Notes

Mark parts (a) and (b) together.

M1: Substitutes \(x = -3\) into \(x = \dfrac{t - 1}{2}\), attempts to find \(t\), and substitutes their \(t\) into \(y = 5(t + 2)^4\)
Condone slips. May be implied by substitution of \(t = -5\) into \(y\) or by 405 or by \(y = 5(-3)^4\)
Alternatively, they may substitute \(x = -3\) into their \(y = \mathrm{f}(x)\)

A1: cao Answer only scores full marks and ISW after seeing 405 e.g. \((-5,\ 405)\)
Accept \((-3,\ 405)\) or e.g. \(P = 405\)

Notes

Mark parts (a) and (b) together.

M1: Attempts to make \(t\) the subject of \(x = \dfrac{t - 1}{2}\) using the correct order of operations and substitutes into \(y = 5(t + 2)^4\) Condone slips e.g. dividing by 2 first: \(t = \dfrac{x}{2} + 1\) or missing the +2 in \(y = 5(t + 2)^4\)
Alt 1: Writes \(t = \left(\dfrac{y}{5}\right)^{\frac{1}{4}} - 2\), substitutes into \(x = \dfrac{t - 1}{2}\) and attempts to make \(y\) the subject using the correct order of operations, i.e., \(2x = \left(\dfrac{y}{5}\right)^{\frac{1}{4}} - 3 \Rightarrow (2x + 3)^4 = \dfrac{y}{5} \Rightarrow y = 5(2x + 3)^4\)
Alt 2: Makes \(t\) the subject and then substitutes into an expanded \(5(t + 2)^4\)
Do not be too concerned about their expansion, but it must contain \(t^4\) and a constant term.

A1: cao and ISW after a correct answer seen. May be scored for \(y = 5(2x + 1 + 2)^4\)
Allow e.g. \(y = 80x^4 + 480x^3 + 1080x^2 + 1080x + 405\) or e.g.
\(y = 5(2x + 1)^4 + 40(2x + 1)^3 + 120(2x + 1)^2 + 160(2x + 1) + 80\)
Their RHS may be unsimplified but do not allow if e.g. binomial coefficients are still present.
Do not accept \(\mathrm{f}(x) = \ldots\) It must be \(y = \ldots\)

Notes

Note: Differentiation seen in (a) or (b) can score marks if used in (c)

M1: Reduces the power of their \((\text{``}2x + 3\text{''})^n\) by one to \(Q(\text{``}2x + 3\text{''})^{n-1}\) where \(Q\) is a constant and could be 1. There should be no other terms using this method.
Alternatively, attempts to expand their \((\text{``}2x + 3\text{''})^n\) (may have been expanded in (b)) and reduces the power of \(x\) by one in at least one term.
They may use parametric differentiation, i.e., \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = a(t + 2)^3\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = b\) where \(a\) and \(b\) are constants, leading to \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{a(t + 2)^3}{b}\) i.e. they must divide the correct way round.
Condone attempts at the chain rule that reach e.g. \(y^{\prime} = \ldots u^3\) but make a slip when substituting back in for \(x\) or \(t\), e.g., \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = (5x + 3)^3\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = (t + 2)^2\), provided the intention is clear.

dM1: Substitutes \(x = -3\) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(x\) which may be implied by their answer.
Using parametric differentiation, substitutes their \(t\) (found from an attempt at substituting \(x = -3\) into \(x = \dfrac{t - 1}{2}\) which may have been seen in (a)) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(t\).
Note for reference, if correct, the parametric differentiation is \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 20(t + 2)^3\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{1}{2}\) leading to \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 40(t + 2)^3\)
They may substitute their value of \(t\) into \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) first before using the chain rule to reach \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) which is acceptable and implies the first M mark.

A1: \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) -1080\)
Correct answer only scores full marks.
May be seen labelled as \(m\) or something else.

