June 2022 Paper 1 Q8

OCR MEICurrent spec10 marksDifferentiation Parametric Equations

8 A particle moves in the \(x\)-\(y\) plane so that its position at time \(t\) s is given by \(x = t^3 - 8t,\ y = t^2\) for \(-3.5 \lt t \lt 3.5\). The units of distance are metres. The graph shows the path of the particle and the direction of travel at the point P \((8, 4)\).

Path of the particle: a curve starting at the origin O, forming a loop to the left and right of the y-axis that crosses itself on the y-axis, with two branches continuing upwards to the left and right; the point P on the lower right part of the loop with an arrow showing the direction of travel down and to the right

(a) Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(t\). [3]

(b) Hence show that the value of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) at P is \(-1\). [2]

(d) Find the cartesian equation of the path, giving \(x^2\) as a function of \(y\). [3]

Mark scheme (a)

Scheme	Marks	AO
\(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 3t^2 - 8,\quad \dfrac{\mathrm{d}y}{\mathrm{d}t} = 2t\)	M1*	1.1a
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2t}{3t^2 - 8}\)	M1 (dep) A1	1.1b 1.1b
	[3]

Notes

M1*: attempt to differentiate both parametric equations
Only allow for a complete method for finding \(\frac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(t\) using the cartesian equation of the curve

M1: Combine their derivatives to find \(\frac{\mathrm{d}y}{\mathrm{d}x}\). Do not allow for reciprocal

A1: cao

Mark scheme (b)

Scheme	Marks	AO
AG \(t^3 - 8t = 8\) and \(t^2 = 4\) gives \(t = -2\)	M1	3.1a
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2(-2)}{3(-2)^2 - 8} = -1\)	E1	1.1b
	[2]

Notes

M1: Attempt to establish the value of \(t\) at \((8, 4)\). Allow for \(\pm 2\) or 2 stated
Allow for \(y = 4\) used in \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2\sqrt{y}}{3y - 8}\) for the M mark only

E1: AG the negativity must be clearly established from correct working

Alternative

Scheme	Marks
When \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2t}{3t^2 - 8} = -1\) giving \(3t^2 + 2t - 8 = 0\) \(t = \frac{4}{3}\) or \(t = -2\)	M1
When \(t = -2\) the coordinates are \(\left((-2)^3 - 8(-2), (-2)^2\right) = (8, 4)\) [which is P]	E1

M1: Uses the value of the derivative to find the value of \(t\) at P.

E1: Allow without reference to \(t = \dfrac{4}{3}\)

Mark scheme (c)

Scheme	Marks	AO
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2t}{3t^2 - 8} = -1\) giving \(3t^2 + 2t - 8 = 0\)	M1	1.1a
\(t = \frac{4}{3}\) or [\(t = -2\) is the point P]	A1	3.2a
	[2]

Notes

M1: Equating their \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to \(-1\) and rearranging to form quadratic equation

A1: allow www
\(-2\) need not be seen but if seen must be rejected

Mark scheme (d)

Scheme	Marks	AO
Substitute \(t^2 = y\)	M1	1.1a
\(x = t^3 - 8t \Rightarrow x^2 = t^6 - 16t^4 + 64t^2\)	A1	1.1b
\(\Rightarrow x^2 = y^3 - 16y^2 + 64y\)	A1	2.1
	[3]

Notes

A1: Allow for \(x^2 = t^2\left(t^2 - 8\right)^2\)

Alternative method

Scheme	Marks
Substitute \(t = \pm y^{\frac{1}{2}}\)	M1
\(x = \pm\left(y^{\frac{3}{2}} - 8y^{\frac{1}{2}}\right)\)	A1
\(x^2 = y(y - 8)^2 = \left[y^3 - 16y^2 + 64y\right]\)	A1

M1: Substituting for \(t\) in their equation for \(x\); allow without \(\pm\)

A1: Soi Allow without \(\pm\)

A1: must be in the form \(x^2 = \ldots\) from fully correct working
Need not be simplified. Do not award if \(\pm\) not seen at all