# Math Help - application problems

1. ## application problems

I have several review word problems from differential equation class and. Hopefully can share (ask) with everyone in here.

1) a nitric acid solution flows at a constant rate of 6L/min into a large tank that initally held 200 L of 0.5% nitric acid solution. The solution inside the tank is kept well stirred and flows out of the tank at a rate of 8 L/min. if the solution entering the tank is 20% nitric acid, determine the volume of nitric acid in the tank after t min. When will the percentage of nitric acid in the tank reach 10%

2) Suppose that at a time t=0,5 thousand people in a city with population 40,000 people have heard a rumor. After 1 week the number P(t) of those who have heard it has increased to P(1)=8000. Assuming that P(t) satisfies a logistic equation, when will 60% of the city's population have heard the rumor?

3) A linear circuit has a resistance of 1 ohm, a capacitance of 0.5 farad a 2 volts battery. Find the current through the loop and the charge q(t) on the capacitor, given that initally there is no charge on the capacitor, q(0) = 0

4) It was noon on a cold December day in Tampa: 16 degree C. Detective Taylor arrived at the crime scence to find the sergent leaning over the body. The sergent said there were several suspects. If they knew the exact time of death, then they could narrow the list. Detective Taylor took out a thermometer and measured temperature of the body. 34.5 degree C. He then left for lunch. Upon returning at 1:00pm, He found the body temperature to be 33.7 degree C. When did the murder occur?

5) A compound C is formed when two chemicals A and B are combined. The resulting reaction between the two is such that for each gram of B, 2 grams of A are used. Initially there are 24grams of A and 18 grams of B. It is observed that 8 grams of the compound C are formed in 15 minutes. Determine the amount of C at any time if the rate of the reaction is proportional to the amounts of A and B remaining. Find the amount after 20 minutes.

6) A white wine at room temperature 70 degree F is chilled in ice (32F). If it take 15 minutes for the wine to chill to 60 degree F, how long will it take for the wine to reach 56 degree F?

2. Originally Posted by ggw
6) A white wine at room temperature 70 degree F is chilled in ice (32F). If it take 15 minutes for the wine to chill to 60 degree F, how long will it take for the wine to reach 56 degree F?
I assume you want this done by Newtonian Cooling (can't really think of another way to do it actually )

Let $T(t)$ (i will write $T$ most of the time) be the temperature of the wine at time $t$.
Let $T'(t)$ (or $T'$) be the rate at which the wine is cooling
Let $T_{amb}$ be the ambient or surrounding temperature.
And let $k$ be an arbitrary constant to be determined later

Then by Newton's Law of Cooling, we have:

$T' = -k(T - T_{amb})$

$\Rightarrow T' = -k(T - 32)$

$\Rightarrow \frac {T'}{T - 32} = -k$

$\Rightarrow \ln (T - 32) = -kt + C$

$\Rightarrow T - 32 = Ae^{-kt}$

$\Rightarrow T = Ae^{-kt} + 32$ ....now there's a formula that allows us to jump right to this step, but i'm horrible at remembering formulas, so i usually work it out.

Now $T(0) = 70$

$\Rightarrow 70 = A + 32$

$\Rightarrow A = 38$

Also, $T(15) = 60$

$\Rightarrow 60 = 38e^{-15k} + 32$

$\Rightarrow \frac {14}{19} = e^{-15k}$

$\Rightarrow \frac { \ln \left( \frac {14}{19} \right)}{-15} = k$

So our equation is: $T = 38e^{ \frac { \ln \left( \frac {14}{19} \right)}{15}t} + 32 \approx 38e^{-0.0204t} + 32$

Now let $T = 56$ and solve for $t$ in minutes

3. problem 4 is done the same way i did problem 6. i think there is some info missing though. what is the temperature of the human body in degrees celcius. that is, what would the body temperature be if the person was alive. you may need to know that, not sure, didn't think about it very long

4. is that the body temperature is 34.5 degree Celcius. one hour after is 33.7 degree celcius. The temperature was measure at noon. Then it is increase after one hour. So should it be the rate of change in time.