Notes

(a) If parametric differentiation is not used in part (a) (e.g. uses Cartesian form) then only the B mark is available but see alternative below.

B1: Uses the given Cartesian coordinates to deduce the correct value for \(t\).
If more than one value for \(t\) e.g. \(t = -5\) is given and \(t = -1\) is not “selected” score B0 but if just \(t = -1\) is used subsequently allow recovery and score B1

M1: Attempts to use \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t} \times \dfrac{\mathrm{d}t}{\mathrm{d}x}\) or equivalent with their differentiated equations.
There must be an attempt to differentiate both parameters, however poor, and divide or multiply correctly so using \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{y}{x}\) scores M0. Both parameters must be “changed”.
Condone confusion with the variables e.g. referring to \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) as \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) if the intention is clear.
This may be implied by e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = -3t^2,\ \dfrac{\mathrm{d}x}{\mathrm{d}t} = 2(t+3), t = -1,\ \dfrac{\mathrm{d}y}{\mathrm{d}t} = -3,\ \dfrac{\mathrm{d}x}{\mathrm{d}t} = 4 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{3}{4}\)

M1: Uses their numerical value of \(t\) (not 4) in their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) to obtain a value.
Condone attempts with different values of \(t\) e.g. \(t = -1\) and \(t = -5\)

ddM1: Applies a correct straight line method with their value of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) which has come from an attempt to use parametric differentiation with their value of \(t\) (not 4) and with \(x = 4\) and \(y = 2\) correctly placed. An attempt at the equation of the normal is M0.
If using \(y = mx + c\) they must reach as far as \(c = \ldots\)
Depends on both previous M marks.

A1*: Correct equation as printed with no errors but condone \(4y + 3x = 20\ *\)
Allow equivalents e.g. \(20 = 4y + 3x\ *\) or \(3x + 4y = 20\ *\)
This is a printed answer so there must be at least one intermediate step as shown in the main scheme.

Alternative for (a) using parametric differentiation but avoids the need for a value for \(t\):

B1: Either a correct expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(x\) and/or \(y\) following a correct \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(t\) or for \(t = -1\) seen anywhere.

M1: Attempts to use \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{d}y}{\mathrm{d}t} \times \dfrac{\mathrm{d}t}{\mathrm{d}x}\) with their differentiated equations.
There must be an attempt to differentiate both parameters, however poor, and divide or multiply correctly so using \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{y}{x}\) scores M0
Condone confusion with the variables e.g. referring to \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) as \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) if the intention is clear.
This may be implied by e.g. \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = -3t^2,\ \dfrac{\mathrm{d}x}{\mathrm{d}t} = 2(t+3), t = -1,\ \dfrac{\mathrm{d}y}{\mathrm{d}t} = -3,\ \dfrac{\mathrm{d}x}{\mathrm{d}t} = 4 \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{3}{4}\)

M1: Attempts to express their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) which is in terms of \(t\), in terms of \(x\) and/or \(y\) and uses \(x = 4\) and \(y = 2\) correctly placed in an attempt to find the gradient of the tangent.

ddM1: Applies a correct straight line method with their value of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) which has come from an attempt to use parametric differentiation with their gradient and with \(x = 4\) and \(y = 2\) correctly placed.
If using \(y = mx + c\) they must reach as far as \(c = \ldots\)
Depends on both previous M marks.

A1*: Correct equation as printed with no errors.
This is a printed answer so there must be at least one intermediate step as shown in the main scheme.

Notes

B1: 9m or equivalent including correct units. Accept e.g. 9 metres, 900cm etc.

Typical solution

Typical solution

Typical solution

The positive square root is used since \(-\dfrac{1}{4} \leqslant t \leqslant 0\)

Typical solution

When \(t = 0.2\), \(y = 10.2\)

\(t = 3\), \(y = \dfrac{11}{3}\)

The slide is safe.

