Chapter 4.6, Problem 16E

(a)

To determine

To find : How fast are the height.

Answer to Problem 16E

The answer: Increasing by $11.19\text{cm}/\text{min}$

Explanation of Solution

Given information:

A conveyor belt at the rate of $10{\text{m}}^3/\text{min}$

Calculation:

A diameter is defined as 2 times the radius

  $d=2r$

  $\begin{array}{c}h=\frac{3}{8}d\\ =\frac{3}{8}(2r)\\ =\frac{3}{4}r\\ r=\frac{4}{3}h\end{array}$

Use the given relationship between $h$ and $r$ to get the cone volume equation $V$ all in terms of $h$ :

  $\begin{array}{c}V=\frac{1}{3}\pi r^{2}h\\ =\frac{1}{3}\pi {\left(\frac{4}{3}h\right)}^{2}h\\ =\frac{16}{27}\pi h^3\end{array}$

So,

  $\begin{array}{c}\frac{dV}{dt}=\frac{(3)16}{27}\pi h^2\frac{dh}{dt}\\ \Rightarrow \frac{dV}{dt}=\frac{16}{9}\pi h^2\frac{dh}{dt}\\ $ \Rightarrow 10=\frac{16}{9}\pi {(4)}^2\frac{dh}{dt}$\Rightarrow \frac{dh}{dt}=10\cdot \frac{9}{256\pi }\\ =\frac{45}{128\pi }\text{m}/\text{min}\\ =\frac{4500}{128\pi }\text{cm}/\text{min}\\ =\frac{1125}{32\pi }\text{cm}/\text{min}\\ \approx 11.19\text{cm}/\text{min}\end{array}$

Therefore the required increasing by $11.19\text{cm}/\text{min}$ fast are the height.

(b)

To determine

To find: How fast are the radius changing when the pile is $4m$ high.

Answer to Problem 16E

The answer: Increasing by $14.92\text{cm}/\text{min}$

Explanation of Solution

Given information:

A conveyor belt at the rate of $10{\text{m}}^3/\text{min}$

Calculation:

Diameter is $2r$

Given: rate of volume increase and relationship between $h$ and $r$

Looking for $\frac{dh}{dt}$ , when $h=4$ ,

Plug in $h=4$ to find $r$ at the same moment for use later.

  $\begin{array}{c}\frac{dV}{dt}=10\\ \Rightarrow h=\frac{3}{8}(2r)\\ =\frac{3}{4}r\\ \Rightarrow (4)=\frac{3}{4}r\\ \Rightarrow r=\frac{4}{3}\cdot 4\\ =\frac{16}{3}\end{array}$

Convert the cone volume equation $\text{V}$ to $\text{r}$

Applying the Power Rule, discriminate implicitly.

  $\begin{array}{c}V=\frac{1}{2}\pi r^{2}h\\ =\frac{1}{3}\pi r^2\left(\frac{3r}{4}\right)\\ =\frac{1}{4}\pi r^3\end{array}$

So,

  $\begin{array}{c}\frac{dV}{dt}=\frac{3}{4}\pi r^2\frac{dr}{dt}\\ \Rightarrow \frac{3}{4}\pi r^2\frac{dr}{dt}=\frac{dV}{dt}\\ \Rightarrow \frac{3}{4}\pi {\left(\frac{16}{3}\right)}^2\frac{dr}{dt}=10\\ \Rightarrow \frac{dr}{dt}=10\cdot \frac{3}{64\pi }\\ =\frac{15}{32\pi }\text{m}/\text{min}\cdot \frac{100\text{cm}}{1\text{m}}\\ =\frac{1500}{32\pi }\text{cm}/\text{min}\\ =\frac{375}{8\pi }\text{cm}/\text{min}\\ \approx 14.92\text{cm}/\text{min}\end{array}$

Therefore the required increasing by $14.92\text{cm}/\text{min}$ fast are the radius changing when the pile is $4m$ high.