The Stoke's Law states that when a spherical object of radius $R$ is falling at velocity $v$ in a fluid, the resistance force $F$ acting on the sphere is given by $F = -6\pi \eta R v ,$ where $\eta$ is the coefficient of viscosity of the fluid. The following table shows the coefficients of viscosity of various fluids at $25 ^\circ\text{C}$. A spherical ball of radius 0.05 m and mass 0.2 kg is released from rest in a huge container filled with corn syrup at $25 ^\circ\text{C}$. It can be shown that as the ball falls through the corn syrup, the velocity $v$ ms^{-1} at time $t$ seconds after release satisfies the differential equation $\frac{dv}{dt} + \frac{6\pi\eta R}{m}v = g ,$ where $g = 9.780\ \text{m s}^{-2}$ is the acceleration due to gravity and $m$ kg is the mass of the ball.

1. Substitute the given values of $\eta$, $R$, $m$, and $g$ into the differential equation.

From the table, the coefficient of viscosity for corn syrup is $\eta = 1.3806$.
Substituting the values $R = 0.05$, $m = 0.2$, and $\eta = 1.3806$:
$\frac{6\pi\eta R}{m} = \frac{6\pi(1.3806)(0.05)}{0.2} = 2.0709\pi \approx 6.5059$
The differential equation becomes:
$\frac{dv}{dt} + 6.5059v = 9.780$

2. Separate the variables $v$ and $t$, then integrate both sides.

$\frac{dv}{dt} = 9.780 - 6.5059v$
$\int \frac{1}{9.780 - 6.5059v} \, dv = \int 1 \, dt$
$-\frac{1}{6.5059} \ln|9.780 - 6.5059v| = t + C$

3. Use the initial condition to find the constant of integration.

Since the ball is released from rest, $v = 0$ when $t = 0$.
$-\frac{1}{6.5059} \ln|9.780 - 0| = 0 + C$
$C = -\frac{1}{6.5059} \ln(9.780)$
Substituting $C$ back:
$-\frac{1}{6.5059} \ln|9.780 - 6.5059v| = t - \frac{1}{6.5059} \ln(9.780)$

4. Rearrange the equation to make $v$ the subject.

Multiply by $-6.5059$:
$\ln|9.780 - 6.5059v| = -6.5059t + \ln(9.780)$
$\ln\left|\frac{9.780 - 6.5059v}{9.780}\right| = -6.5059t$
$\frac{9.780 - 6.5059v}{9.780} = e^{-6.5059t}$
$9.780 - 6.5059v = 9.780 e^{-6.5059t}$
$6.5059v = 9.780 (1 - e^{-6.5059t})$
$v = \frac{9.780}{6.5059} (1 - e^{-6.5059t})$
$v \approx 1.5032(1 - e^{-6.5059t})$
$v = 1.50(1 - e^{-6.51t}) \text{ (to 3 s.f.)}$

1. Evaluate the velocity as $t \to \infty$ by setting the rate of change to zero (or by taking the limit of $v(t)$ from part (i)).

As $t \to \infty$, the velocity approaches a constant value (terminal velocity), so $\frac{dv}{dt} \to 0$.

$\frac{6\pi\eta R}{m}v = g$
$\frac{6\pi(1.3806)(0.05)}{0.2}v = 9.780$
$v = \frac{1.956}{0.41418\pi}$
$v \approx 1.5032$

1. Concisely retrieve the velocity function $v(t)$ obtained from solving the differential equation in part (i).

From part (i), the constant $k = \frac{6\pi\eta R}{m} = \frac{6\pi(1.3806)(0.05)}{0.2} = 2.0709\pi$.
Solving the differential equation $\frac{dv}{dt} = g - kv$ with the initial condition $v(0)=0$ gives:
$v(t) = \frac{g}{k}(1 - e^{-kt}) = \frac{9.780}{2.0709\pi}(1 - e^{-2.0709\pi t})$

2. Set up the definite integral to find the distance covered in 30 seconds using $v = \frac{dx}{dt}$.

Since $v = \frac{dx}{dt}$ and the ball is released from rest, the distance covered is the integral of velocity with respect to time:
$x = \int_0^{30} \frac{9.780}{2.0709\pi}(1 - e^{-2.0709\pi t}) \, dt$

3. Evaluate the definite integral to find the distance.

Integrating analytically (or evaluating directly using a graphing calculator):
$x = \frac{9.780}{2.0709\pi} \left[ t + \frac{e^{-2.0709\pi t}}{2.0709\pi} \right]_0^{30}$
$x = \frac{9.780}{2.0709\pi} \left( 30 + \frac{e^{-62.127\pi}}{2.0709\pi} - \frac{1}{2.0709\pi} \right)$
$x \approx 44.866 \text{ m}$
$x = 44.9 \text{ m} \text{ (to 3 s.f.)}$