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

Answer:

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

Step-by-step explanation:

Given

$V = \frac{4}{3}\pi r^3$

$Radius,\ r = 1.4cm$

$Error = 0.005cm$

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

$V(r) = \frac{4}{3}\pi r^3$ --- Volume as a function of radius

The relationship between the above parameters is as follows;

$\frac{dV}{dr} = V'(r)$

This can be rewritten as

$\frac{dV}{dr} = (V(r))'$

Substitute $\frac{4}{3}\pi r^3$ for $V(r)$

$\frac{dV}{dr} = (\frac{4}{3}\pi r^3)'$

Multiply both sides by dr

$dr * \frac{dV}{dr} = (\frac{4}{3}\pi r^3)' * dr$

$dV = (\frac{4}{3}\pi r^3)' * dr$

$dV = \frac{4}{3}\pi( r^3)' dr$

Differentiate

$dV = \frac{4}{3}\pi( r^3)' dr$

$dV = \frac{4}{3}\pi* 3r^2\ dr$

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

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