Question

Use differential to estimate the amount of tin in a closed tin tan with diameter \(8\,cm\) and height \(12 cm\) if the tin is \(0.04 cm\) thick.

Short answer

The amount of tin in a closed tin tan is \(16\,c{m^3}\).

Step-by-step solution

Given, To Find and Formula Using.

Diameter\( = 8\,cm\)

Height \( = 12 cm\)

Thickness\( = 0.04 cm\)\(\)

The amount of tin in a closed tin tan (To find)

\(v = \pi {r^2}h\)(Formula Using)

Differential of Volume.

\(dv = \frac{{\partial v}}{{\partial r}}dr + \frac{{\partial v}}{{\partial h}}dh\)

Where \(dr\) is thickness of the sides.

\(dh\)is the thickness of top and bottom so

\(\begin{aligned}{l}dh = 0.04 + 0.04\\ = 0.08\;cm\end{aligned}\)

Step 3:- Differentiate volume with respect  to \(r\).

\(\frac{{\partial v}}{{\partial r}} = \frac{\partial }{{\partial r}}\left( {\pi {r^2}h} \right)\)

\( = \pi \left( {2r} \right)h\)

\( = 2\pi rh\)

Differentiate Volume with respect to h.

\(\frac{{\partial v}}{{\partial h}} = \frac{\partial }{{\partial h}}\left( {\pi {r^2}h} \right)\)

\( = \pi {r^2}\)

Volume.

\(dv = \frac{{\partial v}}{{\partial r}}dr + \frac{{\partial v}}{{\partial h}}dh\)

\( = 2\pi rh\left( {dr} \right) + \pi {r^2}\left( {dh} \right)\)

\( = 2\pi rh\left( {0.04} \right) + \pi {r^2}(0.08)\)

\( = 2 \times 3.14 \times 4 \times 12 \times 0.04 + 3.14 \times 4 \times 4 \times 0.08\)

\( = 12.0576 + 4.0192\)

\( = 16.0768\;c{m^3}\)

\( \approx 16\;c{m^3}\)

Therefore, the amount of tin in closed tin can is \(16\;c{m^3}\)