A2 October 2020 Paper 2 Q9

OCR ACurrent spec11 marksHyperbolic Functions

9 Two thin poles, \(OA\) and \(BC\), are fixed vertically on horizontal ground. A chain is fixed at \(A\) and \(C\) such that it touches the ground at point \(D\) as shown in the diagram.

On a coordinate system the coordinates of \(A\), \(B\) and \(D\) are \((0, 3)\), \((5, 0)\) and \((2, 0)\).

Diagram, not to scale: x and y axes with origin O; vertical pole OA on the y-axis and vertical pole BC at B on the x-axis, with C much higher than A; a chain hangs from A down to touch the x-axis at D and rises steeply to C

It is required to find the height of pole \(BC\) by modelling the shape of the curve that the chain forms.

Jofra models the curve using the equation \(y = k\cosh(ax - b) - 1\) where \(k\), \(a\) and \(b\) are positive constants.

(a) Determine the value of \(k\). [2]

(b) Find the exact value of \(a\) and the exact value of \(b\), giving your answers in logarithmic form. [5]

Holly models the curve using the equation \(y = \frac{3}{4}x^2 - 3x + 3\).

(c) Write down the coordinates of the point, \((u, v)\) where \(u\) and \(v\) are both non-zero, at which the two models will agree. [1]

(d) Show that Jofra’s model and Holly’s model disagree in their predictions of the height of pole \(BC\) by 3.32 m to 3 significant figures. [3]

Mark scheme (a)

Scheme	Marks	AO
Min value of cosh is 1 (and point on ground is at the minimum)	M1	2.2a
(so \(0 = k \times 1 - 1 \Rightarrow\)) \(k = 1\)	A1	2.2a
	[2]

Notes

M1: Using minimum point of curve and knowledge of cosh graph. Could be derived by differentiation

A1: If zero scored then sc1 for k=1 www

Mark scheme (b)

Scheme	Marks	AO
Passes through \((0, 3) \Rightarrow 3 = \cosh(-b) - 1\) \(\Rightarrow b = -\cosh^{-1}(3 + 1)\)	*M1	3.3
\(b = (\pm)\ln\left(4 + \sqrt{(4^2 - 1)}\right)\)	dep*M1	3.1a
\(\Rightarrow b = \ln(4 + \sqrt{15})\)	A1	1.1
Passes through \((2, 0) \Rightarrow 0 = \cosh(2a - b) - 1\) \(\Rightarrow b = 2a\)	M1	3.3
\(\Rightarrow a = \frac{1}{2}\ln(4 + \sqrt{15})\)	A1	1.1
	[5]

Notes

*M1: Use of \((0, 3)\) to derive an expression for \(b\). accept \(\cosh(-b) = \frac{4}{k}\)

dep*M1: Correct numerical use of formula. Or rearranges.

M1: Use of \((2, 0)\) to derive \(b = 2a\). Could be from (a). Allow ft

Scheme	Marks	AO
(By symmetry of both;) \((4, 3)\)	B1	2.2a
	[1]

Mark scheme (d)

Scheme	Marks	AO
Holly’s model; \(d_{\mathrm{H}} = 6.75\)	B1	3.4
Jofra’s model: \(d_{\mathrm{J}} = \cosh(5a - b) - 1\)	M1	3.4
AG \(d_{\mathrm{J}} - d_{\mathrm{H}} = 10.067\ldots - 6.75 = 3.32\) (3 sf)	A1	1.1
	[3]

Notes

B1: Condone 27/4

M1: Use of \(x = 5\) with their values of \(a\) and \(b\) to predict \(d\). Must have \(-1\)
\(a = 1.0317\ldots,\ b = 2.0634\ldots,\ d_{\mathrm{J}} = 10.067\ldots\) Condone 10.07 only if clear evidence of production

A1: From correct values