(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 water has decreased to \( W(6)=1 \mathrm{~mm}^{3} \) after 6 hours. (i) Solve the differential equation by separating like variables.
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To solve the differential equation \( \frac{d W}{d t} = -k W^{\frac{2}{3}} \) by separating variables, we can rearrange it as follows: \[ \frac{dW}{W^{\frac{2}{3}}} = -k \, dt \] Next, we integrate both sides. The left side requires the integral of \( W^{-\frac{2}{3}} \): \[ \int W^{-\frac{2}{3}} \, dW = \int -k \, dt \] The integral of \( W^{-\frac{2}{3}} \) is \( \frac{W^{\frac{1}{3}}}{\frac{1}{3}} = 3W^{\frac{1}{3}} + C \), where \( C \) is the constant of integration. The integral of \( -k \, dt \) gives \( -kt + C_1 \), so we can write: \[ 3W^{\frac{1}{3}} = -kt + C_1 \] Now, we will determine the constant using the initial condition \( W(0) = 8 \) mm³: \[ 3(8^{\frac{1}{3}}) = C_1 \implies C_1 = 3 \cdot 2 = 6 \] Thus, our equation becomes: \[ 3W^{\frac{1}{3}} = -kt + 6 \] Next, we apply the second condition \( W(6) = 1 \) mm³ to find \( k \): \[ 3(1^{\frac{1}{3}}) = -6k + 6 \] \[ 3 = -6k + 6 \implies -6k = 3 - 6 \implies -6k = -3 \implies k = \frac{1}{2} \] Now substituting \( k \) back into our equation, we have: \[ 3W^{\frac{1}{3}} = -\frac{1}{2}t + 6 \] To express \( W \), we solve for \( W \): \[ W^{\frac{1}{3}} = -\frac{1}{6}t + 2 \] \[ W = \left(-\frac{1}{6}t + 2\right)^3 \] Thus, the solution to the differential equation is: \[ W(t) = \left(-\frac{1}{6}t + 2\right)^3 \]
