June 2022 Paper 2 Q8

Notes

B1: Correct substitution

M1: NOT if \(h = 50\) or \(r = 50\tan 30\) used

M1: Resulting equation must involve exactly 2 variables

M1: Their equation must involve exactly 2 variables

A1: Ignore limits

M1: Integrals must be of correct forms (see examples below)

Note 1 Candidates who substitute numerical values for \(h\) or \(V\) or \(r\) may be able to score the 2^nd and/or 3^rd M1 marks, but probably nothing else. See the example of this below.

Note 2. There is a special case for candidates who use \(r = h\sin 30\) (answer \(\frac{625\pi}{4}\) or 491).
These can score all 4 M-marks and the final A1

Note 3. The chain rule may be used to find \(\frac{\mathrm{d}V}{\mathrm{d}t}\) or \(\frac{\mathrm{d}h}{\mathrm{d}t}\) or \(\frac{\mathrm{d}V}{\mathrm{d}h}\) or \(\frac{\mathrm{d}V}{\mathrm{d}r}\) or other derivatives. Two of the example methods below illustrate use of \(\frac{\mathrm{d}V}{\mathrm{d}t}\) and \(\frac{\mathrm{d}V}{\mathrm{d}r}\), but use of other derivatives can also lead to correct methods.

Example method 1

Scheme	Marks	AO
\(V = \frac{\pi}{3}(h\tan 30^\circ)^2h\) or \(V = \frac{\pi}{3}\left(\dfrac{h}{\sqrt{3}}\right)^2h\) oe	B1	3.3
\(\dfrac{\mathrm{d}V}{\mathrm{d}h} = \dfrac{\pi}{3}h^2\)	M1	3.4
\(\dfrac{\mathrm{d}V}{\mathrm{d}t} = \dfrac{\pi}{3}h^2\dfrac{\mathrm{d}h}{\mathrm{d}t}\) oe   or \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{3}{\pi h^2}\dfrac{\mathrm{d}V}{\mathrm{d}t}\) \(\left(\text{“}\dfrac{\pi}{3}h^2\dfrac{\mathrm{d}h}{\mathrm{d}t}\text{”} = -2h\right.\) oe   or \(\left.\dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{-6}{\pi h}\right)\)	M1	2.1
\(\displaystyle\pi\int_{50}^{0} h\,\mathrm{d}h = -\int_0^t 6\,\mathrm{d}t\) oe	M1	1.1
\(\left[\dfrac{\pi h^2}{2}\right]_{50}^{0} = \left[-6t\right]_0^t\) oe	A1	2.1
\(-\pi \times \dfrac{50^2}{2} = -6t\)	M1	1.1
Time \(= \dfrac{625\pi}{3}\) secs or 654 secs (3 sf) oe	A1	3.4

B1: or \(V = \frac{\pi}{9}h^3\) oe

M1: Attempt differentiate their \(V\) in terms of \(h\) only
NOT if \(h = 50\) or \(r = 50\tan 30\) used.

M1: Attempt use chain rule for \(\frac{\mathrm{d}V}{\mathrm{d}t}\) or \(\frac{\mathrm{d}h}{\mathrm{d}t}\) in terms of \(t\) & \(h\) only
(Set their \(\frac{\mathrm{d}V}{\mathrm{d}t} = -2h\))

M1: Attempt separate variables in their equation in terms of \(h\) and \(t\) only (not \(V\) or \(r\)). Integral signs not essential

A1: Correct integrals, any limits or none

M1: Substitute correct limits into integrals of forms \(ah^2\) & \(bt\)
OR substitute \(t = 0\) & \(h = 50\) to find \(c\) and substitute \(h = 0\)

A1: Allow without secs   or 10.9 mins or 10 mins 54 secs
or:
SC. Use of \(r = h\sin 30\) (answer \(\frac{625\pi}{4}\) or 491) can score all 4 M-marks and final A1

Example method 2

Scheme	Marks	AO
\(V = \dfrac{\pi}{3}r^2\dfrac{r}{\tan 30^\circ}\) or \(V = \dfrac{\pi}{\sqrt{3}}r^3\) oe	B1
\(\dfrac{\mathrm{d}V}{\mathrm{d}r} = \sqrt{3}\pi r^2\)	M1
\(\dfrac{\mathrm{d}V}{\mathrm{d}t} = \sqrt{3}\pi r^2\dfrac{\mathrm{d}r}{\mathrm{d}t}\) oe \(\left(\text{“}\sqrt{3}\pi r^2\dfrac{\mathrm{d}r}{\mathrm{d}t}\text{”} = -2r\sqrt{3}\text{ oe}\right)\)	M1
\(\displaystyle\pi\int_{\frac{50}{\sqrt{3}}}^{0} r\,\mathrm{d}r = -\int_0^t 2\,\mathrm{d}t\) oe	M1
\(\left[\dfrac{\pi r^2}{2}\right]_{\frac{50}{\sqrt{3}}}^{0} = \left[-2t\right]_0^t\) oe	A1
\(-\dfrac{\pi \times 50^2}{6} = -2t\)	M1
Time \(= \dfrac{625\pi}{3}\) secs or 654 secs (3 sf) oe	A1	3.4