Typical solution

(i)

(ii)

Length of \(RS\) = 2.83 metres

(iii)

Typical solution

Length = 180.5 cm

Typical solution

(i)

(ii)

Width of surfboard = 58 cm

Hence the width is approximately one third of the length

Typical solution

Notes

M1: Attempt to find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\). Correctly combine attempts at algebraic or numerical derivatives

A1: Correct \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\). Algebraic or numerical (ie \(\frac{4}{5}\)) www

B1: Correct \(t\) stated or explicitly used. Ignore any additional \(t\) values

M1: Attempt equation of tangent, using \((2, 3)\) and their numerical gradient (from clearly using their \(t\) value in their attempt at parametric differentiation)
If using \(y = mx + c\) then M1 is awarded when value for \(c\) is obtained
M0 if using \(x = 3\), \(y = 2\)

A1: Obtain given answer www

If trying to start with a Cartesian equation there are a number of different, correct, solutions. To gain any credit it would have to be a valid attempt in obtaining the initial equation, condoning only a slip. Please consult with your TL if seen.

Notes

M1*: Attempt to solve equations of tangent and curve simultaneously. Or valid attempt to obtain an equation in \(x\) or \(y\) only

A1: Obtain correct cubic. Brackets expanded and like terms collected
Some (usually BC) methods may not explicitly state simplified cubic; if M1 awarded and \(t = -\frac{7}{4}\) seen www then allow first A1 BOD, with then possibly SCB1

M1dep*: Attempt to solve cubic to obtain \(t\) value other than \(t = 1\). DR so method needed
Factorise to 3 linear factors and solve oe, eg using division
If using \((t - 1)^2\) as repeated factor, M1 for \((t - 1)^2(4t \pm 7)\)
If using \((t - 1)\) as linear factor then need both attempt at quadratic factor (factorisation or division) and also attempt to solve quadratic – BOD for slips

A1: Obtain correct coordinates, in exact form, www. Ignore \((2, 3)\) if also given, but A0 if other point(s) as well

If M1 A1 awarded for correct cubic, but no method shown to solve either cubic, or quadratic from \((t - 1)\)(quadratic factor), then SC B1 for correct coordinates www ie max of 3 marks

See comment in part (a) with regard to using Cartesian equation

Notes

M1: Attempt \(\displaystyle\int x\frac{\mathrm{d}y}{\mathrm{d}t}\,\mathrm{d}t\) in terms of \(t\), detail required
Clear indication that integrand is given by \(\displaystyle\int x\frac{\mathrm{d}y}{\mathrm{d}t}\,\mathrm{d}t\) (condone just \(\displaystyle\int x\frac{\mathrm{d}y}{\mathrm{d}t}\)), along with \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 4t + 3\) and full substitution into integrand
Condone no d\(t\) in integrand in initial statement and/or when substituting
May instead see integrand as \(\displaystyle\int x\,\mathrm{d}y\) with \(\mathrm{d}y = (4t + 3)\,\mathrm{d}t\)

A1: Obtain correct given integrand
d\(t\) required throughout

B1: Determine correct lower limit from solving equation
Evidence for \(a = 0\) required eg \(2t^2 + 3t = 0\)
\(a = 0\) doesn’t need to be seen explicitly, and could be implied by 0 appearing as the lower limit on an integral sign once sufficient evidence seen
Mark independently of any integrand attempted

M1: Equate expression for \(y\) to 2
Could be implied by \(t = \tfrac{1}{2}\) seen as a limit

A1: Obtain \(b = \tfrac{1}{2}\) as upper limit www
No need for \(t = -2\) to be explicitly rejected
\(b = \tfrac{1}{2}\) doesn’t need to be seen, and may be implied by appearing as the upper limit on an integral sign
Mark independently of any integrand attempted

Notes

M1*: Use \(u = 2t + 1\) to attempt to change entire integrand to a function of \(u\)
Attempt to write numerator, denominator and d\(t\) in terms of \(u\)

