Chapter 5.6, Problem 42E

a.

To determine

To find: how fast is the level in the pot rising when the coffee in the cone in 5 inch deep.

Answer to Problem 42E

The value of $\frac{dh}{dt}$ is $0.35\text{ in}\text{./min}\text{.}$

Explanation of Solution

Given information:

  $h=5\text{ inches}$ .

  $\frac{dv}{dt}=10\text{ in}{\text{.}}^3\text{/min}\text{.}$

From the given diagram, we can see that the conical diameter is 6, cylindrical coffee pot diameter is 6 and the conical filter height is 6.

Calculation :

We have to calculate how fast is the level in the pot rising when the coffee in the cone in 5 inch deep.

Therefore,

Volume of a cylindrical coffee pot is

  $\text{V}=\pi r^{2}h$

Since radius is constant, we put $r=3$ in the above equation.

  $\begin{array}{l}\text{V}=\pi {\left(3\right)}^{2}h\left[\begin{array}{l}\text{Since, D=6}\\ \text{2}r=6\\ r=\frac{6}{2}=3\end{array}\right]\\ \text{V}=9\pi h\end{array}$

Differentiate with respect to t .

  $\frac{dv}{dt}=9\pi \frac{dh}{dt}$

Putting the value of $\frac{dv}{dt}=10$ in above equation,

  $\begin{array}{l}10=9\pi \frac{dh}{dt}\\ \frac{dh}{dt}=\frac{10}{9\pi }\\ \frac{dh}{dt}=\frac{10}{9\times 3.14}\left[\text{Since, }\pi =3.14\right]\\ \frac{dh}{dt}\approx \frac{10}{28.26}\\ \frac{dh}{dt}\approx 0.35\text{ in}\text{./min}\text{.}\end{array}$

Hence, the value of $\frac{dh}{dt}$ is $0.35\text{ in}\text{./min}\text{.}$

b.

To determine

To find: how fast is the level in the cone falling.

Answer to Problem 42E

The value of $\frac{dh}{dt}$ is $0.5095\text{ in}\text{./min}\text{.}$

Explanation of Solution

Given information:

  $h=5\text{ inches}$ .

  $\frac{dv}{dt}=10\text{ in}{\text{.}}^3\text{/min}\text{.}$

From the given diagram, we can see that the conical diameter is 6, cylindrical coffee pot diameter is 6 and the conical filter height is 6.

Calculation :

We have to calculate how fast is the level in the cone falling.

Therefore,

Volume of a cone filter is

  $\text{V}=\frac{1}{3}\pi r^{2}h\mathrm{.....}\left(1\right)$

We use similar triangles of a cone.

We get

  $\begin{array}{l}\frac{r}{h}=\frac{3}{6}\\ r=\frac{3h}{6}\\ r=\frac{h}{2}\end{array}$

Putting the value of $r=\frac{h}{2}$ in equation (1),

  $\begin{array}{l}\text{V}=\frac{1}{3}\pi {\left(\frac{h}{2}\right)}^{2}h\\ \text{V}=\frac{1}{3}\pi \frac{h^2}{4}h\\ \text{V}=\frac{1}{12}\pi h^3\end{array}$

Differentiate with respect to t .

  $\frac{dv}{dt}=\frac{1}{12}\pi \left(3\cdot h^2\frac{dh}{dt}\right)$

Putting the value of $\frac{dv}{dt}=10$ and $h=5$ in above equation,

  $\begin{array}{l}10=\frac{1}{12}\pi \left[3\cdot {\left(5\right)}^2\frac{dh}{dt}\right]\\ 10=\frac{1}{12}\pi \left[3\cdot \left(25\right)\frac{dh}{dt}\right]\\ 10=\frac{1}{12}\pi \left[75\frac{dh}{dt}\right]\\ \frac{dh}{dt}=\frac{10\times 12}{75\pi }\\ \frac{dh}{dt}=\frac{120}{75\times 3.14}\left[\text{Since, }\pi =3.14\right]\\ \frac{dh}{dt}=\frac{120}{235.5}\\ \frac{dh}{dt}=0.5095\text{ in}\text{./min}\text{.}\end{array}$

Hence, value of $\frac{dh}{dt}$ is $0.5095\text{ in}\text{./min}\text{.}$