Annette Sabin

2021-12-29

Assume that the rate of evaporation for water is proportional to the initial amount of water in a bowl. If Jill places 1 cup of water outside and notices that there is $\frac{1}{2}$ cup of water 6 hours later. Can she use a differential equation to find when there is no water left?

Gerald Lopez

Step 1:Define differential equation
Here, given that rate of change of evaporation is proportional to the initial amount of water.
Therefore, we can model a differential equation to find the actual amount of evaporation.
In the following way we can describe the model.
Step 2: Find equation
Let say, rate of change of evaporation is $\frac{de}{dt}$.
Also consider that the initial amount of water was w.
According to question:
$\frac{de}{dt}\mathrm{\infty }w$.
Therefore, by the property of proportion:
$\frac{de}{dt}=k.w$, where k is a constant.
By using integration:
$\int de=k\int wdt$
Solving it we get:
$e=kwt+C$, C is arbitrary const.
So, .
Now, $w=1,e=\frac{1}{2},t=6$,
Therefore, $\frac{\frac{1}{2}}{6}=k.1+{C}_{1}$
$⇒k+{C}_{1}=\frac{1}{12}$.
When $w=1,t=0:e=0$
Therefore, ${C}_{1}=0$
Step 3
Therefore, $k=\frac{1}{12}$.
Therefore, the equation can be written as:
$e=\frac{wt}{12}$.

Annie Gonzalez

We know that rate of change the volume of water w, is proportional to the volume of water currently in the glass, is we know $\frac{dw}{dt}\mathrm{\infty }w$.
$⇒\frac{dw}{dt}=kw$
This is separable differential eq. We separate integrate to find $\frac{dW}{dt}=kW⇒\frac{dW}{W}=kdt⇒\int \frac{dw}{w}=\int kdt$
$⇒\mathrm{ln}w=kt+c⇒w\left(t\right)=A{e}^{kt}$
Where we wrote $A={e}^{c}$ note that is the initial volume of water. We know that half of the water remains after 7 hours, is . And so ${e}^{k\left(7\right)}=\frac{1}{2}$
$⇒7k=\mathrm{ln}\left(\frac{1}{2}\right)$
$⇒k=\frac{1}{7}\mathrm{ln}\left(\frac{1}{2}\right)=-0.09902$
Note that this is negative because this is an example of exponential decay.
We have $A=8oz$, so water eq is
$w\left(t\right)=8{e}^{-0.099oz}$
So after 9hrs, the remaining volume of water is $w\left(9\right)=8{e}^{-0.099oz\left(9\right)}$
$w\left(9\right)=3.28oz$.

Vasquez

$\frac{de}{dt}\mathrm{\infty }w.$
$\frac{de}{dt}=k.w$, where k is a constant.
$\int de=k\int wdt$
$e=kwt+C$, C is arbitrary const.
So,
Now, $w=1,e=\frac{1}{2},t=6$,
Therefore, $\frac{\frac{1}{2}}{6}=k.1+{C}_{1}$
$⇒k+{C}_{1}=\frac{1}{12}$
w=1, t=0: e=0
${C}_{1}=0$
Step 3
$k=\frac{1}{12}.$
$e=\frac{wt}{12}.$

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