Question

Use differentials to estimate the amount of material in a closed cylindrical can that is 30 cm high and 12 cm in diameter if the metal in the top and bottom is 0.2 cm thick, and the metal in the sides is 0.1 cm thick. Note, you are approximating the volume of metal which makes up the can (i.e. melt the can into a blob and measure its volume), not the volume it encloses.

Answer 1

Answer:

the amount of material in a closed cylindrical can is  22.61944 $cm^{3}$

Step-by-step explanation:

Given:

• Height: 30 cm
• Diameter: 12  
• Top and bottom is 0.2 cm thick
• The sides is 0.1 cm thick

The volume of cylindrical can by diameter is:

$V = \frac{\pi}{4}\cdot D^{2}\cdot l$

We use differentials to estimate:

$\Delta V = \frac{\partial V}{\partial D} \cdot \Delta D + \frac{\partial V}{\partial L} \cdot \Delta L$

The partial derivatives are presented hereafter:

$\frac{\partial V}{\partial D} = \frac{\pi}{2}\cdot D\cdot l$

= $\frac{\partial V}{\partial D} = \frac{\pi}{2}\cdot (12\,cm)\cdot (30\,cm)$

=$\frac{\partial V}{\partial D} \approx 565.487\,cm^{2}$

= $\frac{\partial V}{\partial l} = \frac{\pi}{4}\cdot D^{2}$

= $\frac{\partial V}{\partial l} = \frac{\pi}{4}\cdot (12\,cm)^{2}$

= $\frac{\partial V}{\partial l} = \frac{\pi}{4}\cdot (12\,cm)^{2}$

Hence, the estimated amount of material is:

$\Delta V = (565.487\,cm^{2})\cdot (0.02\,cm) + (113.097\,cm^{2})\cdot (0.1\,cm)$

= 22.61944 $cm^{3}$

So the amount of material in a closed cylindrical can is  22.61944 $cm^{3}$