Question

Use differentials to estimate the amount of tin in a closed tin can with diameter 3 inch and height 4 inch, if the top and bottom are 0.02 inch thick and the side is 0.015 inch thick.

Answer 1

Answer:

dv = 1.03 inch^3

Explanation:

given data:

diameter = 3 inch

radius = 1.5 inch

height 4 inch

top and bottom thickness is 0.02 inch

side thickness = 0.015 inch

we know that volume of the cylinder is given as

$v =\pi r^2 h$

by definition of differential we have

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

where dh = -(0.02  + 0.02) = 0.04 inch    [ sum of top and bottom thickness]

the radius is decreased by 0.02 inch, dr = 0.02 inc,

$\frac{\partial v}{\partial r} = 2\pi r h = 37.69$

$\frac{\partial v}{\partial h} = \pi r^2 = 7.06$

dv = 37.69*(0.02) + 7.06*(0.04)

dv = 1.03 inch^3