Question

Camphor is a waxy, colourless solid with strong aroma that evaporates through the process of sublimation if left in the open at room temperature.

(Cylindrical-shaped Camphor tablets) A cylindrical camphor tablet whose height is equal to its radius (r) evaporates when exposed to air such that the rate of reduction of its volume is proportional to its total surface area. Thus, the differential equation \( \frac{dV}{dt} = -kS \) is the differential equation, where \( V \) is the volume, \( S \) is the surface area, and \( t \) is the time in hours.
Based upon the above information, answer the following questions:
(i) Write the order and degree of the given differential equation.}
(ii) Substituting \( V = \pi r^3 \) and \( S = 2 \pi r^2 \), we get the differential equation \( \frac{dr}{dt} = \frac{2}{3}k \). Solve it, given that \( r(0) = 5 \) mm.}
(iii) (a) If it is given that \( r = 3 \) mm when \( t = 1 \) hour, find the value of \( k \). Hence, find \( t \) for \( r = 0 \) mm.}
(iii) (b) If it is given that \( r = 1 \) mm when \( t = 1 \) hour, find the value of \( k \). Hence, find \( t \) for \( r = 0 \) mm.

Hint:

In a differential equation problem, identify the given physical relationships and convert them into mathematical expressions. For cylindrical shapes, remember that volume and surface area are often expressed in terms of the radius.

Answer 1

(i) Order and Degree of the Differential Equation

The differential equation provided is \(\frac{dV}{dt} = kS\). It is classified as a first-order differential equation because it contains only the first derivative of \(V\) with respect to \(t\). The degree of the equation is 1, as the highest power of the derivative in the polynomial form of the equation is 1.

(ii) Solving the Differential Equation

Given the relationships \(V = \pi r^3\) and \(S = 2\pi r^2\), these are substituted into the differential equation \(\frac{dV}{dt} = kS\):

\(\frac{d}{dt}(\pi r^3) = k \cdot 2\pi r^2\)

Applying the chain rule and simplifying yields:

\(3\pi r^2 \frac{dr}{dt} = 2\pi k r^2\)

Assuming \(r eq 0\), division by \(\pi r^2\) results in:

\(3 \frac{dr}{dt} = 2k\)

Rearranging gives:

\(\frac{dr}{dt} = \frac{2k}{3}\)

Separating variables and integrating both sides:

\(\int dr = \int \frac{2k}{3} dt\)

The integration yields the general solution:

\(r = \frac{2k}{3} t + C\)

Using the initial condition \(r(0) = 5 \, \text{mm}\), where \(t = 0\) and \(r = 5\):

\(5 = \frac{2k}{3} \cdot 0 + C\)

This simplifies to \(C = 5\).

Thus, the particular solution is:

\(r = \frac{2k}{3} t + 5\)

(iii)(a) Finding \(k\) and \(t\) for \(r = 0\)

Given that \(r = 3 \, \text{mm}\) when \(t = 1 \, \text{hour}\):

Substituting these values into the particular solution:

\(3 = \frac{2k}{3} \cdot 1 + 5\)

Solving for the term with \(k\):

\(3 - 5 = \frac{2k}{3}\)

\(-2 = \frac{2k}{3}\)

Solving for \(k\):

\(k = -3\)

Now, to find the time \(t\) when \(r = 0\), substitute \(k = -3\) into the particular solution:

\(0 = \frac{2(-3)}{3} t + 5\)

\(0 = -2t + 5\)

Solving for \(t\):

\(2t = 5\)

\(t = 2.5 \, \text{hours}\)

(iii)(b) Finding \(k\) and \(t\) for \(r = 0\)

Given that \(r = 1 \, \text{mm}\) when \(t = 1 \, \text{hour}\):

Substituting these values into the particular solution:

\(1 = \frac{2k}{3} \cdot 1 + 5\)

Solving for the term with \(k\):

\(1 - 5 = \frac{2k}{3}\)

\(-4 = \frac{2k}{3}\)

Solving for \(k\):

\(k = -6\)

Now, to find the time \(t\) when \(r = 0\), substitute \(k = -6\) into the particular solution:

\(0 = \frac{2(-6)}{3} t + 5\)

\(0 = -4t + 5\)

Solving for \(t\):

\(4t = 5\)

\(t = 1.25 \, \text{hours}\)