# Stuck on second part of differentiation problem

• Dec 21st 2011, 05:33 AM
angypangy
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 $\displaystyle P = Ae^{kt}$, 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?
• Dec 21st 2011, 06:30 AM
earboth
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 = $\displaystyle P = 3e^{at-bt^2}$ , 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.<--- ähemm!

...

According to the text $\displaystyle \frac{dP}{dt}=0\ \text{if}\ t=10$

With a = 2 you'll get an equation in b.
• Dec 21st 2011, 06:31 AM
e^(i*pi)
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 = $\displaystyle P = 3e^{at-bt^2}$ , where a and b are constants.

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

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

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

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.

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

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

Not sure if that is correct so far. How would I proceed from here to find b?
You are told that there is a maximum (ie: $\displaystyle \dfrac{dP}{dt} = 0$) when $\displaystyle t=10$

Remember that $\displaystyle e^{f(t)} > 0$
• Dec 21st 2011, 06:46 AM
angypangy
Re: Stuck on second part of differentiation problem
Quote:

Originally Posted by earboth
According to the text $\displaystyle \frac{dP}{dt}=0\ \text{if}\ t=10$

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.
• Dec 22nd 2011, 04:08 AM
angypangy
Re: Stuck on second part of differentiation problem
Quote:

Originally Posted by e^(i*pi)
No problems here.

You are told that there is a maximum (ie: $\displaystyle \dfrac{dP}{dt} = 0$) when $\displaystyle t=10$

Remember that $\displaystyle e^{f(t)} > 0$

ok, so I know a = 2 from above and I know dP/dt = 0 when t=10

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

$\displaystyle (6 - 60b)e^{20-100b} = 0$

$\displaystyle 6e^{-80b} - 60be^{-80b} = 0$

$\displaystyle 6e^{-80b} = 60be^{-80b}$ ??? Is this the way to go?

$\displaystyle e^{-80b} = 10be^{-80b}$ divide by 6

$\displaystyle 1 = 10b$ divide by $\displaystyle e^{-80b}$

b = 1/10

Is that the correct procedure? It does match the answer in the book. I have never divided by a number which includes the number I want to obtain before?
• Dec 22nd 2011, 04:20 AM
e^(i*pi)
Re: Stuck on second part of differentiation problem
Quote:

Originally Posted by angypangy
ok, so I know a = 2 from above and I know dP/dt = 0 when t=10

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

$\displaystyle (6 - 60b)e^{20-100b} = 0$

$\displaystyle 6e^{-80b} - 60be^{-80b} = 0$

$\displaystyle 6e^{-80b} = 60be^{-80b}$ ??? Is this the way to go?

$\displaystyle e^{-80b} = 10be^{-80b}$ divide by 6

$\displaystyle 1 = 10b$ divide by $\displaystyle e^{-80b}$

b = 1/10

Is that the correct procedure? It does match the answer in the book. I have never divided by a number which includes the number I want to obtain before?

No, you cannot say that $\displaystyle e^{20-100b} = e^{-80b}$ because you can only combine like terms and 20 and -100b are not like terms (the former would need to be 20b to do that).

The line $\displaystyle (6-60b)e^{20-100b} = 0$ is correct. Now you can use the null factor law (as if you were solving something like $\displaystyle (x-p)(x-q) =0$). As a consequence therefore we can say that $\displaystyle \text{Either} 6-60b = 0 \text{ OR } e^{20-100b} = 0$

If you use my hint above (or plot the graph of $\displaystyle e^{20-100b}$) and note that the x-axis is an asymptote which means $\displaystyle e^{20-100b} > 0$ for all real b.

Therefore for $\displaystyle (6-60b)e^{20-100b} = 0$ can only be true when $\displaystyle 6-60b = 0$ which gives $\displaystyle b = \dfrac{1}{10}$