C4 January 2008 Q8

EdexcelOld spec13 marksDifferentiation Integration

8. Liquid is pouring into a large vertical circular cylinder at a constant rate of 1600 cm\(^3\) s\(^{-1}\) and is leaking out of a hole in the base, at a rate proportional to the square root of the height of the liquid already in the cylinder. The area of the circular cross section of the cylinder is 4000 cm\(^2\).

(a) Show that at time \(t\) seconds, the height \(h\) cm of liquid in the cylinder satisfies the differential equation\[\frac{\mathrm{d}h}{\mathrm{d}t} = 0.4 - k\sqrt{h}, \text{ where } k \text{ is a positive constant.}\] (3)

When \(h = 25\), water is leaking out of the hole at 400 cm\(^3\) s\(^{-1}\).

(b) Show that \(k = 0.02\) (1)

(c) Separate the variables of the differential equation\[\frac{\mathrm{d}h}{\mathrm{d}t} = 0.4 - 0.02\sqrt{h},\]to show that the time taken to fill the cylinder from empty to a height of 100 cm is given by\[\int_0^{100}\frac{50}{20 - \sqrt{h}}\,\mathrm{d}h.\] (2)

Using the substitution \(h = (20 - x)^2\), or otherwise,

(d) find the exact value of \(\displaystyle\int_0^{100}\frac{50}{20 - \sqrt{h}}\,\mathrm{d}h\). (6)

(e) Hence find the time taken to fill the cylinder from empty to a height of 100 cm, giving your answer in minutes and seconds to the nearest second. (1)

Mark scheme (a)

Scheme	Marks
\(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 1600 - c\sqrt{h}\) or \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 1600 - k\sqrt{h}\),	M1
\(\left(V = 4000h \Rightarrow\right) \dfrac{\mathrm{d}V}{\mathrm{d}h} = 4000\)	M1
\(\dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{\mathrm{d}h}{\mathrm{d}V} \times \dfrac{\mathrm{d}V}{\mathrm{d}t} = \dfrac{\frac{\mathrm{d}V}{\mathrm{d}t}}{\frac{\mathrm{d}V}{\mathrm{d}h}}\)
Either, \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{1600 - c\sqrt{h}}{4000} = \dfrac{1600}{4000} - \dfrac{c\sqrt{h}}{4000} = 0.4 - k\sqrt{h}\) or \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{1600 - k\sqrt{h}}{4000} = \dfrac{1600}{4000} - \dfrac{k\sqrt{h}}{4000} = 0.4 - k\sqrt{h}\)	A1 AG
	(3)

Notes

M1: Either of these statements

M1: \(\dfrac{\mathrm{d}V}{\mathrm{d}h} = 4000\) or \(\dfrac{\mathrm{d}h}{\mathrm{d}V} = \dfrac{1}{4000}\)

A1 AG: Convincing proof of \(\dfrac{\mathrm{d}h}{\mathrm{d}t}\)

Mark scheme (b)

Scheme	Marks
When \(h = 25\) water *leaks out such that* \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 400\)
\(400 = c\sqrt{h} \Rightarrow 400 = c\sqrt{25} \Rightarrow 400 = c(5) \Rightarrow c = 80\)
From above; \(k = \dfrac{c}{4000} = \dfrac{80}{4000} = 0.02\) as required	B1 AG
	(1)

Notes

B1 AG: Proof that \(k = 0.02\)

Aliter (b) Way 2

Scheme	Marks
\(400 = 4000k\sqrt{h}\)
\(\Rightarrow 400 = 4000k\sqrt{25}\)
\(\Rightarrow 400 = k(20000) \Rightarrow k = \tfrac{400}{20000} = 0.02\)	B1 AG
	(1)

B1 AG: Using 400, 4000 and \(h = 25\) or \(\sqrt{h} = 5\). Proof that \(k = 0.02\)

Mark scheme (c)