A1: Obtain correct integrand
Condone no d\(u\)

M1*: Attempt integration to obtain integral of form \(au^{-2} + bu^{-3}\)
M0 if additional terms

M1d*: Attempt use of correct limits: either correct \(t\) limits (ie \(a = 0\) and \(b = \tfrac{1}{2}\)) in a \(t\)-integral or commensurate upper and lower limits in an integral involving a substitution (eg with \(u = 2t + 1\), then upper limit must be 2 and lower limit must be 1)
Dependent on M1 M1
Minimum evidence needed is two terms ie \(\left(-\dfrac{7}{24}\right) - \left(-\dfrac{4}{3}\right)\) or \(\left(\dfrac{3}{4}\right) + \left(\dfrac{7}{24}\right)\)
If these values are not seen then M1 can be awarded for term by term substitution seen (ie 4 terms needed), but allow one error

A1: Obtain correct area, any exact equivalent
Explicit use of limits must be seen in a correct integral for A1

Candidates may mix and match methods eg start with substitution and then try to do the actual integration by parts – the MS allows M1A1 for changing the integrand to useable form; M1A1 for doing the integration; M1A1 for use of limits

Alternative method (integration by parts)

M1*: Attempt integration by parts
Correct parts and correct formula

A1: Obtain correct first step
Allow unsimplified

M1*: Attempt integration to obtain integral of form \(a(8t + 6)(2t + 1)^{-3} + b(2t + 1)^{-2}\)

A1: Obtain fully correct integral
Allow unsimplified

M1d*: Attempt use of correct limits
See guidance in main MS

A1: Obtain correct area, any exact equivalent

Alternative method (separate fractions)

M1*: Attempt to rewrite integrand as separate fractions with constant numerators
Could be informal method, or use of partial fractions (extending expected knowledge)
As far as \(\dfrac{P}{(2t + 1)^3} + \dfrac{Q}{(2t + 1)^4}\), with \(P\) and \(Q\) as constants, and no other fractions

A1: Obtain correct integrand

M1*: Attempt integration to obtain integral of form \(\dfrac{a}{(2t + 1)^2} + \dfrac{b}{(2t + 1)^3}\)

A1: Obtain fully correct integral
Allow unsimplified

M1d*: Attempt use of correct limits
See guidance in main MS

A1: Obtain correct area, any exact equivalent

Alternative method (integrating between curve and \(x\)-axis)

M1*: Attempt integration by substitution / integration by parts on correct expression
Apply the same MS as for integrating between curve and \(y\)-axis

A1: Obtain correct integrand
Using substitution eg \(u = 2t + 1\)
Using integration by parts – first stage required for M1

M1*: Apply the same MS as for integrating between curve and \(y\)-axis

A1: Obtain fully correct integral
Allow unsimplified

M1d*: Attempt use of correct limits, and combine with correct area of rectangle \(\left(= \tfrac{1}{4}\right)\)
Limits must be \(\displaystyle\int_{\frac{1}{2}}^{0}\) or commensurate \(u\)-limits, and used in the correct order
See guidance in main MS, but must also add on the correct area of the rectangle

A1: Obtain correct area, any exact equivalent

Notes

M1: Re-arranges to obtain both \(\sin t = \dfrac{x \pm 4k}{k}\) and \(\cos t = \dfrac{y \pm 2}{4}\) and use \(\sin^2 t + \cos^2 t = 1\) to eliminate \(t\)

A1: Allow any correct un-simplified cartesian form not involving trigonometric terms
ISW once a correct answer seen

SC for part (a): If M0 awarded then SC B1 for \(y = 2 + 4\cos\left(\arcsin\left(\dfrac{x - 4k}{k}\right)\right)\) or \(x = 4k + k\sin\left(\arccos\left(\dfrac{y - 2}{4}\right)\right)\) or \(\arcsin\left(\dfrac{x - 4k}{k}\right) = \arccos\left(\dfrac{y - 2}{4}\right)\) (or any correct form in terms of trig. functions – allow \(\cos^{-1}\) for \(\arccos\) etc.)

