Science:Math Exam Resources/Courses/MATH100/December 2015/Question 08 (b)/Solution 1

Let $r$ be the radius of the circle at the top of the water and $h$ be the height of the water.

Then, using the picture in part (a) with a property of similar triangles gives the ratio between $r$ and $h$ as

${\frac {r}{h}}={\frac {10}{16}}\implies r={\frac {5}{8}}h.$

On the other hand, by the volume formula for a cone, the volume $V$ of water can be written as

$V={\frac {1}{3}}\pi r^{2}h={\frac {1}{3}}\pi \cdot \left({\frac {5}{8}}h\right)^{2}\cdot h={\frac {1}{3}}\pi \cdot \left({\frac {5}{8}}\right)^{2}\cdot h^{3}.$

Therefore, by differentiating with respect to $t$ on both sides, we obtain

${\frac {dV}{dt}}={\frac {1}{3}}\pi \cdot \left({\frac {5}{8}}\right)^{2}\cdot 3h^{2}\cdot {\frac {dh}{dt}}=\pi \cdot \left({\frac {5}{8}}\right)^{2}\cdot h^{2}\cdot {\frac {dh}{dt}}.$

Since the volume changes at a rate of $2\,\mathrm {m} ^{3}/\mathrm {min}$, the rate of change of the height of the water when the height is $10\,\mathrm {m}$ is

$2\,\mathrm {m} ^{3}/\mathrm {min} =\pi \left({\frac {5}{8}}\right)^{2}\cdot (10\,\mathrm {m} )^{2}\cdot {\frac {dh}{dt}}{\bigg |}_{h=10}\implies {\frac {dh}{dt}}{\bigg |}_{h=10}={\frac {2\cdot 8^{2}}{5^{2}\cdot 10^{2}\pi }}\,\mathrm {m} /\mathrm {min} =\color {blue}{{\frac {2^{5}}{5^{4}\pi }}\,\mathrm {m} /\mathrm {min} }.$