \(D(12)\).
\[D(12)=0.16(12)^{3/2}+6\approx 12.651\]