Science:Math Exam Resources/Courses/MATH100/December 2010/Question 03/Solution 1

Let $V_{1}$, $r_{1}$, and $h_{1}$ be the volume, radius, and depth of the water in the large pool and let $V_{2}$, $r_{2}$, and $h_{2}$ be the volume, radius, and depth of the water in the small pool. We know that $r_{1}=8$m and $r_{2}=5$m. Since the pools are being filled at the same rate (in $m^{3}/min$),

${\frac {dV_{1}}{dt}}={\frac {dV_{2}}{dt}}.$

We know that ${\frac {dh_{2}}{dt}}=0.5m/min$ and we want to find ${\frac {dh_{1}}{dt}}$.

By the volume formula for a cylinder,

$V_{1}=\pi \,r_{1}^{2}h_{1}$ and $V_{2}=\pi \,r_{2}^{2}h_{2}$.

Note that $r_{1}$ and $r_{2}$ are constants. The variables of the problem are $V_{1},V_{2},h_{1},$ and $h_{2}.$

Since ${\frac {dV_{1}}{dt}}={\frac {dV_{2}}{dt}}$,

$\pi \,r_{1}^{2}{\frac {dh_{1}}{dt}}=\pi \,r_{2}^{2}{\frac {dh_{2}}{dt}}$.

Hence, the water depth in the the larger pool is increasing at a rate of

${\frac {dh_{1}}{dt}}={\frac {r_{2}^{2}}{r_{1}^{2}}}{\frac {dh_{2}}{dt}}={\frac {(5m)^{2}}{(8m)^{2}}}{\frac {1}{2}}m/min={\frac {25}{128}}m/min$.