5. Originally Posted by ggw
is that the body temperature is 34.5 degree Celcius. one hour after is 33.7 degree celcius. The temperature was measure at noon. Then it is increase after one hour. So should it be the rate of change in time.
yes, but that is the temp of the dead body. we can measure the rate at which the body cools, but we need to know from what temp it started cooling from (that is, the temp of a live person) to pinpoint the time of death.

6. Originally Posted by ggw
I have several review word problems from differential equation class and. Hopefully can share (ask) with everyone in here.

1) a nitric acid solution flows at a constant rate of 6L/min into a large tank that initally held 200 L of 0.5% nitric acid solution. The solution inside the tank is kept well stirred and flows out of the tank at a rate of 8 L/min. if the solution entering the tank is 20% nitric acid, determine the volume of nitric acid in the tank after t min. When will the percentage of nitric acid in the tank reach 10%
.
Let $Q(t)$ be the amount of drinkable nitric acid.
We are given that $Q(0) = 200 \mbox{L} \times .5 \% = 1 \mbox{ L}$

Next we form the differencial equation:
$\frac{dQ}{dt} = \mbox{rate in }-\mbox{rate out}$

The rate in is $6\frac{\mbox{L}}{\mbox{min}} \times 20\% = 1.2 \mbox{ L}/\mbox{min}$

The rate out is more difficult to figure out over here. We need know that 8 liters of mixture are leaving. But if we multiply that by the concentration of nitric acid in this solution we get the amount of nitric acid leaving the tank. The formula for concentration is AMOUNT divided by VOLUME. The AMOUNT is $Q$, that is the function which we let represent the amount at any given time. The VOLUME is $200+6t-8t = 200 - 2t$. Because we have 200 liters to start with another 6 liters come in every minute, so after t minutes we have 6t liters coming in. But we are also losing 8 liters every minute so after t minutes we have 8t liters lost. Thus, $\frac{Q}{200-2t}$ is the concentration at any given time. Multiply that by the amount leaving to get: $\frac{4Q}{100-t}$.

Thus, the differencial equation is:
$\boxed{ \frac{dQ}{dt} = \frac{4Q}{100-t} - 1.2 \mbox{ and }Q(0)=1}$

7. Uhm, all information in the question. I checked the question number 4, and there's nothing has been missing.

Thanks for the number 6 & 1. Thanks a bunch for your help....!

8. Originally Posted by ggw

4) It was noon on a cold December day in Tampa: 16 degree C. Detective Taylor arrived at the crime scence to find the sergent leaning over the body. The sergent said there were several suspects. If they knew the exact time of death, then they could narrow the list. Detective Taylor took out a thermometer and measured temperature of the body. 34.5 degree C. He then left for lunch. Upon returning at 1:00pm, He found the body temperature to be 33.7 degree C. When did the murder occur?
Again we use the Newtonian Cooling law.

we will start the time at noon. so noon will be when $t=0$

$T' = -k(T - T_{amb})$

$\Rightarrow T' = -k(T - 16)$

$\Rightarrow \frac {T'}{T - 16} = -k$

$\Rightarrow \ln (T - 16) = -kt + C$

$\Rightarrow T = Ae^{-kt} + 16$

Since $T(0) = 34.5$

$\Rightarrow 34.5 = A + 16$

$\Rightarrow A = 18.5$

Also, $T(1) = 33.7$

$\Rightarrow 33.7 = 18.5e^{-k} + 16$

$\Rightarrow 0.956756756 = e^{-k}$

$\Rightarrow k \approx 0.04420692$

So our equation is:

$T = 18.5e^{-0.04420692t} + 16$

Now, i did a google search. the normal body temp of a person is 37 degrees C. So now all you need to do is find the time when the body would have been 37 degree C and get back to me with your answer. Note, the way we set up the problem, you will get a negative answer for $t$, don't let that bother you. it just means we are counting backwards in time from noon