Derivative Problem

[Jason76]

Maybe this is actually an integral one:

The population P of Bacteria grows according to the equation: \(\displaystyle \dfrac{dP}{dt} = kP\) where \(\displaystyle k\) is a constant, and \(\displaystyle t\) is measured in hours. If the population of bacteria doubles every 24 hours, then what is the value of \(\displaystyle k\)?

Any starting hints?

 

[Bob Brown MSEE]

2=e^(r 24)

This exponential e^(rt) goes from 1 to 2 in 24hrs. (choosing the most convenient P)

(this assumes that you already know the solution is an exponential)

 

[Deleted member 4993]

Maybe this is actually an integral one:

The population P of Bacteria grows according to the equation: \(\displaystyle \dfrac{dP}{dt} = kP\) where \(\displaystyle k\) is a constant, and \(\displaystyle t\) is measured in hours. If the population of bacteria doubles every 24 hours, then what is the value of \(\displaystyle k\)?

Any starting hints?

\(\displaystyle \displaystyle \frac{dP}{dt} = kP \)

\(\displaystyle \displaystyle \frac{dP}{P} = k \ dt \) ..... integrating

\(\displaystyle \displaystyle ln (P) = kt + C_1\)

\(\displaystyle \displaystyle P = P_o*e^{kt}\)

at t = 24 → \(\displaystyle \dfrac{P}{P_o} \ = \ 2\) → \(\displaystyle \displaystyle 2 = e^{k*24}\) → solve for k

 

[DrPhil]

Maybe this is actually an integral one: Not really . . .

The population P of Bacteria grows according to the equation: \(\displaystyle \dfrac{dP}{dt} = kP\) where \(\displaystyle k\) is a constant, and \(\displaystyle t\) is measured in hours. If the population of bacteria doubles every 24 hours, then what is the value of \(\displaystyle k\)?

Any starting hints?

After you "separate the variables" the left side is \(\displaystyle \dfrac{dP}{P}\), which is \(\displaystyle P^{-1}\), the one power that doesn't use the power rule - it integrates to \(\displaystyle \ln P\). The right side is \(\displaystyle k\ dt\), which is the 0th power of \(\displaystyle t\). Generally I just think of a constant integrating to the 1st power, without consciously invoking the power law.

EDIT - I assumed your first comment referred to previous discussion of using a power law when integrating. That rule works for ANY power - integer or not - EXCEPT the power \(\displaystyle -\)1. So "integer" power rule is not a good name anyhow.

example: \(\displaystyle \displaystyle \int \sqrt{x}\ dx = \int x^{1/2}\ dx = \dfrac{2}{3}x^{3/2} + C\)

 

[MarkFL]

Maybe this is actually an integral one:

The population P of Bacteria grows according to the equation: \(\displaystyle \dfrac{dP}{dt} = kP\) where \(\displaystyle k\) is a constant, and \(\displaystyle t\) is measured in hours. If the population of bacteria doubles every 24 hours, then what is the value of \(\displaystyle k\)?

Any starting hints?

This is what's called an ordinary differential equation (ODE). You have already been shown how to solve it using separation of variables, so here's another technique. If we write the ODE in standard linear form, we have:

\(\displaystyle \dfrac{dP}{dt}-kP=0\)

Multiply through by \(\displaystyle e^{-kt}\) (the integrating factor):

\(\displaystyle e^{-kt}\dfrac{dP}{dt}-ke^{-kt}P=0\)

Now, observe that the left side may be written as the differentiation of a product:

\(\displaystyle \dfrac{d}{dt}\left(e^{-kt}P \right)=0\)

Integrating with respect to \(\displaystyle t\) we get:

\(\displaystyle e^{-kt}P=C\)

Solve for \(\displaystyle P(t)\):

\(\displaystyle P(t)=Ce^{kt}\)

Now, if we define \(\displaystyle P(0)=P_0\), we find:

\(\displaystyle P(0)=C=P_0\)

and so we have:

\(\displaystyle P(t)=P_0e^{kt}\)

Now you may solve for \(\displaystyle k\) using the hints above.

