C4 June 2018 Q4

Figure 1: inverted cone of height 50 cm with semi-vertical angle 30 degrees, containing water of depth h cm with surface radius r cm; diagram not drawn to scale — Figure 1

A water container is made in the shape of a hollow inverted right circular cone with semi-vertical angle of 30°, as shown in Figure 1. The height of the container is 50 cm.

When the depth of the water in the container is \(h\) cm, the surface of the water has radius \(r\) cm and the volume of water is \(V\) cm³.

(a) Show that \(V = \dfrac{1}{9}\pi h^3\)
[You may assume the formula \(V = \dfrac{1}{3}\pi r^2 h\) for the volume of a cone.] (2)

Given that the volume of water in the container increases at a constant rate of 200 cm³ s⁻¹,

(b) find the rate of change of the depth of the water, in cm s⁻¹, when \(h = 15\)
Give your answer in its simplest form in terms of \(\pi\). (4)

Mark scheme (a)

Scheme	Marks
\(\dfrac{r}{h} = \tan 30 \Rightarrow r = h\tan 30\ \left\{\Rightarrow r = \dfrac{h}{\sqrt{3}} \text{ or } r = \dfrac{\sqrt{3}}{3}h\right\}\) or \(\dfrac{h}{r} = \tan 60 \Rightarrow r = \dfrac{h}{\tan 60}\ \left\{\Rightarrow r = \dfrac{h}{\sqrt{3}} \text{ or } r = \dfrac{\sqrt{3}}{3}h\right\}\) or \(\dfrac{r}{\sin 30} = \dfrac{h}{\sin 60} \Rightarrow r = \dfrac{h\sin 30}{\sin 60}\ \left\{\Rightarrow r = \dfrac{h}{\sqrt{3}} \text{ or } r = \dfrac{\sqrt{3}}{3}h\right\}\) or \(h^2 + r^2 = (2r)^2 \Rightarrow r^2 = \dfrac{1}{3}h^2\) Correct use of trigonometry to find \(r\) in terms of \(h\) or correct use of Pythagoras to find \(r^2\) in terms of \(h^2\)	M1
\(\left\{V = \dfrac{1}{3}\pi r^2 h \Rightarrow\right\}\ V = \dfrac{1}{3}\pi\left(\dfrac{h}{\sqrt{3}}\right)^2 h \Rightarrow V = \dfrac{1}{9}\pi h^3\ \) Correct proof of \(V = \dfrac{1}{9}\pi h^3\) or \(V = \dfrac{1}{9}h^3\pi\). Or shows \(\dfrac{1}{9}\pi h^3\) or \(\dfrac{1}{9}h^3\pi\) with some reference to \(V =\) in their solution*	A1 *
	(2)

Notes

Note: Allow M1 for writing down \(r = h\tan 30\)

Note: Give M0 A0 for writing down \(r = \dfrac{h\sqrt{3}}{3}\) or \(r = \dfrac{h}{\sqrt{3}}\) with no evidence of using trigonometry on \(r\) and \(h\) or Pythagoras on \(r\) and \(h\)

Note: Give M0 (unless recovered) for evidence of \(\dfrac{1}{3}\pi r^2 h = \dfrac{1}{9}\pi h^3\) leading to either \(r^2 = \dfrac{1}{3}h^2\) or \(r = \dfrac{h\sqrt{3}}{3}\) or \(r = \dfrac{h}{\sqrt{3}}\)

Mark scheme (b)

Scheme	Marks
Way 1
\(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 200\)
\(\dfrac{\mathrm{d}V}{\mathrm{d}h} = \dfrac{1}{3}\pi h^2\) \(\dfrac{1}{3}\pi h^2\) o.e.	B1
Either • \(\left\{\dfrac{\mathrm{d}V}{\mathrm{d}h} \times \dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{\mathrm{d}V}{\mathrm{d}t} \Rightarrow\right\}\ \left(\dfrac{1}{3}\pi h^2\right)\dfrac{\mathrm{d}h}{\mathrm{d}t} = 200\) • \(\left\{\dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{\mathrm{d}V}{\mathrm{d}t} \div \dfrac{\mathrm{d}V}{\mathrm{d}h} \Rightarrow\right\}\ \dfrac{\mathrm{d}h}{\mathrm{d}t} = 200 \times \dfrac{1}{\frac{1}{3}\pi h^2}\) either \(\left(\text{their } \dfrac{\mathrm{d}V}{\mathrm{d}h}\right) \times \dfrac{\mathrm{d}h}{\mathrm{d}t} = 200\) or \(200 \div \left(\text{their } \dfrac{\mathrm{d}V}{\mathrm{d}h}\right)\)	M1
When \(h = 15,\ \dfrac{\mathrm{d}h}{\mathrm{d}t} = 200 \times \dfrac{1}{\frac{1}{3}\pi(15)^2}\ \left\{= \dfrac{200}{75\pi} = \dfrac{600}{225\pi}\right\}\) dependent on the previous M mark	dM1
\(\dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{8}{3\pi}\ (\text{cm}\,\text{s}^{-1})\) \(\dfrac{8}{3\pi}\)	A1 cao
	(4)
	(6 marks)

Notes

Way 2

Scheme	Marks
\(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 200 \Rightarrow V = 200t + c \Rightarrow \dfrac{1}{9}\pi h^3 = 200t + c\)
\(\left(\dfrac{1}{3}\pi h^2\right)\dfrac{\mathrm{d}h}{\mathrm{d}t} = 200\) \(\dfrac{1}{3}\pi h^2\) o.e. as in Way 1	B1 M1
When \(h = 15,\ \dfrac{\mathrm{d}h}{\mathrm{d}t} = 200 \times \dfrac{1}{\frac{1}{3}\pi(15)^2}\ \left\{= \dfrac{200}{75\pi} = \dfrac{600}{225\pi}\right\}\) dependent on the previous M mark	dM1
\(\dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{8}{3\pi}\ (\text{cm}\,\text{s}^{-1})\) \(\dfrac{8}{3\pi}\)	A1 cao
	(4)

B1: Correct simplified or un-simplified differentiation of \(V\). E.g. \(\dfrac{1}{3}\pi h^2\) or \(\dfrac{3}{9}\pi h^2\)

Note: \(\dfrac{\mathrm{d}V}{\mathrm{d}h}\) does not have to be explicitly stated, but it should be clear that they are differentiating their \(V\)

M1: \(\left(\text{their } \dfrac{\mathrm{d}V}{\mathrm{d}h}\right) \times \dfrac{\mathrm{d}h}{\mathrm{d}t} = 200\) or \(200 \div \left(\text{their } \dfrac{\mathrm{d}V}{\mathrm{d}h}\right)\)

dM1: dependent on the previous M mark
Substitutes \(h = 15\) into an expression which is a result of either \(200 \div \left(\text{their } \dfrac{\mathrm{d}V}{\mathrm{d}h}\right)\) or \(200 \times \dfrac{1}{\left(\text{their } \frac{\mathrm{d}V}{\mathrm{d}h}\right)}\)

A1: \(\dfrac{8}{3\pi}\) (units are not required)

Note: Give final A0 for using \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = -200\) to give \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = -\dfrac{8}{3\pi}\), unless recovered to \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{8}{3\pi}\)