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- Homework Statement
- Let p(t) denote the number of bacteria at time t measured so that the carrying capacity is reached when p = 1. The model of p(t) is:

$$\frac{dp}{dt}=\frac{k}{c} p(t)(1-p(t)^{c})$$

where k and c > 0

What is the value ##p_{max}## of ##p## does ##\frac{dp}{dt}## is maximum? What does changing the value of c do to the ##p_{max}##?

- Relevant Equations
- Integration and Differentiation

To find maximum of ##\frac{dp}{dt} \rightarrow \frac{d^{2} p}{dt^{2}}=0##

$$\frac{k}{c}p'(t)(1-p(t)^{c})-\frac{k}{c}p(t)(c.p(t)^{c-1}).p'(t)=0$$

$$p'(t)=0~\text{or}~p(t)=\left(\frac{1}{c+1}\right)^{\frac{1}{c}}$$

For p'(t) = 0 → p(t) = 0 or p(t) = 1

So I have 3 values of p(t), how to know which one will result in maximum ##\frac{dp}{dt}##?

I tried using sign diagram for 2nd derivative but the term ##\frac{k}{c}p'(t)(1-p(t)^{c})## and ##\frac{k}{c}p(t)(c p(t)^{c-1}).p'(t)## both are always positive so I can't determine the sign of their subtraction

Thanks

$$\frac{k}{c}p'(t)(1-p(t)^{c})-\frac{k}{c}p(t)(c.p(t)^{c-1}).p'(t)=0$$

$$p'(t)=0~\text{or}~p(t)=\left(\frac{1}{c+1}\right)^{\frac{1}{c}}$$

For p'(t) = 0 → p(t) = 0 or p(t) = 1

So I have 3 values of p(t), how to know which one will result in maximum ##\frac{dp}{dt}##?

I tried using sign diagram for 2nd derivative but the term ##\frac{k}{c}p'(t)(1-p(t)^{c})## and ##\frac{k}{c}p(t)(c p(t)^{c-1}).p'(t)## both are always positive so I can't determine the sign of their subtraction

Thanks

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