 

[Jason76]

This is what's called an ordinary differential equation (ODE). You have already been shown how to solve it using separation of variables, so here's another technique. If we write the ODE in standard linear form, we have:

\(\displaystyle \dfrac{dP}{dt}-kP=0\)

Multiply through by \(\displaystyle e^{-kt}\) (the integrating factor):

\(\displaystyle e^{-kt}\dfrac{dP}{dt}-ke^{-kt}P=0\)

Now, observe that the left side may be written as the differentiation of a product:

\(\displaystyle \dfrac{d}{dt}\left(e^{-kt}P \right)=0\)

Integrating with respect to \(\displaystyle t\) we get:

\(\displaystyle e^{-kt}P=C\)

Solve for \(\displaystyle P(t)\):

\(\displaystyle P(t)=Ce^{kt}\)

Now, if we define \(\displaystyle P(0)=P_0\), we find:

\(\displaystyle P(0)=C=P_0\)

and so we have:

\(\displaystyle P(t)=P_0e^{kt}\)

Now you may solve for \(\displaystyle k\) using the hints above.

Thanks a lot. But I'm wondering why they need differential equations on an exam for testing out of Calculus I

 

[Deleted member 4993]

Thanks a lot. But I'm wondering why they needdifferential equations on an exam for testing out of Calculus I

You do not needDE - you can solve it by simple integration, as I have shown above.

The method shown by Mark - through DE - is more general way of solving the problem.

 

[Jason76]

\(\displaystyle \displaystyle \frac{dP}{dt} = kP \)

\(\displaystyle \displaystyle \frac{dP}{P} = k \ dt \) ..... integrating

\(\displaystyle \displaystyle ln (P) = kt + C_1\)

\(\displaystyle \displaystyle P = P_o*e^{kt}\)

at t = 24 → \(\displaystyle \dfrac{P}{P_o} \ = \ 2\) → \(\displaystyle \displaystyle 2 = e^{k*24}\) → solve for k

\(\displaystyle \displaystyle \frac{dP}{P} = k \ dt \) ..... integrating

So \(\displaystyle \int \dfrac{dp}{p} = \int (dp) * \int \dfrac{1}{P} = (1) * \ln P + C = \ln P + C \) - on the left side

\(\displaystyle \int k dt = kt + C\) on the right side

\(\displaystyle \displaystyle ln (P) = kt + C_1\)

\(\displaystyle \displaystyle P = P_o*e^{kt}\) - taking the \(\displaystyle e\) on both sides to get rid of the \(\displaystyle \ln\) on the left side

Where did \(\displaystyle P_{0}\) come from

 

[DrPhil]

So \(\displaystyle \int \dfrac{dP}{P} = \int (dp) * \int \dfrac{1}{P} = (1) * \ln P + C = \ln P + C \) - on the left side

\(\displaystyle \int k dt = kt + C\) on the right side

\(\displaystyle \displaystyle ln (P) = kt + C_1\)

\(\displaystyle \displaystyle P = P_o*e^{kt}\) - taking the \(\displaystyle e\) on both sides to get rid of the \(\displaystyle \ln\) on the left side

Where did \(\displaystyle P_{0}\) come from

Never write an integral sign without an increment such as \(\displaystyle dx\). In your first line, there is only one integration, with respect to \(\displaystyle dP\).

\(\displaystyle \displaystyle ln (P) = kt + C_1\)

Now you are going to change from log to exponentiating both sides. On the right side, the arbitrary constant C_1 must be treated as the log of something. An additive constant in the logarithmic form becomes a multiplicative constant when exponentials are taken. Thus \(\displaystyle \displaystyle P_0 = \mathrm e^{C_1}\) represents the arbitrary constant of integration. In many problems you will be given some initial condition that lets you assign a value to the constant.

 

[Jason76]

Never write an integral sign without an increment such as \(\displaystyle dx\). In your first line, there is only one integration, with respect to \(\displaystyle dP\).

