Question

Use differentials to estimate the amount of metal in a closed cylindrical can that is 26 cm high and 10 cm in diameter if the metal in the top and the bottom is 0.3 cm thick and the metal in the sides is 0.05 cm thick. (Round your answer to two decimal places.)

Answer 1

Answer:

The estimated amount of metal in the can is 87.96 cubic cm

Step-by-step explanation:

We can find the differential of volume from the volume of a cylinder equation given by

$V= \pi r^2 h$

Thus that way we will find the amount of metal that makes up the can.

Finding the differential.

A small change in volume is given by:

$dV =\cfrac{\partial V}{\partial h} dh + \cfrac{\partial V}{\partial r} dr$

So finding the partial derivatives we get

$dV =\pi r^2 dh + \pi 2r h dr$

$dV =\pi r^2 dh + 2\pi r h dr$

Evaluating the differential at the given information.

The height of the can is h = 26 cm, the diameter is 10 cm, which means the radius is half of it, that is r = 5 cm.

On the other hand the thickness of the side is 0.05 cm that represents dr = 0.05 cm, and the thickness on both top and bottom is 0.3 cm, thus dh = 0.3 cm +0.3 cm which give us 0.6 cm.

Replacing all those values on the differential we get

$dV =\pi 5^2 (0.6) + 2\pi (5) (26) (0.05)$

That give us

$V= 28 \pi \, cm^3$

Or in decimal value

$\boxed{dV= 87.96 \, cm^3}$

Thus the volume of metal in the can is 87.96 cubic cm.