Answer:
the amount of material in a closed cylindrical can is 22.61944 [tex]cm^{3}[/tex]
Step-by-step explanation:
Given:
The volume of cylindrical can by diameter is:
[tex]V = \frac{\pi}{4}\cdot D^{2}\cdot l[/tex]
We use differentials to estimate:
[tex]\Delta V = \frac{\partial V}{\partial D} \cdot \Delta D + \frac{\partial V}{\partial L} \cdot \Delta L[/tex]
The partial derivatives are presented hereafter:
[tex]\frac{\partial V}{\partial D} = \frac{\pi}{2}\cdot D\cdot l[/tex]
= [tex]\frac{\partial V}{\partial D} = \frac{\pi}{2}\cdot (12\,cm)\cdot (30\,cm)[/tex]
=[tex]\frac{\partial V}{\partial D} \approx 565.487\,cm^{2}[/tex]
= [tex]\frac{\partial V}{\partial l} = \frac{\pi}{4}\cdot D^{2}[/tex]
= [tex]\frac{\partial V}{\partial l} = \frac{\pi}{4}\cdot (12\,cm)^{2}[/tex]
= [tex]\frac{\partial V}{\partial l} = \frac{\pi}{4}\cdot (12\,cm)^{2}[/tex]
Hence, the estimated amount of material is:
[tex]\Delta V = (565.487\,cm^{2})\cdot (0.02\,cm) + (113.097\,cm^{2})\cdot (0.1\,cm)[/tex]
= 22.61944 [tex]cm^{3}[/tex]
So the amount of material in a closed cylindrical can is 22.61944 [tex]cm^{3}[/tex]
We have that for the Question "" it can be said that the amount of material in a closed cylindrical can is
[tex]dv=158.34[/tex]
From the question we are told
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.
Generally the equation for the volume of the cylinder is mathematically given as
[tex]V=\pir^2h\\\\Therefore\\\\dv=\frac{dv}{dr}dr+\frac{dv}{dh}dh\\\\dv=(2\pirh)dr+(\pir^2} dh\\\\Therefore\\\\dv=(2*\pi * 6*30)(0.1)+(\pir^2} (0.2+0.2)\\\\[/tex]
[tex]dv=158.34[/tex]
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