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 $m^{3} / m i n$ ),

\frac{d V_{1}}{d t} = \frac{d V_{2}}{d t} .

We know that $\frac{d h_{2}}{d t} = 0.5 m / m i n$ and we want to find $\frac{d h_{1}}{d t}$ .

By the volume formula for a cylinder,

V_{1} = π r_{1}^{2} h_{1}

and

V_{2} = π 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 $\frac{d V_{1}}{d t} = \frac{d V_{2}}{d t}$ ,

π r_{1}^{2} \frac{d h_{1}}{d t} = π r_{2}^{2} \frac{d h_{2}}{d t}

.

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

\frac{d h_{1}}{d t} = \frac{r_{2}^{2}}{r_{1}^{2}} \frac{d h_{2}}{d t} = \frac{(5 m)^{2}}{(8 m)^{2}} \frac{1}{2} m / m i n = \frac{25}{128} m / m i n

.