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

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

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 $2{\mathrm{m}}^3/\mathrm{m}\mathrm{i}\mathrm{n}$, the rate of change of the height of the water when the height is $10\mathrm{m}$ is

$2{\mathrm{m}}^3/\mathrm{m}\mathrm{i}\mathrm{n}=\pi {\left(\frac{5}{8}\right)}^2\cdot (10\mathrm{m}{)}^2\cdot \frac{dh}{dt}{|}_{h=10}⟹\frac{dh}{dt}{|}_{h=10}=\frac{2\cdot 8^2}{5^2\cdot 10^2\pi }\mathrm{m}/\mathrm{m}\mathrm{i}\mathrm{n}=\frac{2^5}{5^4\pi }\mathrm{m}/\mathrm{m}\mathrm{i}\mathrm{n}.$