Question

Problem PageQuestion The area covered by a certain population of bacteria increases according to a continuous exponential growth model. Suppose that a sample culture has an initial area of and an observed doubling time of days. (a)Let be the time (in days) passed, and let be the area of the sample at time . Write a formula relating to . Use exact expressions to fill in the missing parts of the formula. Do not use approximations. (b)What will the area of the sample be in days

Answer 1

Answer:

a) The relation between y and t is

$y(t)=8.6e^{ln(2)\cdot t/5}$

b) y(6) = 19.8 mm^2

Step-by-step explanation:

The question is incomplete: "The area covered by a certain population of bacteria increases according to a continuous exponential growth model. Suppose that a sample culture has an initial area of 8.6 mm^2 and an observed doubling time of 5 days. (a)Let be the time (in days) passed, and let y be the area of the sample at time . Write a formula relating y to t. Use exact expressions to fill in the missing parts of the formula. Do not use approximations. (b)What will the area of the sample be in 6 days?

As the area covered increases according to a continous growth model, we start with these relation:

$y(t)=Ce^{kt}$

We know that the initial area y(0) is 8.6. Then:

$y(0)=Ce^0=C=8.6\\\\C=8.6$

We also know that the area duplicates after 5 days.

$y(t+5)=2y(t)\\\\\frac{y(0+5)}{y(0)} =2=\frac{e^{0+5k}}{e^0} =e^{5k}\\\\5k=ln(2)\\\\k=ln(2)/5\approx0.14$

Then, the model becomes:

$y(t)=8.6e^{ln(2)\cdot t/5}$

b) We have to calculate y(6)

$y(6)=8.6e^{ln(2)\cdot 6/5}=8.6e^{ 0.832}=8.6*2.297=19.758$