Stuck on second part of differentiation problem

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 , where A and k are constants.

(a) write down in terms of A, k and t. Find the values of A and k.

let u = kt du/dt = k

To find values of A and k.

P = Ae^kt, when P=3, t=0 so

A=3, k=2

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

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 = , where a and b are constants.

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

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.

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

Re: Stuck on second part of differentiation problem

Quote:

Originally Posted by

**angypangy** ...

(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 =

, where a and b are constants.

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

When t=0, P=3, and dP/dt = 6 6 = 3ae^0,

**a = 5/3 = 2.**<--- ähemm!
...

According to the text

With a = 2 you'll get an equation in b.

Re: Stuck on second part of differentiation problem

Quote:

Originally Posted by

**angypangy** (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 =

, where a and b are constants.

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

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

No problems here.

Quote:

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

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

You are told that there is a maximum (ie: ) when

Remember that

Re: Stuck on second part of differentiation problem

Quote:

Originally Posted by

**earboth** According to the text

With a = 2 you'll get an equation in b.

Ah yes I missed the maximum when t = 10 bit. Ah yes that makes it easier to solve. Thanks. 5/3 was a type.

Re: Stuck on second part of differentiation problem

Re: Stuck on second part of differentiation problem