Scheme	Marks
\(\dfrac{\mathrm{d}h}{\mathrm{d}t} = 0.4 - k\sqrt{h} \Rightarrow \displaystyle\int\frac{\mathrm{d}h}{0.4 - k\sqrt{h}} = \int\mathrm{d}t\)	M1 oe
\(\therefore\) time required \(= \displaystyle\int_0^{100}\frac{1}{0.4 - 0.02\sqrt{h}}\,\mathrm{d}h \quad \frac{\div 0.02}{\div 0.02}\)
time required \(= \displaystyle\int_0^{100}\frac{50}{20 - \sqrt{h}}\,\mathrm{d}h\)	A1 AG
	(2)

Notes

M1 oe: Separates the variables with \(\displaystyle\int\frac{\mathrm{d}h}{0.4 - k\sqrt{h}}\) and \(\displaystyle\int\mathrm{d}t\) on either side with integral signs not necessary.

A1 AG: Correct proof

Mark scheme (d)

Scheme	Marks
\(\displaystyle\int_0^{100}\frac{50}{20 - \sqrt{h}}\,\mathrm{d}h\) with substitution \(h = (20 - x)^2\)
\(\dfrac{\mathrm{d}h}{\mathrm{d}x} = 2(20 - x)(-1)\) or \(\dfrac{\mathrm{d}h}{\mathrm{d}x} = -2(20 - x)\)	B1 aef
\(h = (20 - x)^2 \Rightarrow \sqrt{h} = 20 - x \Rightarrow x = 20 - \sqrt{h}\)
\(\displaystyle\int\frac{50}{20 - \sqrt{h}}\,\mathrm{d}h = \int\frac{50}{x}.-2(20 - x)\,\mathrm{d}x\)	M1
\(= 100\displaystyle\int\frac{x - 20}{x}\,\mathrm{d}x\)
\(= 100\displaystyle\int\left(1 - \frac{20}{x}\right)\mathrm{d}x\)
\(= 100(x - 20\ln x)\ \ (+c)\)	M1 A1
change limits: when \(h = 0\) then \(x = 20\) and when \(h = 100\) then \(x = 10\)
\(\displaystyle\int_0^{100}\frac{50}{20 - \sqrt{h}}\,\mathrm{d}h = \big[100x - 2000\ln x\big]_{20}^{10}\) or \(\displaystyle\int_0^{100}\frac{50}{20 - \sqrt{h}}\,\mathrm{d}h = \Big[100\left(20 - \sqrt{h}\right) - 2000\ln\left(20 - \sqrt{h}\right)\Big]_0^{100}\)
\(= (1000 - 2000\ln 10) - (2000 - 2000\ln 20)\)	ddM1
\(= 2000\ln 20 - 2000\ln 10 - 1000\)
\(= 2000\ln 2 - 1000\)	A1 aef
	(6)

Notes

B1 aef: Correct \(\dfrac{\mathrm{d}h}{\mathrm{d}x}\)

M1: \(\pm\lambda\displaystyle\int\frac{20 - x}{x}\,\mathrm{d}x\) or \(\pm\lambda\displaystyle\int\frac{20 - x}{20 - (20 - x)}\,\mathrm{d}x\) where \(\lambda\) is a constant

M1: \(\pm\alpha x \pm \beta\ln x\); \(\alpha, \beta \ne 0\) A1: \(100x - 2000\ln x\)

ddM1: Correct use of limits, ie. putting them in the correct way round. Either \(x = 10\) and \(x = 20\) or \(h = 100\) and \(h = 0\)

A1 aef: Combining logs to give… \(2000\ln 2 - 1000\) or \(-2000\ln\left(\tfrac{1}{2}\right) - 1000\)

Mark scheme (e)

Scheme	Marks
Time required \(= 2000\ln 2 - 1000 = 386.2943611\ldots\) sec
\(= 386\) seconds (nearest second)
\(=\) 6 minutes and 26 seconds (nearest second)	B1
	(1)
	(13 marks)

Notes

B1: 6 minutes, 26 seconds