Question

When a bactericide is added to a nutrient broth in which bacteria are​ growing, the bacterium population continues to grow for a​ while, but then stops growing and begins to decline. The size of the population at time t​ (hours) is b equals 8 Superscript 6 Baseline plus 8 Superscript 5 Baseline t minus 8 Superscript 4 Baseline t squared .b=86+85t−84t2. Find the growth rates at t equals 0 hours commat=0 hours, t equals 4 hours commat=4 hours, and t equals 8 hours.t=8 hours.

Answer 1

Answer:

The growth rates at t = 0 is 8⁵.

The growth rates at t = 4 is 0.

The growth rates at t = 8 is -8⁵.

Step-by-step explanation:

The expression representing the size of the population at time t​ hours is:

$b(t)=8^{6}+8^{5}t-8^{4}t^{2}$

Differentiate b (t) with respect to t to determine the growth rate as follows:

$\frac{db(t)}{dt}=\frac{d}{dt} (8^{6}+8^{5}t-8^{4}t^{2})$

       $=0+8^{5} (1)-8^{4}(2t)\\=8^{4}(8-2t)$

The growth rate is:

R (t) = 8⁴ (8 - 2t)

Compute the growth rates at t = 0 as follows:

$R (0) = 8^{4} (8 - 2\times 0)\\=8^{4}\times 8\\=8^{5}$

Thus, the growth rates at t = 0 is 8⁵.

Compute the growth rates at t = 4 as follows:

$R (4) = 8^{4} (8 - 2\times 4)\\=8^{4}\times 0\\=0$

Thus, the growth rates at t = 4 is 0.

Compute the growth rates at t = 8 as follows:

$R (4) = 8^{4} (8 - 2\times 8)\\=8^{4}\times (-8)\\=-8^{5}$

Thus, the growth rates at t = 8 is -8⁵.