Question

The volume of a spherical cancerous tumor is given by the following equation. V(r) =(4/3)pi r^3 If the radius of a tumor is estimated at 1.4 cm, with a maximum error in measurement of 0.005 cm, determine the error that might occur when the volume of the tumor is calculated:___________.

Answer 1

The error that might occur is
`\±0.123cm^3`

Explanation:

Given

`V = (4)/(3)\pi r^3`

`Radius,\ r = 1.4cm`

`Error = 0.005cm`

Required

Determine the error that might occur in the volume of the tumor

Given that there's an error in measurement, this question will be solved using the concept of differentiation;

First, we'll rewrite the given parameters in differentiation notations;

`r = 1.4` --- Radius

`dV = \±0.005` --- Change in measurement

`V(r) = (4)/(3)\pi r^3` --- Volume as a function of radius

The relationship between the above parameters is as follows;

`(dV)/(dr) = V'(r)`

This can be rewritten as

`(dV)/(dr) = (V(r))'`

Substitute
`(4)/(3)\pi r^3` for
`V(r)`

`(dV)/(dr) = ((4)/(3)\pi r^3)'`

Multiply both sides by dr

`dr * (dV)/(dr) = ((4)/(3)\pi r^3)' * dr`

`dV = ((4)/(3)\pi r^3)' * dr`

`dV = (4)/(3)\pi( r^3)' dr`

Differentiate

`dV = (4)/(3)\pi( r^3)' dr`

`dV = (4)/(3)\pi* 3r^2\ dr`

`dV = 4\pi* r^2\ dr`

Substitute the values of r, dr and take
`\pi` as 3.142

`dV = 4 * 3.142* 1.4^2 * \±0.005`

`dV = \±0.1231664`

`dV = \±0.123` (Approximated)

Hence, the error that might occur is ±0.123