Question

The volume of a cylindrical can with a radius r and height h is given by
`V=\pi r^(2) h`

You are wondering how inaccuracies in the radius or height of your can will effect the volume of that can. Use differentials to answer the questions below.
An error of p percent in the radius will cause an error of approximately _______ percent in the volume of the can.
An error of p percent in the height will cause an error of approximately _____ percent in the volume of the can.

answer will depend on p
.

Answer 1

An error of p percent in the radius will cause an error of approximately 2p percent in the volume of the can.

To see why, we use differentials. We have:

v = pi r^2 h

Taking the differential with respect to r, we get:

dv = 2pi rh dr + pi r^2 dh

Dividing both sides by v, we get:

dv/v = 2(r/h)dr + (r^2/h)dh

If we assume that the percentage error in the radius is p, then dr = (p/100)r. Substituting this into the previous equation, we get:

dv/v = 2(p/100)(r/h) + (r^2/h)dh

The first term represents the error due to the radius, and the second term represents the error due to the height.

If we assume that the percentage error in the height is p, then dh = (p/100)h. Substituting this into the previous equation, we get:

dv/v = 2(r/h)dr + (p/100)(r^2/h)

= (2r/h)(p/100)r + (r^2/h)(p/100)

= (p/100)(2r^2/h + r^2/h)

= (p/100)(3r^2/h)

Therefore, an error of p percent in the height will cause an error of approximately (3p/100) percent in the volume of the can