Question

The edge of a cube was found to be 30 cm with a possible error in measurement of 0.4 cm. Use differentials to estimate the maximum possible error, relative error, and percentage error in computing the volume of the cube and the surface area of the cube. (Round your answers to four decimal places.)(a) the volume of the cube (b) the surface area of the cube

Answer 1

Explanation:

Hello

The volume of a cube is the cube of the length of one of the sides

a)
`V = x^3` where x is the length of the sides for such cube:

if we differentiate we get :
`(dV)/(dx) = 3*x^2`, now if we want to find the maximum error we approximate
`(dV)/(dx)` to
`(\Delta V)/(\Delta x)` which is a rude approximation.

now if we solve for
`\Delta V` we get:
`\Delta V = 3x^2*\Delta x`

now we replace x with 30 and
`\Delta x` with 0.4 and we get
`\Delta V = 3*(30)^2*0.4 = 1080`cm^3.

the relative error is given by dividing our maximum error by the volume V:
`(1080)/(30^3)= 0.0400`

the percentage error will be the product of this value with 100 (in order to get a percentage)
`0.0400*100 = 4\%`

So having this value, the volume of the cube will be 27000cm³
`\pm` 1080 cm³

b)

The area is six times the area of a single square:

`A = 6*x^2` where x is the length of the sides for such cube:

if we differentiate we get :
`(dA)/(dx) = 6*(2)*x = 12x`, now if we want to find the maximum error we approximate
`(dA)/(dx)` to
`(\Delta A)/(\Delta x)` which is a rude approximation.

now if we solve for
`\Delta A` we get:
`\Delta A = 12x*\Delta x`

now we replace x with 30 and
`\Delta x` with 0.4 and we get
`\Delta A = 12*(30)*0.4 = 144`cm^2.

the relative error is given by dividing our maximum error by the surface area A:
`(144)/(6*30^2)  = 0.8`

the percentage error will be the product of this value with 100 (in order to get a percentage)
`0.8*100 = 80\%`

Therefore, the area will be 5400cm^2
`\pm` 144cm^2