Question

Use differentials to estimate the amount of metal in a closed cylindrical can that is 30 cm high and 6 cm in diameter if the metal in the top and the bottom is 0.2 cm thick and the metal in the sides is 0.05 cm thick. (Round your answer to two decimal places.)

Answer 1

Answer:

$dV=113.55 \ cm^3$

Explanation:

The Differential of Multivariable Functions

Given a multivariable function V(r,h), the total differential of V is computed by

$dV=\displaystyle \frac{\partial V}{\partial r} dr+\frac{\partial V}{\partial h} dh$

The volume of a cylinder of radius r and height h is

$V=\pi\cdot r^2\cdot h$

Let's compute the partial derivatives

$\frac{\partial V}{\partial r}=2\pi rh$

$\displaystyle \frac{\partial V}{\partial h}=\pi r^2$

The total differential is

$dV=(2\pi rh)dr+(\pi r^2)dh$

The differential dr is approximated to $\Delta r$ and the differential dh is approximated to $\Delta h$. We can see the increment of radius is the thick of the metal in the sides, and the increment of the height is the thick of the metal in the top and bottom. Thus dr=0.05 cm, dr=0.2 cm

$dV=(2\pi \cdot 3\ cm\cdot 30\ cm)0.2\ cm+(\pi\ (3\ cm)^2)0.05\ cm$

$dV=113.55 \ cm^3$