Question

Use differentials to estimate the amount of tin in a closed tin can with diameter 12 cm and height 11 cm if the tin is 0.04 cm thick.

Answer 1

Answer:

16.65 cm³

Step-by-step explanation:

Given that :

the diameter of the closed tin can = 12 cm

the height of the closed tin can = 11 cm

The thickness of the closed tin can  = 0.04 cm

The main task required to be carried out is to use differentials to estimate the amount of tin in a closed tin can.

We all know that the volume of a tin can = $\pi r ^ 2 h$

The differential of the volume   is :

$dV = \dfrac{\partial V}{\partial r}dr + \dfrac{\partial V}{\partial h} dh \ ----- \ (1)$

Differentiating V  with respect to r; we have:

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

Differentiating V with respect to h ; we have:

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

Hence; the estimate of the amount of tin in a closed tin can is:

= $2 \pi rh \ dr+ \pi r ^2 \ dh$

= $\pi r(2h \ dr+ r \ dh)$

= $\pi *6 (2 * 11 * 0.04 + 0.04 * 0.08)$

where;

dr = increase in radius = 0.04cm (since thickness of the radius = 0.04 cm)

dh = increase in the height = 0.04 + 0.04 = 0.08

= 16.65 cm³

Hence; the amount of tin in a closed tin can  is = 16.65 cm³