Hello Kavina,
Let $V(t)$ be the volume in $\text{cm}^3$ of water in the cylindrical barrel at time $t$ in minutes, and $h$ be the depth in $\text{cm}$ of the water in the barrel.
We are given:
$\displaystyle \frac{dV}{dt}=-k\sqrt{h}$ where $0<k$ in the constant of proportionality.
The formula for the volume of a cylinder is:
$\displaystyle V=\pi r^2h$
Differentiating this with respect to time (and observing that $\pi r^2$ will remain constant) we find:
$\displaystyle \frac{dV}{dt}=\pi r^2\frac{dh}{dt}$
Now, equating the two expressions for $\displaystyle \frac{dV}{dt}$ we have:
$\displaystyle \pi r^2\frac{dh}{dt}=-k\sqrt{h}$
$\displaystyle \frac{dh}{dt}=-\frac{k}{\pi r^2}\sqrt{h}$
Now, we may redefine the constant of proportionality and write the IVP:
$\displaystyle \frac{dh}{dt}=-k\sqrt{h}$ where $h(0)=29,\,h(3)=25$.
The ODE associated with the IVP is separable:
$\displaystyle h^{-\frac{1}{2}}\,dh=-k\,dt$
Integrate:
$\displaystyle \int h^{-\frac{1}{2}}\,dh=-k\int\,dt$
$\displaystyle 2h^{\frac{1}{2}}=-kt+C$
Use initial conditions to find parameter $C$
$\displaystyle 2(29)^{\frac{1}{2}}=-k(0)+C$
$\displaystyle C=2\sqrt{29}$
and so we have:
$\displaystyle 2h^{\frac{1}{2}}=-kt+2\sqrt{29}$
Now, use other given point to determine the constant of proportionality $k$:
$\displaystyle 2(25)^{\frac{1}{2}}=-k(3)+2\sqrt{29}$
$\displaystyle k=\frac{2(\sqrt{29}-5)}{3}$
Hence, we may write:
$\displaystyle t=\frac{3(\sqrt{29}-\sqrt{h})}{\sqrt{29}-5}$
Now, to find when the barrel will be empty, we may let $h=0$ and we find:
$\displaystyle t(0)=\frac{3\sqrt{29}}{\sqrt{29}-5}$
Since "now" is given to be $t=3$ we must subtract 3 from this to find the answer to the question:
$\displaystyle \frac{3\sqrt{29}}{\sqrt{29}-5}-3=\frac{3\sqrt{29}-3(\sqrt{29}-5)}{\sqrt{29}-5}=\frac{15}{\sqrt{29}-5}$
