Question

the volume of a solid can be expressed as ( v = 7x^{3} ). the volume is to be calculated with an error of no more than 1% of the true value. find approximately the greatest error that can be tolerated in the measurement of x, expressed as a percentage of x.

the greatest tolerated error in the measurement of x is (square%).
(type an integer or a simplified fraction.)

Explanation:

Step1: Differentiate the volume formula

Given \( V = 7x^{3} \), using the power rule \( (x^{n})^\prime=nx^{n - 1} \), we get \( dV=21x^{2}dx \).

Step2: Find the relative error formula

The relative error in \( V \) is \( \frac{dV}{V} \), and the relative error in \( x \) is \( \frac{dx}{x} \). Substitute \( V = 7x^{3} \) and \( dV = 21x^{2}dx \) into \( \frac{dV}{V} \):

$$ \frac{dV}{V}=\frac{21x^{2}dx}{7x^{3}} = 3\frac{dx}{x} $$

Step3: Solve for \( \frac{dx}{x} \)

We know that \( \frac{dV}{V}\leq 0.01 \). Since \( \frac{dV}{V}=3\frac{dx}{x} \), then \( 3\frac{dx}{x}\leq 0.01 \). Solving for \( \frac{dx}{x} \), we get \( \frac{dx}{x}\leq\frac{0.01}{3}=\frac{1}{300}\approx 0.0033\).

Answer:

\(\frac{1}{3}\%\)