\(\displaystyle \displaystyle ln (P) = kt + C_1\)

Now you are going to change from log to exponentiating both sides. On the right side, the arbitrary constant C_1 must be treated as the log of something. An additive constant in the logarithmic form becomes a multiplicative constant when exponentials are taken. Thus \(\displaystyle \displaystyle P_0 = \mathrm e^{C_1}\) represents the arbitrary constant of integration. In many problems you will be given some initial condition that lets you assign a value to the constant.

I saw that (\(\displaystyle e^{c}\) stuff) in a video a long time ago, but forgot about it. It makes sense.

Any hints on this line? at t = 24 → \(\displaystyle \dfrac{P}{P_o} \ = \ 2\) → \(\displaystyle \displaystyle 2 = e^{k*24}\) → solve for k

What is the \(\displaystyle 2\) doing in there?

Here is the whole problem again:

The population P of Bacteria grows according to the equation: \(\displaystyle \dfrac{dP}{dt} = kP\) where \(\displaystyle k\) is a constant, and \(\displaystyle t\) is measured in hours. If the population of bacteria doubles every 24 hours, then what is the value of \(\displaystyle k\)?

\(\displaystyle \displaystyle \frac{dP}{P} = k \ dt \) ..... integrating

\(\displaystyle \displaystyle ln (P) = kt + C_1\)

\(\displaystyle \displaystyle P = P_o*e^{kt}\)

at t = 24 → \(\displaystyle \dfrac{P}{P_o} \ = \ 2\) → \(\displaystyle \displaystyle 2 = e^{k*24}\) → solve for k

 

[Bob Brown MSEE]

This is what's called an ordinary differential equation (ODE). You have already been shown how to solve it using separation of variables, so here's another technique. If we write the ODE in standard linear form, we have:

....

and so we have:

\(\displaystyle P(t)=P_0e^{kt}\)

Now you may solve for \(\displaystyle k\) using the hints above.

1) Mark showed, what I ONLY assumed.
2) The problem implies that P(t) doubles for every 24hrs
3) The problem implies that P(t) doubles for ANY \(\displaystyle P_0\).
4) The problem implies that P(t) doubles for ANY interval (t, t+24).

I claimed that, to get the simplest example for \(\displaystyle P(t)=P_0e^{kt}\)
3) choose \(\displaystyle 1=P_0\)
4) choose interval (t, t+24) = (0,24)

P(t) = e^(kt) <=============This is from \(\displaystyle 1=P_0\)
P(0) = 1
P(24)=2P(0)=2 <=============This is where the 2 comes from
P(24) = e^(k24) <=============This is from P(t) = e^(kt)
so
2 = e^(k24) <================This follows

 

[Bob Brown MSEE]

Specifically

I saw that (\(\displaystyle e^{c}\) stuff) in a video a long time ago, but forgot about it. It makes sense.

Any hints on this line? at t = 24 → \(\displaystyle \dfrac{P}{P_o} \ = \ 2\) → \(\displaystyle \displaystyle 2 = e^{k*24}\) → solve for k

What is the \(\displaystyle 2\) doing in there?

Here is the whole problem again:

\(\displaystyle \dfrac{P(24)}{P_o} \ = \ 2\)

\(\displaystyle \dfrac{P_o e^{k*24}}{P_o} \ = \ 2\)

 

[HallsofIvy]

I saw that (\(\displaystyle e^{c}\) stuff) in a video a long time ago, but forgot about it. It makes sense.

Any hints on this line? at t = 24 → \(\displaystyle \dfrac{P}{P_o} \ = \ 2\) → \(\displaystyle \displaystyle 2 = e^{k*24}\) → solve for k

What is the \(\displaystyle 2\) doing in there?

Here is the whole problem again:

And the original problem said "If the population of bacteria doubles every 24 hours". Do you know what "doubles" means?

 

[Jason76]

And the original problem said "If the population of bacteria doubles every 24 hours". Do you know what "doubles" means?

