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

Beginner2021-12-30Added 29 answers

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,$\frac{e}{t}=kw+{C}_{1},where\text{}{C}_{1}=\frac{C}{t}$ .

Now,$w=1,e=\frac{1}{2},t=6$ ,

Therefore,$\frac{\frac{1}{2}}{6}=k.1+{C}_{1}$

$\Rightarrow k+{C}_{1}=\frac{1}{12}$ .

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:

$e=\frac{wt}{12}$ .

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

Also consider that the initial amount of water was w.

According to question:

Therefore, by the property of proportion:

By using integration:

Solving it we get:

So,

Now,

Therefore,

When

Therefore,

Step 3

Therefore,

Therefore, the equation can be written as:

Annie Gonzalez

Beginner2021-12-31Added 41 answers

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 $\frac{dw}{dt}\mathrm{\infty}w$ .

$\Rightarrow \frac{dw}{dt}=kw$

This is separable differential eq. We separate integrate to find$\frac{dW}{dt}=kW\Rightarrow \frac{dW}{W}=kdt\Rightarrow \int \frac{dw}{w}=\int kdt$

$\Rightarrow \mathrm{ln}w=kt+c\Rightarrow w\left(t\right)=A{e}^{kt}$

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 $\frac{w}{A}=\frac{1}{2},when\text{}t=7$ . And so $e}^{k\left(7\right)}=\frac{1}{2$

$\Rightarrow 7k=\mathrm{ln}\left(\frac{1}{2}\right)$

$\Rightarrow k=\frac{1}{7}\mathrm{ln}\left(\frac{1}{2}\right)=-0.09902$

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$w\left(9\right)=8{e}^{-0.099oz\left(9\right)}$

$w\left(9\right)=3.28oz$ .

This is separable differential eq. We separate integrate to find

Where we wrote

Note that this is negative because this is an example of exponential decay.

We have

So after 9hrs, the remaining volume of water is

Vasquez

Expert2022-01-09Added 669 answers

So,

Now,

Therefore,

w=1, t=0: e=0

Step 3

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