Question

The number of bacteria in a culture decreases according to a continuous exponential decay model. The initial population in a study is 6200 bacteria, and there are 2914 bacteria left after 7 minutes.

a. Let t be the time (in minutes) since the beginning of the study, and let y be the number of bacteria at time t. Write a formula relating y to t Use exact expressions to fill in the missing parts of the formula. Do not use approximations.
y= ___e
b. How many bacteria are there 18 minutes after the beginning of the study?

Answer 1

Answer:

a) $y(t) = 6200e^{-0.1079t}$

b) There are 889 bacteria left 18 minutes after the beginning of the study.

Step-by-step explanation:

The continuous exponential decay model is as follows:

$y(t) = y_{0}e^{-rt}$

In which $y_{0}$ is the initial population and r is the rate that the population decreases.

a. Let t be the time (in minutes) since the beginning of the study, and let y be the number of bacteria at time t. Write a formula relating y to t Use exact expressions to fill in the missing parts of the formula. Do not use approximations.

The initial population in a study is 6200 bacteria, which means that $y_{0} = 6200$.

There are 2914 bacteria left after 7 minutes, so $y(7) = 2914$.

We use this to find the value of r.

$y(t) = y_{0}e^{-rt}$

$2914 = 6200e^{-7r}$

$e^{-7r} = \frac{2914}{6200}$

$e^{-7r} = 0.47$

Applying ln to both sides

$\ln{e^{-7r}} = \ln{0.47}$

$-7r = -0.755$

Multiplying by -1

$7r = 0.755$

$r = \frac{0.755}{7}$

$r = 0.1079$

So

$y(t) = 6200e^{-0.1079t}$

b. How many bacteria are there 18 minutes after the beginning of the study?

This is y(18).

$y(t) = 6200e^{-0.1079t}$

$y(18) = 6200e^{-0.1079*18} = 889$

There are 889 bacteria left 18 minutes after the beginning of the study.

Answer 2

We want to find an equation that models the population of the given bacteria. The solutions are:

a) y = 6200*( 0.8978)^t

b) 890.

We know that the population decreases exponentially, this means that the population can be written as:

y = a*b^t

Where a is the initial population, in this case 6200, and b is the rate at which the population decreases.

We also know that after 7 minutes, the population is 2914, then we have:

2914 = 6200*b^7

Now we can solve this for b.

(2914/6200) = b^7

(2914/6200)^(1/7) = b = 0.8978

a) The equation is:

y = 6200*( 0.8978)^t

b) The population after 18 minutes is given by evaluating in t = 18

y = 6200*( 0.8978)^18 = 890.48

Rounding to the next whole number, we conclude that the population after 18 minutes is 890.

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