If considering one branch only e.g. \(y\left(= 2 + 4\sqrt{1 - \sin^2 t}\right) = 2 + 4\sqrt{1 - \left(\dfrac{x - 4k}{k}\right)^2}\) this scores M1 only (or equivalent expression for \(x\))
If considering both branches e.g. \(y\left(= 2 \pm 4\sqrt{1 - \sin^2 t}\right) = 2 \pm 4\sqrt{1 - \left(\dfrac{x - 4k}{k}\right)^2}\) then this scores M1 A1 (or equivalent expression for \(x\))
Note a (common) correct answer for M1 A1 is: \(x^2 + y^2 = 16k^2 + 8k(x - 4k) + (x - 4k)^2 + 4 + 4(y - 2) + (y - 2)^2\)

Notes

(b)(i)

M1: Setting \(c\) and \(d\) equal in their \(\dfrac{(x \pm a)^2}{c} + \dfrac{(y \pm b)^2}{d} = 1\) or stating that \(y_{\min} = -2\) and \(y_{\max} = 6\)
Possibly implied by correct value for \(r\) www

A1: \(r = \pm 4\) is A0 unless replaced with positive \(r\) only

(b)(ii) B2: B1 for either one correct centre or both \(x\)-coordinates correct
These marks are dependent on \(r = 4\) from correct working or www

Notes

B1*: B1 for \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0.2 \pm \sin t\)

M1dep*: Uses \(\displaystyle\int y\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}t\) and replaces \(\sin^2 t\) with \(1 - \cos^2 t\) to obtain an expression involving \(\cos^2 t\) and \(\sin t\)
No limits required for this mark

A1: AG (so must be checked carefully for any errors e.g. must contain relevant brackets around the \(1 - \cos^2 t\) term(s)). A correct expression e.g. \(\displaystyle\int_0^{\pi} k(1 - \cos^2 t)(0.2 + \sin t)\,\mathrm{d}t\) followed by the correct given answer can score this mark – limits and d\(t\) must be seen at least once (but need not be on the final integral)
Do not need to see \(= 1\) anywhere in their solution (and condone lack of brackets around the integrand)

Notes

B1: Correctly applies \(2\cos^2 t \equiv 1 + \cos 2t\) (so not just stating this identity) - implied by seeing \(-0.1t - 0.05\sin 2t\) after integration
e.g. applies could be for an attempt to integrate \(\frac{1}{2}(1 + \cos 2t)\) or stating this identity in an integral

B1: First two terms integrated correctly (Look out for those that have \(0.1t\) only from combining \(0.2t\) with \(-0.1t\))
This mark should be awarded if 0.2 is combined with another constant term and integrated correctly

M1*: M1 for an answer of the form \(\pm pt \pm q\sin 2t \pm r\cos^3 t\) or an answer of the form \(\pm pt \pm q\sin 2t \pm r\sin t\sin 2t \pm u\cos t\cos 2t\) or an answer of the form \(\pm pt \pm q\cos t \pm r\sin 2t \pm u\sin t\sin 2t \pm v\cos t\cos 2t\) for non-zero \(p\), \(q\), \(r\) (and \(u\), \(v\)) from integrating \(-0.2\cos^2 t - \sin t\cos^2 t\)

A1: A1 for the correct remaining three/four terms
(Alternatives: \(-0.1t - 0.05\sin 2t + \frac{1}{6}\sin t\sin 2t + \frac{1}{3}\cos t\cos 2t\) or \(-0.1t + 0.5\cos t - 0.05\sin 2t - \frac{1}{3}\sin t\sin 2t - \frac{1}{6}\cos t\cos 2t\))

