Integral question about depreciation rate in percentage.

[hobomath]

This is my first time tackling this kind of problem.

The rate in which a student forgets about what he studies is proportional to the difference of the percentage of what he remembers right now and the percentage of minimal retention of that student. Student A has a rate of minimal retention of 50%. A the end of a class, student A remembers 95% of his studies. One week later, he remembers 80% of his studies of that class.

The exam is in 3 weeks after that class. If student A stops studying, what is the percentage of what he remembers?

1. Let $Q$ be the percentage of what he remembers.
2. Let $t$ be the time in weeks.
3. Initial condition 1: $Q=0.95 $ and $t=0$
5. Initial condition 2: $Q=0.80 $ and $t=1$

Now I'm stuck about finding the rate. From what I understand from the text:
$$ dQ-0.5 = kQ $$After that, I'm not sure what I have to do. I imagine I'd have to find the value of the constant $k$ by plugging the initial conditions 1 and/or 2. But I don't see where I should put the time (t) anywhere?

I am so confused with this problem... so any hints or help to solve is much appreciated.

 

[MarkFL]

This is mathematically equivalent to Newton's Law of Cooling. The IVP we want is:

\(\displaystyle \d{Q}{t}=-k(Q-M)\) where $0<k$ and \(\displaystyle Q(0)=0.95,\,Q(1)=0.80\)

The ODE is separable...can you proceed?

 

[hobomath]

Hello MarkFL, thank you for the reply and the lead. I'm looking into the Newton's law of Cooling right now... I'll reply back if I find anything.

 

[MarkFL]

Hello MarkFL, thank you for the reply and the lead. I'm looking into the Newton's law of Cooling right now... I'll reply back if I find anything.

My comment about Newton's Law of Cooling was just meant as an observation, in case you are familiar with problems involving that law...I didn't mean for you to necessarily look it up. You have all the information you need to solve the problem as given.

You want to separate variables, and use the boundaries as the limits of integration, and then apply the FTOC. :)

 

[MarkFL]

To follow up, we can write:

\(\displaystyle \int_{Q_0}^{Q(t)}\frac{1}{u-M}\,du=-k\int_0^t\,dv\)

Applying the FTOC, there results:

\(\displaystyle \ln\left(\frac{Q_0-M}{Q(t)-M}\right)=kt\)

Now, using this, we may determine $k$ given $Q_1=Q(1)$:

\(\displaystyle k=\ln\left(\frac{Q_0-M}{Q_1-M}\right)\)

Hence:

\(\displaystyle \ln\left(\frac{Q_0-M}{Q(t)-M}\right)=\ln\left(\left(\frac{Q_0-M}{Q_1-M}\right)^t\right)\)

Or:

\(\displaystyle \frac{Q_0-M}{Q(t)-M}=\left(\frac{Q_0-M}{Q_1-M}\right)^t\)

Solving for $Q(t)$, we find:

\(\displaystyle Q(t)=\left(Q_0-M\right)\left(\frac{Q_0-M}{Q_1-M}\right)^{-t}+M\)

Now we can answer the question:

The exam is in 3 weeks after that class. If student A stops studying, what is the percentage of what he remembers?

Plugging in the given data, we have:

\(\displaystyle Q(t)=\left(0.95-0.50\right)\left(\frac{0.95-0.50}{0.80-0.50}\right)^{-t}+0.50\)

\(\displaystyle Q(t)=0.45\left(\frac{0.45}{0.30}\right)^{-t}+0.50\)

\(\displaystyle Q(t)=\frac{1}{20}\left(9\left(\frac{2}{3}\right)^{t}+10\right)\)

And so we finally find:

\(\displaystyle Q(3)=\frac{1}{20}\left(9\left(\frac{2}{3}\right)^{3}+10\right)=\frac{1}{20}\left(\frac{8}{3}+10\right)=\frac{19}{30}=63\tfrac{1}{3}\%\)

 

[hobomath]

Hey again, thanks you were a great help!Sorry for the lack of follow-up, but I'm glad I found the same answer as yours. It was truly helpful.

 

[MarkFL]

Hey again, thanks you were a great help!Sorry for the lack of follow-up, but I'm glad I found the same answer as yours. It was truly helpful.

I am glad to be of help, and that you got the same answer I did (presuming we are correct of course). :)

I posted a follow-up just in case other people find this thread via a search engine, then the working and solution is available making MHB more useful and encouraging a return visit and perhaps even a new member posting their own questions.