Q1
Answer
An inverted conical funnel is full of salt. The salt is allowed to leave by a small hole in the vertex. It leaves at a constant rate of 6 cm$^3$ s$^{-1}$. Given that the angle of the cone between the slanting edge and the vertical is 30$^{\circ}$, show that the volume of the salt is $\dfrac{1}{9}\pi h^3$, where $h$ is the height of the salt at time $t$ seconds, and hence show that the rate of change of the height of the salt in the funnel is inversely proportional to $h^2$.
Use $\tan 30 = \dfrac{r}{h}$ and $V = \dfrac{1}{3}\pi r^2h$ to find $\dfrac{\mathrm{d}V}{\mathrm{d}h}$ and $\dfrac{\mathrm{d}h}{\mathrm{d}t}$
Q2
Answer
  1. Given that $\displaystyle\int_a^{2a}10-6x\ \mathrm{d}x = 1$, find the two possible values of $a$.
  2. Labelling all axes intercepts, sketch the graph of $y=10-6x$ for $0 leq x leq 2$.
  3. With reference to the integral in part a and the sketch in part b, explain why the larger value of a found in part a produces a solution for which the actual area under the graph between a and 2a is not equal to 1. State whether the area is greater than 1 or smaller than 1.
  1. $1, \dfrac{1}{9}$
  2. Straight line crosses at $(0,10)$ and $\left(\frac{10}{6},0\right)$
  3. Integral involves areas above and below the axis, expect actual area to be bigger
Q3
Answer
Solve $\mathrm{e}^{\frac{3}{2}x} = \mathrm{e}^{3x}-2$, giving your answer(s) in exact form.
$\dfrac{2}{3}\ln 2$