M1dep*: Uses correct limits correctly \(F(\pi) - F(0)\) - condone limits the wrong way round only if the sign of their answer is subsequently changed
Must be using exact values (so must include term(s) in \(\pi\))

A1: www oe e.g. \(k = \dfrac{1}{0.1\pi + \frac{4}{3}}\) - isw once a correct expression for \(k\) seen
Correct exact value must be seen at some stage for the final mark – check first A mark carefully

Notes

M1: Attempt correct process to find gradient in terms of \(t\) or \(p\)
Correctly combine attempts at two derivatives
Need \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = kt^{-2}\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 2\)
SC B1 for gradient of \(-2x^{-2}\) if it is never seen in terms of \(t\) or \(p\)

A1: Obtain correct gradient
In terms of \(t\) or \(p\)

M1: Attempt equation of tangent
Condone still working in terms of \(t\)
Allow mixture of \(t\) and \(p\) as long as convincingly recovered
Using their gradient from a differentiation attempt, but not dependent on first M1
Substitution into \(y - y_1 = m(x - x_1)\) or equation involving \(c\) from \(y = mx + c\)

A1: Obtain given answer
Must now be in terms of \(p\)
Expand brackets and simplify to given answer, or find \(c\) and substitute back into equation

Notes

B1FT: Correct (unsimplified) gradient of normal, following their derivative
Gradient in terms of \(t\) or \(p\), but not \(x\)
Could either FT on their incorrect derivative or deduce the gradient from the equation given in (a)

M1: Attempt equation of normal
Attempt to use their gradient and \(P\)
Allow mixture of \(t\) and \(p\) as long as convincingly recovered
Substitution into \(y - y_1 = m(x - x_1)\) or equation involving \(c\) from \(y = mx + c\)

M1: Using their attempt at normal equation
As far as finding an expression for \(x\)

A1: Correct \(x\)-coordinate for \(B\)
Any equivalent form

B1: Correct \(x\)-coordinate for \(A\)
Any equivalent form

M1: Attempt length of \(PA\) or \(PB\)
Or M1 for attempting one of \((PA)^2\) or \((PB)^2\)
Must correct distance formula
Using the given \(P\), and their coordinates for \(A\) and/or \(B\), which must involve a function of \(p\)

A1: Or correct \((PA)^2\) and \((PB)^2\)

A1: Simplify ratio to obtain given answer
Must show clear method, such as same expression in each square root before cancelling
Could also consider fraction and then cancel to deduce given ratio
Could simplify \((PA)^2 : (PB)^2\), and then square root to obtain ratio

Notes

B1: Or show that \(2\mathrm{e}^{-3t} = 0\) has no solutions
Need to see \(y \neq 0\) or \(y \gt 0\) and reason relating to exponential (or logarithmic) function
Must clearly be referring to \(y\) or \(2\mathrm{e}^{-3t}\)

Notes

M1: Equate to 0 and attempt to solve disguised quadratic
‘determine’ so some evidence of method needed

A1: Obtain both correct values
A0 for \(\ln 1\) and not 0

Notes

B1: Correct \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\)
Mark derivative and condone no/wrong label

B1: Correct \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\)
Mark derivative and condone no/wrong label

M1: Attempt correct method to combine derivatives
Combine their derivatives correctly

A1: Show manipulation to given answer
Need to see some evidence of how \(\mathrm{e}^{-3t}\) is dealt with
AG so method must be fully correct

Notes

M1: Equate denominator to 0
Or \(\dfrac{\mathrm{d}x}{\mathrm{d}y} = 0\) or \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0\)

A1: Solve for \(t\) to obtain \(t = \ln 2\)
No need to see \(\mathrm{e}^{4t} = 0\) discounted

A1: Obtain correct coordinate
Or \(x = -1\), \(y = \frac{1}{4}\)

