Edexcel A level Maths (9MA0) · Pure mathematics › Coordinate geometry
Practise Parametric equations in modelling. 2 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy6 marks
A stone is thrown from the top of a vertical cliff. The position of the stone, \(t\) seconds after it is released, is modelled by the parametric equations \(x = 12t, \quad y = 30 + 5t - 5t^2, \quad t \ge 0\) where \(x\) metres is the horizontal distance of the stone from the foot of the cliff and \(y\) metres is its height above the surface of the sea, which is modelled as horizontal.
(a) Find the height of the stone above the sea when it is 12 m horizontally from the foot of the cliff.[2]
(b) Find the time taken for the stone to reach the sea.[3]
(c) Find the horizontal distance of the stone from the foot of the cliff when it reaches the sea.[1]
Show the answer and mark scheme
(a)Answer: \(30\text{ m}\)
M1 for \(t = \frac{12}{12} = 1\) substituted into the equation for \(y\)
A1 for \(30\) (m)
Worked solution: \(12t = 12 \Rightarrow t = 1\), so \(y = 30 + 5 \times 1 - 5 \times 1^2 = 30\). The height is 30 m.
(b)Answer: \(t = 3\text{ s}\)
M1 for setting \(y = 0\) to obtain \(5t^2 - 5t - 30 = 0\) oe
M1 for solving their quadratic
A1 for \(t = 3\) (s), rejecting the negative root
Worked solution: \(5t^2 - 5t - 30 = 0\) factorises as \(5(t - 3)(t + 2) = 0\). Since \(t \ge 0\), \(t = 3\) seconds.
(c)Answer: \(36\text{ m}\)
B1ft for \(12 \times 3 = 36\) (m), following through their time
Worked solution: \(x = 12 \times 3 = 36\), so 36 m.
Question 2Medium9 marks
A flare is fired from the top of a lighthouse. The position of the flare, \(t\) seconds after it is released, is modelled by the parametric equations \(x = 6t, \quad y = 9 + 12t - 5t^2, \quad t \ge 0\) where \(x\) metres is the horizontal distance of the flare from the foot of the lighthouse and \(y\) metres is its height above the surface of the sea, which is modelled as horizontal.
(a) Show that a Cartesian equation for the path of the flare is \(y = 9 + 2x - \frac{5}{36}x^2\).[3]
(b) Find the maximum height of the flare above the sea.[3]
(c) Find the horizontal distance of the flare from the foot of the lighthouse when it reaches the sea.[3]
A1* for \(y = 9 + 2x - \frac{5}{36}x^2\) with no errors (cso)
Worked solution: \(t = \frac{x}{6}\), so \(y = 9 + 12\left(\frac{x}{6}\right) - 5\left(\frac{x}{6}\right)^2 = 9 + 2x - \frac{5}{36}x^2\).
(b)Answer: \(16.2\text{ m}\)
M1 for a valid method, e.g. \(\frac{\mathrm{d}y}{\mathrm{d}t} = 12 - 10t = 0\) or completing the square in \(t\) or \(x\)
A1 for \(t = \frac{6}{5}\) (or \(x = 7.2\))
A1 for \(16.2\) (m)
Worked solution: \(y = 9 + 12t - 5t^2 = 9 + \frac{36}{5} - 5\left(t - \frac{6}{5}\right)^2\). The maximum is at \(t = \frac{6}{5}\), giving \(y = 16.2\). The maximum height is 16.2 m.
(c)Answer: \(18\text{ m}\)
M1 for setting \(y = 0\) in the Cartesian equation or \(5t^2 - 12t - 9 = 0\)
M1 for solving to find the positive root
A1 for \(18\) (m)
Worked solution: \(5t^2 - 12t - 9 = 0\) gives \(t = 3\) (the other root is negative), so \(x = 6 \times 3 = 18\) m.
Question 3Hard11 marks
A reflector is fixed to the rim of a bicycle wheel of radius 0.35 m. The bicycle moves in a straight line along horizontal ground and the wheel rolls without slipping. The reflector starts at its lowest point, touching the ground. The position of the reflector is modelled by the parametric equations \(x = 0.35(\theta - \sin\theta), \quad y = 0.35(1 - \cos\theta), \quad \theta \ge 0\) where \(\theta\) radians is the angle through which the wheel has turned, \(x\) metres is the horizontal distance travelled by the reflector and \(y\) metres is its height above the ground. The diagram shows the path for \(0 \le \theta \le 2\pi\).
[object Object]
(a) Find the maximum height of the reflector above the ground and the horizontal distance it has travelled when it first reaches this height. Give the distance to 3 significant figures.[3]
(b) Find the horizontal distance travelled by the reflector between the first two occasions on which it touches the ground. Give your answer to 3 significant figures.[2]
(c) Find the values of \(\theta\), in the interval \(0 \le \theta \le 2\pi\), for which the reflector is 0.525 m above the ground.[3]
(d) Hence find the horizontal distance travelled by the reflector while it is more than 0.525 m above the ground, giving your answer to 3 significant figures.[3]
Show the answer and mark scheme
(a)Answer: 0.7 m, after a horizontal distance of \(0.35\pi = 1.10\) m
B1 for maximum height 0.7 m (when \(\cos\theta = -1\))
M1 for \(\theta = \pi\) substituted into \(x\)
A1 for \(0.35\pi = 1.10\) m (awrt)
Worked solution: \(y\) is greatest when \(\cos\theta = -1\), i.e. \(\theta = \pi\), giving \(y = 0.35 \times 2 = 0.7\) m. Then \(x = 0.35(\pi - \sin\pi) = 0.35\pi = 1.0996\ldots \approx 1.10\) m.
Worked solution: \(y = 0\) when \(\cos\theta = 1\), i.e. \(\theta = 0\) and \(\theta = 2\pi\). The distance is \(0.35(2\pi - \sin 2\pi) - 0 = 0.7\pi = 2.1991\ldots \approx 2.20\) m.