ice is forming on a pond at a rate given by \n\\(\\frac{dy}{dt}=\\frac{\\sqrt{t}}{2}\\) inches per hour…

Explanation:

Step1: Integrate to find thickness

We know that $\frac{dy}{dt}=\frac{\sqrt{t}}{2}=\frac{t^{\frac{1}{2}}}{2}$. Integrating with respect to $t$ gives $y=\int\frac{t^{\frac{1}{2}}}{2}dt$. Using the power - rule for integration $\int t^n dt=\frac{t^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $y=\frac{1}{2}\times\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C=\frac{1}{2}\times\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{1}{3}t^{\frac{3}{2}}+C$. Since at $t = 0$, $y=0$, then $C = 0$.

Step2: Calculate thickness at $t = 13$

Substitute $t = 13$ into $y=\frac{1}{3}t^{\frac{3}{2}}$. So $y=\frac{1}{3}\times13^{\frac{3}{2}}=\frac{1}{3}\times13\times\sqrt{13}\approx\frac{1}{3}\times13\times3.606\approx15.62$ inches.

Step3: Find rate of increase at $t = 13$

The rate of change of the thickness is given by $\frac{dy}{dt}=\frac{\sqrt{t}}{2}$. Substitute $t = 13$ into $\frac{dy}{dt}$. Then $\frac{dy}{dt}=\frac{\sqrt{13}}{2}\approx\frac{3.606}{2}\approx1.80$ inches per hour.

Answer:

(a) 15.62 inches (b) 1.80 inches per hour