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 given differential equation is \(\frac{dV}{dt} = kS\). This is a first-order differential equation since it involves the first derivative of \(V\) with respect to \(t\). The degree is 1 as the equation is polynomial in the derivative with the highest power being 1.

(ii) Solving the Differential Equation

Given \(V = \pi r^3\) and \(S = 2\pi r^2\), substitute into \(\frac{dV}{dt} = kS\):

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

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

Assuming \(r \neq 0\), divide by \(\pi r^2\):

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

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

Separate variables and integrate:

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

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

Given \(r(0) = 5 \, \text{mm}\), substitute \(t = 0\), \(r = 5\):

\(5 = C\)

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

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

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

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

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

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

\(k = -3\)

Now, find \(t\) when \(r = 0\):

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

\(0 = -2t + 5\)

\(2t = 5\)

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

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

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

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

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

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

\(k = -6\)

Now, find \(t\) when \(r = 0\):

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

\(0 = -4t + 5\)

\(4t = 5\)

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