Here is the whole problem again:


The population P of Bacteria grows according to the equation: \(\displaystyle \dfrac{dP}{dt} = kP\) where \(\displaystyle k\) is a constant, and \(\displaystyle t\) is measured in hours. If the population of bacteria doubles every 24 hours, then what is the value of \(\displaystyle k\)?

\(\displaystyle \displaystyle \int \frac{dP}{P} = \int k \ dt \) ..... integrating

\(\displaystyle \displaystyle ln (P) = kt + C_1\)

\(\displaystyle \displaystyle e^{ln (P)} = e^{kt}+e^{C}\)

\(\displaystyle e^{C} = p_{0}\)

\(\displaystyle \displaystyle P = P_o*e^{kt}\)

\(\displaystyle P = 2\)

\(\displaystyle 2 = e^{24k}\)

\(\displaystyle \ln(2) = \ln(e^{24k})\)

\(\displaystyle \ln(2) = 24k\)

\(\displaystyle k = \dfrac{\ln(2)}{24}\)

 

[DrPhil]

Here is the whole problem again:


The population P of Bacteria grows according to the equation: \(\displaystyle \dfrac{dP}{dt} = kP\) where \(\displaystyle k\) is a constant, and \(\displaystyle t\) is measured in hours. If the population of bacteria doubles every 24 hours, then what is the value of \(\displaystyle k\)?

\(\displaystyle \displaystyle \int \frac{dP}{P} = \int k \ dt \) ..... integrating

\(\displaystyle \displaystyle ln (P) = kt + C\)

\(\displaystyle \displaystyle e^{ln (P)} = e^{kt}+e^{C}\)X exponent of sum is not sum of exponents \(\displaystyle \displaystyle e^{kt + C} \)

\(\displaystyle e^{C} = p_{0}\)

\(\displaystyle \displaystyle P = P_o*e^{kt}\)

\(\displaystyle P = 2\)

\(\displaystyle 2 = e^{24k}\)

\(\displaystyle \ln(2) = \ln(e^{24k})\)

\(\displaystyle \ln(2) = 24k\)

\(\displaystyle k = \dfrac{\ln(2)}{24}\)

ok - units of \(\displaystyle k\) should be per hour

 

[Deleted member 4993]

Here is the whole problem again:


The population P of Bacteria grows according to the equation: \(\displaystyle \dfrac{dP}{dt} = kP\) where \(\displaystyle k\) is a constant, and \(\displaystyle t\) is measured in hours. If the population of bacteria doubles every 24 hours, then what is the value of \(\displaystyle k\)?

\(\displaystyle \displaystyle \int \frac{dP}{P} = \int k \ dt \) ..... integrating

\(\displaystyle \displaystyle ln (P) = kt + C_1\)

\(\displaystyle \displaystyle e^{ln (P)} = e^{kt}+e^{C}\).... Incorrect

\(\displaystyle \displaystyle e^{ln (P)} = e^{kt} \ * \ e^{C}\)

\(\displaystyle e^{C} = p_{0}\)

\(\displaystyle \displaystyle P = P_o*e^{kt}\)

\(\displaystyle P = 2\)

\(\displaystyle 2 = e^{24k}\)

\(\displaystyle \ln(2) = \ln(e^{24k})\)

\(\displaystyle \ln(2) = 24k\)

\(\displaystyle k = \dfrac{\ln(2)}{24}\)

.

 

[Jason76]

.

But I'm wondering shouldn't a doubled population be \(\displaystyle 2P\), not \(\displaystyle 2\)?

 

[MarkFL]

But I'm wondering shouldn't a doubled population be \(\displaystyle 2P\), not \(\displaystyle 2\)?

What you want to write is:

\(\displaystyle P(24)=P_0e^{24k}=2P_0\)

Now you can divide through by \(\displaystyle P_0\ne0\) to get:

\(\displaystyle e^{24k}=2\)

and then from here, solve for \(\displaystyle k\).