(d) Suppose that a cell dehydration model satisfies the differential equation, $$ \frac{d W}{d t}=-k W^{\frac{2}{3}} $$ Assume that the volume of water in the cell initially is $$ W(0)=8 \mathrm{~mm}^{3} $$. Also, assume that the volume of water has decreased to $$ W(6)=1 \mathrm{~mm}^{3} $$ after 6 hours. (i) Solve the differential equation by separating like variables. (ii) Determine the value of k and the time when the cell will be completely dry.

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### Problem (d)(i): Solving the Differential Equation by Separating Variables

We are given the differential equation governing the volume of water $$ W(t) $$ in a cell:

$$
\frac{dW}{dt} = -k W^{\frac{2}{3}}
$$

with the initial condition $$ W(0) = 8 \, \text{mm}^3 $$.

**Step 1: Separate Variables**

To solve the differential equation, we separate the variables $$ W $$ and $$ t $$:

$$
\frac{dW}{W^{\frac{2}{3}}} = -k \, dt
$$

**Step 2: Integrate Both Sides**

Next, we integrate both sides of the equation:

$$
\int W^{-\frac{2}{3}} \, dW = \int -k \, dt
$$

**Integral on the Left Side:**

$$
\int W^{-\frac{2}{3}} \, dW = \frac{3}{1} W^{\frac{1}{3}} + C_1 = 3 W^{\frac{1}{3}} + C_1
$$

**Integral on the Right Side:**

$$
\int -k \, dt = -k t + C_2
$$

Combining both integrals:

$$
3 W^{\frac{1}{3}} + C_1 = -k t + C_2
$$

For simplicity, let $$ C = C_2 - C_1 $$:

$$
3 W^{\frac{1}{3}} = -k t + C
$$

**Step 3: Apply Initial Condition**

Using the initial condition $$ W(0) = 8 \, \text{mm}^3 $$:

$$
3 (8)^{\frac{1}{3}} = 3 \times 2 = 6 = -k \times 0 + C \implies C = 6
$$

**Step 4: Solve for $$ W(t) $$**

Substituting $$ C = 6 $$ into the equation:

$$
3 W^{\frac{1}{3}} = 6 - k t
$$

Divide both sides by 3:

$$
W^{\frac{1}{3}} = \frac{6 - k t}{3} = 2 - \frac{k t}{3}
$$

Cubing both sides to solve for $$ W(t) $$:

$$
W(t) = \left(2 - \frac{k t}{3}\right)^3
$$

**Final Solution:**

$$
W(t) = \left(2 - \frac{k t}{3}\right)^3 \quad \text{mm}^3
$$

Alternatively, it can be written as:

$$
W(t) = \left(\frac{6 - k t}{3}\right)^3 \quad \text{mm}^3
$$

Both forms are equivalent and valid representations of the solution.

### Answer

Problem i Answer:

After separating and integrating, a general solution is

W(t) = [(6 – k t) ⁄ 3]³.

Thus,

W(t) = ((6 − kt)/3)^3 mm³.
The solution to the differential equation is: $$ W(t) = \left(\frac{6 - k t}{3}\right)^3 \quad \text{mm}^3 $$

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