Chapter 5.6, Problem 41E

To determine

To find: the cylinder volume increasing when the diameter is $3.800\text{ in}\text{.}$

Answer to Problem 41E

The cylinder volume is $0.0238\text{ in}{\text{.}}^3\text{/min}\text{.}$

Explanation of Solution

Given information :

Height is 6 inch deep.

  $\frac{dr}{dt}=\frac{0.001}{3}$

  $h=6$

  $D=3.800\text{ in}.$

Assume all variables are differentiable function of t .

Calculation:

We have to calculate cylinder volume increasing when the diameter is $3.800\text{ in}\text{.}$

Therefore,

Volume of a cylinder is

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

Since

  $D=3.800\text{ in}.$

Then,

  $\begin{array}{l}2r=3.800\text{ in}.\\ r=\frac{3.800}{2}\\ r=1.9\end{array}$

Putting the value of $h=6$ in equation (1),

  $\text{V}=6\pi r^2$

Differentiate with respect to t using product rule.

  $\frac{dv}{dt}=6\pi \left[2r\frac{dr}{dt}\right]$

Putting the value of $r=1.9$ and $\frac{dr}{dt}=\frac{0.001}{3}$ in above equation,

  $\begin{array}{l}\frac{dv}{dt}=6\pi \left[2\left(1.9\right)\frac{0.001}{3}\right]\\ \frac{dv}{dt}=6\pi \left[3.8\times \frac{0.001}{3}\right]\\ \frac{dv}{dt}=6\pi \left[1.26\times 0.001\right]\\ \frac{dv}{dt}=6\pi \left[0.00126\right]\\ \frac{dv}{dt}=0.00756\pi \\ \frac{dv}{dt}=0.00756\times 3.14\left[\pi =3.14\right]\\ \frac{dv}{dt}=0.0238\text{ in}{\text{.}}^3/\min .\end{array}$

Hence, the cylinder volume is $0.0238\text{ in}{\text{.}}^3\text{/min}\text{.}$