C4 June 2013 (R) Q2
2. The curve \(C\) has equation \[3^{x - 1} + xy - y^2 + 5 = 0\]
Show that \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) at the point \((1, 3)\) on the curve \(C\) can be written in the form \(\dfrac{1}{\lambda}\ln(\mu\mathrm{e}^3)\), where \(\lambda\) and \(\mu\) are integers to be found. (7)
| Scheme | Marks |
|---|---|
| \(3^{x - 1} + xy - y^2 + 5 = 0\) | |
| \(\left\{\cancel{\dfrac{\mathrm{d}y}{\mathrm{d}x}}\cancel{\times}\right\}\quad 3^{x - 1}\ln 3 + \left(y + x\dfrac{\mathrm{d}y}{\mathrm{d}x}\right) - 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) (ignore) \(3^{x - 1} \to 3^{x - 1}\ln 3\) Differentiates implicitly to include either \(\pm\lambda x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\pm ky\dfrac{\mathrm{d}y}{\mathrm{d}x}\). \(xy \to +\,y + x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) \(\ldots + y + x\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) | B1 oe M1* B1 A1 |
| \(\{(1, 3) \Rightarrow\}\ 3^{(1 - 1)}\ln 3 + 3 + (1)\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2(3)\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) Substitutes \(x = 1,\ y = 3\) into their differentiated equation or expression. | dM1* |
| \(\ln 3 + 3 + \dfrac{\mathrm{d}y}{\mathrm{d}x} - 6\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow 3 + \ln 3 = 5\dfrac{\mathrm{d}y}{\mathrm{d}x}\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3 + \ln 3}{5}\) | dM1* |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{5}\left(\ln\mathrm{e}^3 + \ln 3\right) = \dfrac{1}{5}\ln\left(3\mathrm{e}^3\right)\) Uses \(3 = \ln\mathrm{e}^3\) to achieve \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{5}\ln\left(3\mathrm{e}^3\right)\) | A1 cso |
| (7) | |
| (7 marks) |
Notes
B1: Correct differentiation of \(3^{x - 1}\). I.e. \(3^{x - 1} \to 3^{x - 1}\ln 3\) or \(3^{x - 1} = \dfrac{1}{3}\left(3^x\right) \to \dfrac{1}{3}\left(3^x\right)\ln 3\)
or \(3^{x - 1} = \mathrm{e}^{(x - 1)\ln 3} \to \ln 3\ \mathrm{e}^{(x - 1)\ln 3}\) or \(3^{x - 1} = \dfrac{1}{3}\left(3^x\right) = \dfrac{1}{3}\mathrm{e}^{x\ln 3} \to \dfrac{1}{3}(\ln 3)\mathrm{e}^{x\ln 3}\)
M1: Differentiates implicitly to include either \(\pm\lambda x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\pm ky\dfrac{\mathrm{d}y}{\mathrm{d}x}\). (Ignore \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right)\)).
B1: \(xy \to +\,y + x\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
1st A1: \(\ldots + y + x\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) Note: The 1st A0 follows from an award of the 2nd B0.
Note: The "\(= 0\)" can be implied by rearrangement of their equation.
ie: \(3^{x - 1}\ln 3 + y + x\dfrac{\mathrm{d}y}{\mathrm{d}x} - 2y\dfrac{\mathrm{d}y}{\mathrm{d}x}\) leading to \(3^{x - 1}\ln 3 + y = 2y\dfrac{\mathrm{d}y}{\mathrm{d}x} - x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) will get A1 (implied).
2nd M1: Note: This method mark is dependent upon the 1st M1* mark being awarded.
Substitutes \(x = 1,\ y = 3\) into their differentiated equation or expression. Allow one slip.
3rd M1: Note: This method mark is dependent upon the 1st M1* mark being awarded.
Candidate has two differentiated terms in \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) and rearranges to make \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) the subject.
Note: It is possible to gain the 3rd M1 mark before the 2nd M1 mark.
Eg: Candidate may write \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{y + 3^{x - 1}\ln 3}{2y - x}\) before substituting in \(x = 1\) and \(y = 3\)
2nd A1: cso. Uses \(3 = \ln\mathrm{e}^3\) to achieve \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{5}\ln\left(3\mathrm{e}^3\right)\), \(\left(= \dfrac{1}{\lambda}\ln\left(\mu\mathrm{e}^3\right),\ \lambda = 5 \text{ and } \mu = 3\right)\)
Note: \(3 = \ln\mathrm{e}^3\) needs to be seen in their proof.
Aliter Way 2 – Alternative Method: Multiplying both sides by 3
| Scheme | Marks |
|---|---|
| \(3^{x - 1} + xy - y^2 + 5 = 0\) \(3^x + 3xy - 3y^2 + 15 = 0\) | |
| \(\left\{\cancel{\dfrac{\mathrm{d}y}{\mathrm{d}x}}\cancel{\times}\right\}\quad 3^x\ln 3 + \left(3y + 3x\dfrac{\mathrm{d}y}{\mathrm{d}x}\right) - 6y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) (ignore) \(3^x \to 3^x\ln 3\) Differentiates implicitly to include either \(\pm\lambda x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or \(\pm ky\dfrac{\mathrm{d}y}{\mathrm{d}x}\). \(3xy \to +\,3y + 3x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) \(\ldots + 3y + 3x\dfrac{\mathrm{d}y}{\mathrm{d}x} - 6y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) | B1 M1* B1 A1 |
| \(\{(1, 3) \Rightarrow\}\ 3^1\ln 3 + 3(3) + (3)(1)\dfrac{\mathrm{d}y}{\mathrm{d}x} - 6(3)\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0\) Substitutes \(x = 1,\ y = 3\) into their differentiated equation or expression. | dM1* |
| \(3\ln 3 + 9 + 3\dfrac{\mathrm{d}y}{\mathrm{d}x} - 18\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0 \Rightarrow 9 + 3\ln 3 = 15\dfrac{\mathrm{d}y}{\mathrm{d}x}\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{9 + 3\ln 3}{15}\ \left\{= \dfrac{3 + \ln 3}{5}\right\}\) | dM1* |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{5}\left(\ln\mathrm{e}^3 + \ln 3\right)\) | |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{5}\left(\ln\mathrm{e}^3 + \ln 3\right) = \dfrac{1}{5}\ln\left(3\mathrm{e}^3\right)\) Uses \(3 = \ln\mathrm{e}^3\) to achieve \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{5}\ln\left(3\mathrm{e}^3\right)\) | A1 cso |
| (7) |
NOTE: Only apply this scheme if the candidate has multiplied both sides of their equation by 3.
NOTE: For reference, \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{3y + 3^x\ln 3}{6y - 3x}\)
NOTE: If the candidate applies this method then \(3xy \to +\,3y + 3x\dfrac{\mathrm{d}y}{\mathrm{d}x}\) must be seen for the 2nd B1 mark.