Question

The rate of change of the mass, A, of salt at time t is proportional to the square of the mass of salt present at time t. Initially there is 10 grams of salt and 10 hour later there are 4 grams of salt. (a) Find the initial value problem (the differential equation and the initial condition) that fits this physical description. (b) Solve the I.V.P (initial value problem) from part (a) for the specific solution. (c) Based on your solution find the time when A(t)< 1.

Answer 1

Answer:

(a) -dA/dt = kA², A₀ = 10

(b) A =10/(1+ kt)

(c) t > 60 h

Step-by-step explanation:

(a) Find the IVP

A differential equation with an initial condition y₀ = f(x₀) is called an initial value problem.

The rate of decrease of A is proportional to A², and A₀ = 10, so the IVP is

-dA/dt = kA², A₀ = 10

(b) Solve the IVP

$\begin{array}{rcl}-\dfrac{\text{d}A}{\text{d}t} & = & kA^{2}\\\\\dfrac{\text{d}A}{A^{2}} & =&-k\text{d}t\\\\-\dfrac{1}{A} & = & -kt + C\\\\\dfrac{1}{A} & = & kt + C\\\\\end{array}$

Apply the initial condition: A₀ = 10 (when t = 0)

$\dfrac{1}{10} = C\\\\\dfrac{1}{A} = kt + \dfrac{1}{10}\\\\\dfrac{10}{A} = 10kt + 1\\\\\mathbf{A} = \mathbf{\dfrac{10}{1+ kt}}$

(c) Find the time when A(t) < 1

(i) Find the value of k (A₁₀ = 4)

$\begin{array}{rcl}A& =&\dfrac{10}{1+ kt}\\\\4 & =& \dfrac{10}{1 + 10k}\\\\4 + 40 k & = & 10\\\\40k & = & 6\\k & = & 0.15\\\end{array}\\\\\mathbf{A} =\mathbf{ \dfrac{10}{0.15t + 1}}$

(ii) Find t when A < 1

$\begin{array}{rcl}A(t) & < & 1\\\dfrac{10}{0.15t + 1} & < & 1\\\\10 & < & 0.15t + 1\\9 & < & 0.15t\\t & > & 60\\\\\end{array}\\\mathbf{A < 1} \textbf{ when }\mathbf{ t >60}$

The figure below shows the graph of A vs t.