Stuck on second part of differentiation problem

**angypangy** · December 21st 2011, 05:33 AM

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 $P = Ae^{kt}$ , where A and k are constants.

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

let u = kt du/dt = k

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

To find values of A and k.

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

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

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

A=3, k=2

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

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

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

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

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 = $P = 3e^{at-bt^2}$ , where a and b are constants.

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

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

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

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.

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

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

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

**earboth** · December 21st 2011, 06:30 AM

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 = $P = 3e^{at-bt^2}$ , where a and b are constants.

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

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

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

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

...

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

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

**e^(i*pi)** · December 21st 2011, 06:31 AM

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 = $P = 3e^{at-bt^2}$ , where a and b are constants.

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

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

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

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

No problems here.

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

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

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

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: $\dfrac{dP}{dt} = 0$ ) when $t=10$

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

**angypangy** · December 21st 2011, 06:46 AM

Originally Posted by earboth

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

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.

**angypangy** · December 22nd 2011, 04:08 AM

Originally Posted by e^(i*pi)

No problems here.

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

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

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

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

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

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

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

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

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

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?

**e^(i*pi)** · December 22nd 2011, 04:20 AM

Originally Posted by angypangy

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

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

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

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

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

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

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

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 $e^{20-100b} = e^{-80b}$ 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 $(6-60b)e^{20-100b} = 0$ is correct. Now you can use the null factor law (as if you were solving something like $(x-p)(x-q) =0$ ). As a consequence therefore we can say that $\text{Either} 6-60b = 0 \text{ OR } e^{20-100b} = 0$

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

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

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