Notes

B1: Triangle POB drawn, where B is the point on the x- axis such that PB is perpendicular to the x-axis.
P may be anywhere on curve in 1st quadrant.
The points P and B need not be labelled
Vertical and horizontal sides should be labelled y and x OR \(x\) and \(y\) clearly indicated on the axes OR Point on curve labelled \((x, y)\)
Allow angle \(\theta\) not labelled
Condone good freehand – mark intent

Notes

B1: AG
Correct completion to given result.
Must see at least one step of working and no errors or omissions

Notes

B1: Correct expression in terms of \(a\) and \(t\)
Must be simplified

Notes

M1: Dividing the correct expressions for \(3a - x\) and \(a + x\)

A1: AG
Simplifying to \(t^2\)
Using \(t = \frac{y}{x}\) to give a correct completion to given result
Must have at least one line of working and no incorrect work seen

Alternative method 1

M1: Substituting correct expressions for \(3a - x\) and \(a + x\) into \(y^2 = \frac{x^2(3a-x)}{a+x}\)

A1: AG
Simplifying to \(x^2t^2\)
Using \(t = \frac{y}{x}\) to give a correct completion to given result
Must have at least one line of working and no incorrect work seen

Alternative method 2

M1: Substituting \(x = \frac{a(3-t^2)}{1+t^2}\) and \(3a - x = \frac{4at^2}{1+t^2}\) into \(y^2 = \frac{x^2(3a-x)}{a+x}\)

A1: AG
Simplifying and showing \(y^2 = (xt)^2\) to give a correct completion to given result
Must have at least one line of working and no incorrect work seen

Alternative method 3

M1: Making \(t^2\) the subject of \(x = \frac{a(3-t^2)}{1+t^2}\) as far as \(t^2 = \ldots\)
Condone algebraic or sign slips

A1: AG
Using \(t = \frac{y}{x}\) to give a correct completion to given result
No incorrect work seen

Notes

M1: For differentiation of \(y\) wrt. \(\theta\), may be as part of \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

M1: For \(\dfrac{\mathrm{d}y}{\mathrm{d}\theta} = 0\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) and use of sine or cosine double angle formula.
If \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta}\) is used incorrectly then give M0

*M1: All on one side and factorised.
If they have divided through by \(\sin\theta\) then \(\boldsymbol{1 = -2\cos\theta}\) or \(\boldsymbol{1 + 2\cos\theta = 0}\) will earn this mark

DM1: Dependent on the previous method mark
At least one trig value
Condone if divides through by sin.

A1 A1: www
A marks dependent on M4
At least 2 correct values
All four values correct with no extras

A1: www
At least 2 correct coordinates
oe, must be exact form

A1: www
All four coordinates correct with no extras
oe, must be exact form

Additional guidance

The first M1 is for differentiating y wrt \(\theta\). It might be seen on it’s own as \(\mathrm{d}y/\mathrm{d}\theta\) or as the numerator of \(\mathrm{d}y/\mathrm{d}x\). We are ignoring \(\mathrm{d}x/\mathrm{d}\theta\) unless they try to use it incorrectly in their subsequent work.

The second M1 is for putting their expression equal to 0 AND using either \(\sin 2\theta = 2\sin\theta\cos\theta\) or \(\cos 2\theta\) = one of the correct forms. But if \(\mathrm{d}x/\mathrm{d}\theta\) is used incorrectly here then give M0.

The third M1 is for getting terms on one side and factorised. However, some divide through by sin at this point – they can still get the M1 and also the fourth M1. The 3^rd and 4^th M1s can be implied (BOD) by seeing 2 correct solutions \(\sin\theta = 0\) and \(\cos\theta = -\frac{1}{2}\).

The 4^th M1 is dependent on the previous M1. (We have seen some candidates obtain \(\sin\theta = 0\) (and possibly \(\cos\theta = -\frac{1}{2}\)) from completely wrong working but this does not score).

