Question

the population of bacteria in a petri dish is given by p(t)=16cdot3^{t - 2} where t represents the number of hours after the initial number of bacteria was measured. if there is an error in measuring time of 0.1 minutes, what is the error in measuring the population of bacteria 10 hours after the initial population was measured? round your final answer to two decimal places. error in measuring bacteria = bacteria

Explanation:

Step1: Find the actual population at \(t = 10\) hours

Substitute \(t = 10\) into the population - growth formula \(P(t)=16\cdot3^{t - 2}\).
\[

$$\begin{align*} P(10)&=16\cdot3^{10 - 2}\\ &=16\cdot3^{8}\\ &=16\times6561\\ &=104976 \end{align*}$$

\]

Step2: Find the measured population

The time - measurement error is \(\Delta t=0.1\) minutes. Since \(1\) hour \( = 60\) minutes, \(\Delta t=\frac{0.1}{60}=\frac{1}{600}\) hours. The measured time \(t_{m}=10+\frac{1}{600}\) hours.
\[

$$\begin{align*} P(t_{m})&=16\cdot3^{(10+\frac{1}{600})- 2}\\ &=16\cdot3^{8+\frac{1}{600}}\\ &=16\cdot3^{8}\cdot3^{\frac{1}{600}} \end{align*}$$

\]
We know that \(a^{x + y}=a^{x}\cdot a^{y}\), and using the approximation \(a^{h}\approx1 + h\ln(a)\) for small \(h\). Here \(a = 3\) and \(h=\frac{1}{600}\), so \(3^{\frac{1}{600}}\approx1+\frac{1}{600}\ln(3)\).
\[

$$\begin{align*} P(t_{m})&\approx16\cdot3^{8}\cdot(1+\frac{1}{600}\ln(3))\\ &=104976\cdot(1+\frac{\ln(3)}{600}) \end{align*}$$

\]
\(\ln(3)\approx1.0986\), then \(\frac{\ln(3)}{600}\approx\frac{1.0986}{600}=0.001831\).
\[

$$\begin{align*} P(t_{m})&\approx104976\times(1 + 0.001831)\\ &=104976+104976\times0.001831\\ &=104976 + 192.211\\ &\approx105168.21 \end{align*}$$

\]

Step3: Calculate the error

The error \(\text{Error}=|P(t_{m})-P(10)|\).
\[

$$\begin{align*} \text{Error}&\approx105168.21 - 104976\\ &=192.21 \end{align*}$$

\]

Answer:

\(192.21\)