Start of problem I can do but I have to explain this part to show variable values.

In a biological experiment, bacteria are being grown in a culture. The mass of the bacteria at time t hours is P milligrams. At time t = 0, P=3 and dy/dx = 6.

(i) A standard model for this situation is given by $\displaystyle P = Ae^{kt}$, where A and k are constants.

(a) write down $\displaystyle \frac{dP}{dt}$ in terms of A, k and t. Find the values of A and k.

let u = kt du/dt = k

$\displaystyle \frac{dP}{dt} = \frac{dP}{du} * \frac{dP}{du} = Ake^{kt}$

To find values of A and k.

P = Ae^kt, when P=3, t=0 so $\displaystyle 3 = Ae^0 A = 3$

$\displaystyle \frac{dP}{dt} = 3ke^{kt}$

$\displaystyle 6 = 3ke^0 Therefore k = \frac{6}{3} = 2$

A=3, k=2

(b) Find the value of t for which P=12.

$\displaystyle P = 12 = 3e^{2t}$

$\displaystyle 12 = 3e^{2t} = \frac{12}{3} = 4$

$\displaystyle 2t = ln4 t = \frac{ln4}{2} = \frac{ln2^2}{2} = \frac{2 * ln2}{2} = ln2$

t = ln2.

(c) According to this model, what happens to the mass of the bacteria after a long time?

After a long time the mass of bacteria would get infinitely large.

OK, that is the first part of the problem and all my answers agree with book answers. So now onto the second part.

(ii) As the experiment progresses it is found that the value of P increases to a maximum at time t = 10 and then begins to decrease. A new model is proposed in which P = $\displaystyle P = 3e^{at-bt^2}$ , where a and b are constants.

(a) Express dP/dt in terms of a, b and t. Find the values of a and b.

$\displaystyle let u = at-bt^2 \frac{du}{dt} = a - 2bt$

$\displaystyle \frac{dP}{du} = 3e^u \frac{dP}{dt} = (3a - 6bt)e^{at-bt^2}$

When t=0, P=3, and dP/dt = 6 6 = 3ae^0, a = 5/3 = 2.

So the way I thought to find b was to use the previous result of t=ln2 when dP/dt = 12.

$\displaystyle 12 = (6 - 6bln2)e^{2ln2-b(ln2)^2}$

$\displaystyle \frac{12}{6 - 6ln2} = \frac{2}{1-ln2} =e^{2ln2-b(ln2)^2}$

Not sure if that is correct so far. How would I proceed from here to find b?