Chapter 5.6, Problem 16E

a.

To determine

Tofind the height of the pile.

Answer to Problem 16E

The height is increasing by $11.19cm/\min$ .

Explanation of Solution

Given information:

Sand falls from a conveyor at the rate of $10m^3/\min$ onto the top of a conical pile. The height of the pile is always three-eights of the base diameter.

Formula:

  $1m=100cm$

Let diameter is defined as two times the radius $d=2r$ .

  $\begin{array}{l}h=\frac{3}{8}d=\frac{3}{8}\left(2r\right)=\frac{3}{4}r\\ \Rightarrow r=\frac{4}{3}h\end{array}$ ,

By using the given relationship between $h$ and $r$ to get the cone volume equation $V$ all in terms of $h$ ,

Volume of cone is $\pi r^2\frac{h}{3}$

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

Implicitly differentiate the cylinder volume equation $V$ with respect to $t$ ,

  $\begin{array}{l}\frac{dV}{dt}=\frac{\left(3\right)16}{27}\pi h^2\frac{dh}{dt}\\ \frac{dV}{dt}=\frac{16}{9}\pi h^2\frac{dh}{dt}\end{array}$

Substitute the given values,

  $\begin{array}{l}\frac{dV}{dt}=10\\ h=4\end{array}$

  $\begin{array}{l}10=\frac{16}{9}\pi {\left(4\right)}^2\frac{dh}{dt}\\ \frac{dh}{dt}=10\cdot \frac{9}{256\pi }\\ \frac{dh}{dt}=\frac{45}{128\pi }m/\min \\ \frac{dh}{dt}=\frac{4500}{128\pi }cm/\min \\ \frac{dh}{dt}=\frac{1125}{32\pi }cm/\min \\ \frac{dh}{dt}\approx 11.19cm/\min \end{array}$

Therefore,the height is increasing by $11.19cm/\min$ .

b.

To determine

To find the change in radius whenthe pile $4m$ is high.

Answer to Problem 16E

The change in radius is increased by $14.92cm/\min$ .

Explanation of Solution

Given information:

Sand falls from a conveyor at the rate of $10m^3/\min$ onto the top of a conical pile. The height of the pile is always three-eights of the base diameter.

Formula:

  $1m=100cm$

Let the diameter is $d=2r$ .

Rate of volume increased is given by $\frac{dV}{dt}=10$ and $h=4$ ,

  $\begin{array}{l}h=\frac{3}{8}d=\frac{3}{8}\left(2r\right)=\frac{3}{4}r\\ \Rightarrow r=\frac{4}{3}\left(4\right)=\frac{16}{3}\end{array}$ ,

By using the given relationship between $h$ and $r$ to get the cone volume equation $V$ all in terms of $h$ ,

Volume of cone is $\pi r^2\frac{h}{3}$

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

Implicitly differentiate the cylinder volume equation $V$ with respect to $t$ ,

  $\frac{dV}{dt}=\frac{3}{4}\pi r^2\frac{dr}{dt}$

Substitute the given values,

  $\begin{array}{l}\frac{dV}{dt}=10\\ h=4\end{array}$

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

Therefore,the change in radius is increased by $14.92cm/\min$ .