C4 June 2014 Q7

Notes

1^st M1: Applies their \(\dfrac{\mathrm{d}y}{\mathrm{d}\theta}\) divided by their \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta}\) or applies \(\dfrac{\mathrm{d}y}{\mathrm{d}\theta}\) multiplied by their \(\dfrac{\mathrm{d}\theta}{\mathrm{d}x}\)

SC: Award Special Case 1^st M1 if both \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta}\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}\theta}\) are both correct.

1^st A1: Correct \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) i.e. \(\dfrac{-8\cos\theta\sin\theta}{3\sec^2\theta}\) or \(-\dfrac{8}{3}\cos^3\theta\sin\theta\) or \(-\dfrac{4}{3}\sin 2\theta\cos^2\theta\) or any equivalent form.

2^nd M1: Some evidence of substituting \(\theta = \dfrac{\pi}{4}\) or \(\theta = 45^\circ\) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)

Note: For 3^rd M1 and 4^th M1, \(m(\mathbf{T})\) must be found by using \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\).

3^rd M1: applies \(m(\mathbf{N}) = \dfrac{-1}{m(\mathbf{T})}\). Numerical value for \(m(\mathbf{N})\) is required here.

4^th M1:

• Applies \(y - 2 = (\text{their } m_N)(x - 3)\), where \(\mathrm{m}(\mathbf{N})\) is a numerical value,
• or finds c by solving \(2 = (\text{their } m_N)3 + c\), where \(\mathrm{m}(\mathbf{N})\) is a numerical value,

and \(m_N = -\dfrac{1}{\text{their m}(\mathbf{T})}\) or \(m_N = \dfrac{1}{\text{their m}(\mathbf{T})}\) or \(m_N = -\text{their m}(\mathbf{T})\).

Note: This mark can be implied by subsequent working.

2^nd A1: \(x = \dfrac{5}{3}\) or \(1\dfrac{2}{3}\) or awrt 1.67 from a correct solution only.

Working with a Cartesian Equation

A cartesian equation for \(C\) is \(y = \dfrac{36}{x^2 + 9}\)

1^st M1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \pm\lambda x\left(\pm\alpha x^2 \pm \beta\right)^{-2}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\pm\lambda x}{\left(\pm\alpha x^2 \pm \beta\right)^2}\)

1^st A1: \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -36(x^2 + 9)^{-2}(2x)\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-72x}{(x^2 + 9)^2}\) un-simplified or simplified.

2^nd dM1: Dependent on the 1^st M1 mark if a candidate uses this method
For substituting \(x = 3\) into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
i.e. at \(P(3, 2)\), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{-72(3)}{(3^2 + 9)^2}\ \left\{= -\dfrac{2}{3}\right\}\)
From this point onwards the original scheme can be applied.

Another cartesian equation for \(C\) is \(x^2 = \dfrac{36}{y} - 9\)

1^st M1: \(\pm\alpha x = \pm\dfrac{\beta}{y^2}\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\pm\alpha x\dfrac{\mathrm{d}x}{\mathrm{d}y} = \pm\dfrac{\beta}{y^2}\)

1^st A1: \(2x = -\dfrac{36}{y^2}\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(2x\dfrac{\mathrm{d}x}{\mathrm{d}y} = -\dfrac{36}{y^2}\)

2^nd dM1: Dependent on the 1^st M1 mark if a candidate uses this method
For substituting \(x = 3\) to find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
i.e. at \(P(3, 2)\), \(2(3) = -\dfrac{36}{4}\dfrac{\mathrm{d}y}{\mathrm{d}x} \Rightarrow \dfrac{\mathrm{d}y}{\mathrm{d}x} = \ldots\)
From this point onwards the original scheme can be applied.

Notes

1^st M1: Applying \(\displaystyle\int y^2\mathrm{d}x\) as \(y^2\dfrac{\mathrm{d}x}{\mathrm{d}\theta}\) with their \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta}\). Ignore \(\pi\) or \(\dfrac{1}{3}\pi\) outside integral.

Note: You can ignore the omission of an integral sign and/or \(\mathrm{d}\theta\) for the 1^st M1.

Note: Allow 1^st M1 for \(\displaystyle\int (\cos^2\theta)^2 \times \text{"their } 3\sec^2\theta\text{"}\,\mathrm{d}\theta\) or \(\displaystyle\int 4(\cos^2\theta)^2 \times \text{"their } 3\sec^2\theta\text{"}\,\mathrm{d}\theta\)

1^st A1: Correct expression \(\left\{\pi\displaystyle\int y^2\mathrm{d}x\right\} = \pi\displaystyle\int (4\cos^2\theta)^2\,3\sec^2\theta\,\{\mathrm{d}\theta\}\) (Allow the omission of \(\mathrm{d}\theta\))

Note: IMPORTANT: The \(\pi\) can be recovered later, but as a correct statement only.

2^nd A1: \(\left\{\displaystyle\int y^2\mathrm{d}x\right\} = \displaystyle\int 48\cos^2\theta\{\mathrm{d}\theta\}\). (Ignore \(\mathrm{d}\theta\)). Note: 48 can be written as 24(2) for example.

2^nd M1: Applies \(\cos 2\theta = 2\cos^2\theta - 1\) to their integral. (Seen or implied.)

3^rd dM1*: which is dependent on the 1^st M1 mark.
Integrating \(\cos^2\theta\) to give \(\pm\alpha\theta \pm \beta\sin 2\theta,\ \alpha \neq 0,\ \beta \neq 0\), un-simplified or simplified.

3^rd A1: which is dependent on the 3^rd M1 mark and the 1^st M1 mark.
Integrating \(\cos^2\theta\) to give \(\dfrac{1}{2}\theta + \dfrac{1}{4}\sin 2\theta\), un-simplified or simplified.
This can be implied by \(k\cos^2\theta\) giving \(\dfrac{k}{2}\theta + \dfrac{k}{4}\sin 2\theta\), un-simplified or simplified.

4^th dM1: which is dependent on the 3^rd M1 mark and the 1^st M1 mark.
Some evidence of applying limits of \(\dfrac{\pi}{4}\) and 0 (0 can be implied) to an integrated function in \(\theta\)

5^th M1: Applies \(V_{\text{cone}} = \dfrac{1}{3}\pi(2)^2\left(3 - \text{their part (a) answer}\right)\).

Note: Also allow the 5^th M1 for \(V_{\text{cone}} = \pi\displaystyle\int_{\text{their } \frac{5}{3}}^{3} \left(\dfrac{3}{2}x - \dfrac{5}{2}\right)^2\{dx\}\), which includes the correct limits.

4^th A1: \(\dfrac{92}{9}\pi + 6\pi^2\) or \(10\dfrac{2}{9}\pi + 6\pi^2\)

Note: A decimal answer of 91.33168464... (without a correct exact answer) is A0.

Note: The \(\pi\) in the volume formula is only needed for the 1^st A1 mark and the final accuracy mark.

Working with a Cartesian Equation

1^st M1: For \(\displaystyle\int \left(\dfrac{\pm\lambda}{\pm\alpha x^2 \pm \beta}\right)^2\{\mathrm{d}x\}\) (\(\pi\) not required for this mark)

A1: For \(\pi\displaystyle\int \left(\dfrac{36}{x^2 + 9}\right)^2\{\mathrm{d}x\}\) (\(\pi\) required for this mark)

To integrate, a substitution of \(x = 3\tan\theta\) is required which will lead to \(\displaystyle\int 48\cos^2\theta\,\mathrm{d}\theta\) and so from this point onwards the original scheme can be applied.