Question

the volume v of a spherical cell of radius r is given by $v(r)=\frac{4}{3}pi r^{3}$. if you can determine the radius to within an accuracy of 9%, how accurate is your calculation of the volume? use linear approximations. the calculation of the volume is accurate to within $square$%.

Explanation:

Step1: Find the derivative of \(V(r)\)

The volume function is \(V(r)=\frac{4}{3}\pi r^{3}\). Using the power rule \((x^{n})^\prime = nx^{n - 1}\), the derivative \(V^\prime(r)=\frac{4}{3}\pi\times3r^{2}=4\pi r^{2}\).

Step2: Use the linear - approximation formula \(\Delta V\approx V^\prime(r)\Delta r\)

Divide both sides by \(V(r)\):
\(\frac{\Delta V}{V}\approx\frac{V^\prime(r)\Delta r}{V(r)}\).
Substitute \(V(r)=\frac{4}{3}\pi r^{3}\) and \(V^\prime(r) = 4\pi r^{2}\) into the right - hand side:
\(\frac{V^\prime(r)\Delta r}{V(r)}=\frac{4\pi r^{2}\Delta r}{\frac{4}{3}\pi r^{3}}\).
Simplify the fraction \(\frac{4\pi r^{2}\Delta r}{\frac{4}{3}\pi r^{3}}=\frac{3\Delta r}{r}\).

Step3: Substitute the relative error of \(r\)

Given that \(\frac{\Delta r}{r}=0.09\) (since the radius is determined to within an accuracy of \(9\%\), i.e., the relative error of \(r\) is \(9\% = 0.09\)).
Then \(\frac{\Delta V}{V}\approx3\times\frac{\Delta r}{r}\).

Answer:

The calculation of the volume is accurate to within \(27\%\).