B1: Subst \(h = \dfrac{r}{\tan 30^\circ}\) into correct formula for \(V\)

M1: Attempt use chain rule to find \(\frac{\mathrm{d}V}{\mathrm{d}t}\) or \(\frac{\mathrm{d}r}{\mathrm{d}t}\) in terms of \(t\) and \(r\)
(Set their \(\frac{\mathrm{d}V}{\mathrm{d}t} = -2r\sqrt{3}\) oe)

M1: Attempt separate variables in their equation in terms of \(r\) and \(t\) only (not \(V\) or \(h\)). Integral signs not essential

A1: Correct integrals, any limits or none

M1: Substitute correct limits into integrals of the form \(ar^2\) & \(bt\)
OR substitute \(t = 0\) & \(r = \frac{50}{\sqrt{3}}\) to find \(c\) and substitute \(r = 0\)

A1: Allow without secs   or 10.9 mins or 10 mins 54 secs
SC. Use of \(r = h\sin 30\) (answer 491) can score M4A1

Example method 3 (NOT using chain rule)

This method is different from the summary method above

Scheme	Marks	AO
\(V = \frac{\pi}{3}(h\tan 30^\circ)^2h\) or \(V = \frac{\pi}{3}\left(\dfrac{h}{\sqrt{3}}\right)^2h\) oe	B1	3.3
\(h = \sqrt[3]{\dfrac{9V}{\pi}}\)	M1
\(\dfrac{\mathrm{d}V}{\mathrm{d}t} = -2 \times \sqrt[3]{\dfrac{9V}{\pi}}\)	M1
\(\displaystyle\sqrt[3]{\frac{\pi}{9}}\int_{\frac{\pi 50^3}{9}}^{0} V^{-1/3}\,\mathrm{d}V = -2\left[t\right]_0^t\)	M1
\(\displaystyle\sqrt[3]{\frac{\pi}{9}} \times \frac{3}{2}\left[V^{2/3}\right]_{\frac{\pi 50^3}{9}}^{0} = -2t\)	A1
\(\displaystyle -\sqrt[3]{\frac{\pi}{9}} \times \frac{3}{2} \times \left(\frac{\pi 50^3}{9}\right)^{2/3} = -2t\)	M1
Time \(= \dfrac{625\pi}{3}\) secs or 654 secs (3 sf) oe	A1

B1: or \(V = \frac{\pi}{9}h^3\) oe

M1: Allow \(h = kV^{1/3}\)

M1: \(\frac{\mathrm{d}V}{\mathrm{d}t} = -2 \times\)(their \(h\) in terms of \(V\))

M1: Attempt separate variables in their equation in terms of \(V\) and \(t\) only (not \(h\) or \(r\)). Integral signs not essential

A1: Correct integrals, any limits or none

M1: Substitute correct limits into integrals of forms \(aV^{2/3}\) & \(bt\)
OR substitute \(t = 0\) & \(V = \frac{\pi 50^3}{9}\) to find \(c\) and substitute \(V = 0\)

A1: Allow without secs   or 10.9 mins or 10 mins 54 secs
or:
SC. Use of \(r = h\sin 30\) (answer \(\frac{625\pi}{4}\) or 491) can score all 4 M-marks and final A1

Example incorrect method

Scheme	Marks
\(r = 50/\sqrt{3}\)   \(V = \dfrac{\pi}{3} \times 2500 \times \dfrac{h}{3}\)	B0
\(\dfrac{\mathrm{d}V}{\mathrm{d}h} = \dfrac{2500\pi}{9}\)	M0
\(\dfrac{\mathrm{d}V}{\mathrm{d}h} = \dfrac{\mathrm{d}V}{\mathrm{d}t} \times \dfrac{\mathrm{d}t}{\mathrm{d}h}\) \(\dfrac{2500\pi}{9} = -2h\dfrac{\mathrm{d}t}{\mathrm{d}h}\)	M1
\(\dfrac{\mathrm{d}h}{h} = -\dfrac{18}{2500\pi}\,\mathrm{d}t\)	M1