Question

An oil tank in the form of a right circular cylinder of radius of r has a height h of 34 meters and a volume of V =34πr2 . The radius is measured as 15 meters with a maximum possible error of ±0.15 meters. Estimate the propogated and relative errors in the calculated volume of the tank.

Answer 1

Assuming that the height is a fixed value

`V = \pi r^2 h`

Since h = 34 then we have
`V = 34 \pi r^2`

We can find the derivate of V respect to the radius like this:

`dV = 68 \pi r dr`

And if we replace when r = 15 m we got this:

`dV= 68 \pi (15) (\pm 0.15) = \pm 480.664 m^3`

Assuming that the height is variable and assuming dr=dh =0.15

`V = \pi r^2 h`

We can find the derivate of V respect to the radius like this:

`dV = (2\pi rh) dr+ (\pi r^2) dh`

And if we replace when r = 15 m, h = 14m , dr=dh =0.15 we got this:

`dV = (2\pi 15m*34 m) (\pm 0.15m)+ (\pi (15m)^2) (\pm 0.15m)= 153\pi +33.75 \pi = 586.692 m^3`

Explanation:

Data given

h = 34 represent the height for th cone

r = 15m represent the radius for the cylinder

`dr = \pm 0.15` represent the error for the radius

Solution for the problem

We want to estimate the propagated and relative error in the calculated volume of the tank. For this first we need to remember that the volume for a cylinder is given by:

Assuming that the height is a fixed value

`V = \pi r^2 h`

Since h = 34 then we have
`V = 34 \pi r^2`

We can find the derivate of V respect to the radius like this:

`dV = 68 \pi r dr`

And if we replace when r = 15 m we got this:

`dV= 68 \pi (15) (\pm 0.15) = \pm 480.664 m^3`

Assuming that the height is variable and assuming dr=dh =0.15

`V = \pi r^2 h`

We can find the derivate of V respect to the radius like this:

`dV = (2\pi rh) dr+ (\pi r^2) dh`

And if we replace when r = 15 m, h = 14m , dr=dh =0.15 we got this:

`dV = (2\pi 15m*34 m) (\pm 0.15m)+ (\pi (15m)^2) (\pm 0.15m)= 153\pi +33.75 \pi = 586.692 m^3`