Question

Use differentials to estimate the amount of metal in a closed cylindrical can that is \(10\)cm high and \(4\)cm is diameter if the metal in the top and bottom is \(0.1\)cm thick and the metal in the side is \(0.05\) thick.

Short answer

The amount if metal in the closed cylindrical can required is \(8.792c{m^3}\).

Step-by-step solution

Diff \(V\) and then partial differentiating \(V\) w.r.t \(r\) and \(h\)

Let \(r\) and \(h\) be the radius and height of cylinder

Diff (1) both sides

\(dV = \frac{{\partial V}}{{\partial r}}dr + \frac{{\partial V}}{{\partial h}}dh\)

So, \(\frac{{\partial V}}{{\partial r}} = 2\pi rh\)

\(\frac{{\partial V}}{{\partial h}} = \pi {r^2}\)

So,

Substituting \(r\), \(h\), \(dr\)and \(dh\)in (2):

\(dV = 2\pi rh.dr + \pi {r^2}.dh\)

Given ,

\(\begin{aligned}{l}r = 2,h = 10\\dr = 0.05,dh = 0.1 + 0.1 = 0.2\end{aligned}\)

(As can is thick from both top and bottom)

Hence, the amount if metal required in cylindrical can is \(8.792\)\(c{m^3}\).