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 = 6^5 + 6^4t - 6^3t^2. find the growth rates at t = 0 hours, t = 3 hours, and t = 6 hours.

Explanation:

Step1: Find the derivative of the population function

The population function is $b = 6^{5}+6^{4}t - 6^{3}t^{2}$. The derivative $b^\prime(t)$ using the power - rule $\frac{d}{dt}(x^n)=nx^{n - 1}$ and $\frac{d}{dt}(C)=0$ (where $C$ is a constant) is $b^\prime(t)=\frac{d}{dt}(6^{5})+\frac{d}{dt}(6^{4}t)-\frac{d}{dt}(6^{3}t^{2})$. Since $\frac{d}{dt}(6^{5}) = 0$, $\frac{d}{dt}(6^{4}t)=6^{4}$, and $\frac{d}{dt}(6^{3}t^{2})=2\times6^{3}t$, we have $b^\prime(t)=6^{4}-2\times6^{3}t$.

Step2: Evaluate at $t = 0$

Substitute $t = 0$ into $b^\prime(t)$: $b^\prime(0)=6^{4}-2\times6^{3}\times0=6^{4}$.

Step3: Evaluate at $t = 3$

Substitute $t = 3$ into $b^\prime(t)$: $b^\prime(3)=6^{4}-2\times6^{3}\times3=1296 - 2\times216\times3=1296-1296 = 0$.

Step4: Evaluate at $t = 6$

Substitute $t = 6$ into $b^\prime(t)$: $b^\prime(6)=6^{4}-2\times6^{3}\times6=1296-2\times216\times6=1296 - 2592=- 6^{4}$.

Answer:

The growth rate at $t = 0$ hours is $6^{4}$ bacteria per hour.
The growth rate at $t = 3$ hours is $0$ bacteria per hour.
The growth rate at $t = 6$ hours is $-6^{4}$ bacteria per hour.