User:Xenocidius/Specialist/Q5

i.
$$ \begin{align} \frac{dx}{dt} = \frac{p}{d} \end{align} $$

$$ \begin{align} d = q(T + t) \end{align} $$

$$ \begin{align} \frac{dx}{dt} & = \frac{p}{q(T + t)} \\ & = \frac{k}{T + t} \end{align} $$

ii.
$$ \begin{align} x & = \int \! \frac{k}{T + t} \,dt \\ & = k \ln{(T + t)} + c \\ \end{align} $$

$$ \begin{align} & \therefore 0 = k \ln{(T)} + c \\ & \therefore c = -k \ln{(T)} \end{align} $$

$$ \begin{align} & \therefore 2 = k \ln{(T + 1)} - k \ln{(T)} \\ & \therefore 2 = k \ln{(\frac{T + 1}{T})} \\ & \therefore \frac{6}{k} = 3 \ln{(\frac{T + 1}{T})} \\ & \therefore \frac{6}{k} = \ln{((\frac{T + 1}{T})^3)} \end{align} $$

$$ \begin{align} & \therefore 3 = k \ln{(T + 2)} - k \ln{(T)} \\ & \therefore 3 = k \ln{(\frac{T + 2}{T})} \\ & \therefore \frac{6}{k} = 2 \ln{(\frac{T + 2}{T})} \\ & \therefore \frac{6}{k} = \ln{((\frac{T + 2}{T})^2)} \\ \end{align} $$

$$ \begin{align} & \therefore \ln{((\frac{T + 1}{T})^3)} = \ln{((\frac{T + 2}{T})^2)} \\ & \therefore (\frac{T + 1}{T})^3 = (\frac{T + 2}{T})^2 \end{align} $$

iii.
$$ \begin{align} & \therefore \frac{T^3 + 3T^2 + 3T + 1}{T^3} = \frac{T^2 + 4T + 4}{T^2} \\ & \therefore T^3 + 3T^2 + 3T + 1 = T(T^2 + 4T + 4) \\ & \therefore T^3 + 3T^2 + 3T + 1 = T^3 + 4T^2 + 4T \\ & \therefore T^2 + T = 1 \\ & \therefore T^2 + T + \frac{1}{4} = 1 + \frac{1}{4} \\ & \therefore (T + \frac{1}{2})^2 = \frac{5}{4} \\ & \therefore T + \frac{1}{2} = \pm \frac{\sqrt 5}{2} \\ & \therefore T = -\frac{1}{2} \pm \frac{\sqrt 5}{2} \\ & \therefore T = -\frac{1}{2} + \frac{\sqrt 5}{2}, T > 0 \\ & \therefore T = \frac{\sqrt 5 - 1}{2} \\ & \therefore T \approx 0.618 \\ \end{align} $$