The A marks are dependent on getting M4 and are for finding exact values of \(\theta\) and the coordinates. Note that the A marks are all www marks (without wrong working), so dividing through by \(\sin\theta\) will cost all the A marks. Allow equivalent correct forms (e.g. \(\frac{4-\sqrt{3}}{2}\) for \(2 - \frac{\sqrt{3}}{2}\), \(\frac{4+\sqrt{3}}{2}\) for \(2 + \frac{\sqrt{3}}{2}\)) for the x-values.

Additional guidance

\(x=2\) is the only acceptable answer for B1.

Notes

M1: allow sign errors in their expressions for \(\sin\theta\) and \(\cos\theta\); allow if just see brackets expanded, but must be 3 terms in each case

A1: allow SC2 for \((x+3)^2 + (y-1)^2 = 4\) unsupported

Alternatively

M1: allow one sign error in bracket;

Notes

M1: allow sign errors

M1: or \(\left(\frac{x-2}{5}\right)^2 + \left(\frac{y-1}{5}\right)^2 = \cos^2\theta + \sin^2\theta\) oe

A1: may see eg \(x^2 - 4x + y^2 - 2y = 20\)
if M0M0 allow SC1 for \(y = 1 + 5\sin\left(\cos^{-1}\left(\frac{x-2}{5}\right)\right)\) or \(x = 2 + 5\cos\left(\sin^{-1}\left(\frac{y-1}{5}\right)\right)\)

Alternatively

M1: if only seen in expanded form, allow one coefficient error; allow sign errors

M1: must have terms in \(\cos\theta\) and \(\sin\theta\)

Alternatively

M1: allow sign error in coordinates of centre

M1: FT their centre

A1: all correct

Notes

M1: FT \(1 \div\) their \(-\frac{4}{3}\)

M1: allow one sign error; FT their \(\frac{3}{4}\)
may see \(-3 = \frac{3}{4} \times 5 + c\)

Alternatively

B1: or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{2-x}{y-1}\) oe
eg \(2(x - 2) + 2(y - 1)\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\)

M1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3/5}{-(-4/5)}\) or \(\dfrac{2-5}{-3-1}\) oe; allow one sign error;

M1: allow one sign error; FT their \(\frac{3}{4}\)
may see \(-3 = \frac{3}{4} \times 5 + c\)

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

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

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

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

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

Notes

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

Alternative method

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

Notes

M1: allow sign error

Alternative

A1: if M0M0 allow SC1 for
\(y = 5\sin\left\{\cos^{-1}\left(\dfrac{x-7}{5}\right)\right\} - 3\)
\(x = 5\cos\left\{\sin^{-1}\left(\dfrac{y+3}{5}\right)\right\} + 7\)

Notes

B1: FT their \((x-7)^2 + (y+3)^2 = 25\)

Notes

M1: Attempt to differentiate both equations

M1: Combining derivatives for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
Note that \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \left(-\dfrac{3}{t^4} + \dfrac{1}{t^2}\right) \times \left(-\dfrac{t^3}{2}\right)\)
(corrected from the printed mark scheme: the first bracket is printed as \(\left(-\frac{3}{t^4} - \frac{1}{t^2}\right)\))

A1: AG Correct derivative in required form.

Notes

B1: Establishing gradient \(-\dfrac{1}{4}\)
\(y = -\dfrac{1}{4}x + \dfrac{1}{4}\) not sufficient on its own

M1: Forming and solving quadratic equation.

A1: Using the value of \(t\) for both coordinates
Ignore any point based on \(t = -\dfrac{3}{2}\)

Notes

B1: or equivalent eg \(\dfrac{1}{t} = \sqrt{x}\)
\(y = \dfrac{1}{\left(\frac{1}{\sqrt{x}}\right)^3} - \dfrac{1}{\left(\frac{1}{\sqrt{x}}\right)}\)

M1: Attempt to eliminate \(t\)

A1: factorised form. Allow surd or index form
Do not allow for \(= \pm(x-1)\sqrt{x}\)