Science:Math Exam Resources/Courses/MATH104/December 2016/Question 13/Solution 1

Let ${\textstyle h}$ be the height of the water and ${\textstyle r}$ the corresponding radius. Then the volume would be given by:

$$V={\frac {1}{3}}\pi h(r^{2}+10r+100).$$

Since we know that ${\frac {dV}{dt}}=1000{\text{cm}}^{3}/{\text{min}}$, we’ll need to replace ${\textstyle h}$ or ${\textstyle r}$. Let’s find an equation relating them:

From which we see that

$$h=2r-20.$$

Thus, our equation becomes:

$$V={\frac {1}{3}}\pi h(r^{2}+10r+100)={\frac {\pi }{3}}(2r-20)(r^{2}+10r+100)={\frac {\pi }{3}}(2r^{3}-2000)$$

Then, applying the chain rule, we have

$${\frac {dV}{dt}}={\frac {\pi }{3}}\cdot 6r^{2}{\frac {dr}{dt}},$$

and then we plug ${\frac {dV}{dt}}=1000$ and solve for ${\frac {dr}{dt}}$ to get

$$1000=2\pi r^{2}{\frac {dr}{dt}}\implies {\frac {dr}{dt}}{\bigg |}_{r=25,h=30}={\frac {500}{25^{2}\cdot \pi }}={\frac {4}{5\pi }}$$

Here, $r=25$ when $h=30$ follows from the relation $h=2r-20.$.

Since ${\textstyle h=2r-20}$, we finally get

$${\frac {dh}{dt}}{\bigg |}_{r=25,h=30}=2{\frac {dr}{dt}}{\bigg |}_{r=25,h=30}={\frac {8}{5\pi }}$$

Answer: $\color {blue}{\frac {8}{5\pi }}{\text{cm}}/{\text{min}}$