C4 June 2015 Q5
5. A curve \(C\) has parametric equations \[x = 4t + 3, \quad y = 4t + 8 + \frac{5}{2t}, \quad t \neq 0\]
| Scheme | Marks |
|---|---|
| Note: You can mark parts (a) and (b) together. | |
| \(x = 4t + 3,\ y = 4t + 8 + \dfrac{5}{2t}\) | |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 4,\ \dfrac{\mathrm{d}y}{\mathrm{d}t} = 4 - \dfrac{5}{2}t^{-2}\) Both \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 4\) or \(\dfrac{\mathrm{d}t}{\mathrm{d}x} = \dfrac{1}{4}\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 4 - \dfrac{5}{2}t^{-2}\) | B1 |
| So, \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{4 - \frac{5}{2}t^{-2}}{4}\ \left\{= 1 - \dfrac{5}{8}t^{-2} = 1 - \dfrac{5}{8t^2}\right\}\) Candidate’s \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) divided by a candidate’s \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) | M1 o.e. |
| \(\{\text{When } t = 2,\}\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{27}{32}\) \(\dfrac{27}{32}\) or 0.84375 cao | A1 |
| (3) |
Notes
B1: \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 4\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 4 - \dfrac{5}{2}t^{-2}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = \dfrac{8t^2 - 5}{2t^2}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}t} = 4 - 5(2t)^{-2}(2)\), etc.
Note: \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) can be simplified or un-simplified.
Note: You can imply the B1 mark by later working.
M1: Candidate’s \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) divided by a candidate’s \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) or \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) multiplied by a candidate’s \(\dfrac{\mathrm{d}t}{\mathrm{d}x}\)
Note: M1 can be also be obtained by substituting \(t = 2\) into both their \(\dfrac{\mathrm{d}y}{\mathrm{d}t}\) and their \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) and then dividing their values the correct way round.
A1: \(\dfrac{27}{32}\) or 0.84375 cao
Way 2: Cartesian Method
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1 - \dfrac{10}{(x - 3)^2}\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 1 - \dfrac{10}{(x - 3)^2}\), simplified or un-simplifed. \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \pm\lambda \pm \dfrac{\mu}{(x - 3)^2},\ \lambda \neq 0, \mu \neq 0\) | B1 M1 |
| \(\{\text{When } t = 2, x = 11\}\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{27}{32}\) \(\dfrac{27}{32}\) or 0.84375 cao | A1 |
| (3) |
Way 3: Cartesian Method
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{(2x + 2)(x - 3) - (x^2 + 2x - 5)}{(x - 3)^2}\) Correct expression for \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\), simplified or un-simplified. | B1 |
| \(\left\{= \dfrac{x^2 - 6x - 1}{(x - 3)^2}\right\}\) \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{\mathrm{f}'(x)(x - 3) - 1\mathrm{f}(x)}{(x - 3)^2}\), where \(\mathrm{f}(x)\) = their "\(x^2 + ax + b\)", \(\mathrm{g}(x) = x - 3\) | M1 |
| \(\{\text{When } t = 2, x = 11\}\ \dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{27}{32}\) \(\dfrac{27}{32}\) or 0.84375 cao | A1 |
| (3) |
| Scheme | Marks |
|---|---|
| \(\left\{t = \dfrac{x - 3}{4} \Rightarrow\right\}\ y = 4\left(\dfrac{x - 3}{4}\right) + 8 + \dfrac{5}{2\left(\frac{x - 3}{4}\right)}\) Eliminates \(t\) to achieve an equation in only \(x\) and \(y\) | M1 |
| \(y = x - 3 + 8 + \dfrac{10}{x - 3}\) | |
| \(y = \dfrac{(x - 3)(x - 3) + 8(x - 3) + 10}{x - 3}\) or \(y(x - 3) = (x - 3)(x - 3) + 8(x - 3) + 10\) or \(y = \dfrac{(x + 5)(x - 3) + 10}{x - 3}\) or \(y = \dfrac{(x + 5)(x - 3)}{x - 3} + \dfrac{10}{x - 3}\) See notes | dM1 |
| \(\Rightarrow y = \dfrac{x^2 + 2x - 5}{x - 3},\ \{a = 2 \text{ and } b = -5\}\) Correct algebra leading to \(y = \dfrac{x^2 + 2x - 5}{x - 3}\) or \(a = 2\) and \(b = -5\) | A1 cso |
| (3) | |
| (6 marks) |
Notes
M1: Eliminates \(t\) to achieve an equation in only \(x\) and \(y\).
dM1: dependent on the first method mark being awarded.
Either: (ignoring sign slips or constant slips, noting that \(k\) can be 1)
- Combining all three parts of their \(\underline{x - 3} + \underline{\underline{8}} + \underline{\underline{\underline{\left(\dfrac{10}{x - 3}\right)}}}\) to form a single fraction with a common denominator of \(\pm k(x - 3)\). Accept three separate fractions with the same denominator.
- Combining both parts of their \(\underline{x + 5} + \underline{\underline{\left(\dfrac{10}{x - 3}\right)}}\), (where \(\underline{x + 5}\) is their \(4\left(\dfrac{x - 3}{4}\right) + 8\)), to form a single fraction with a common denominator of \(\pm k(x - 3)\). Accept two separate fractions with the same denominator.
- Multiplies both sides of their \(y = \underline{x - 3} + \underline{\underline{8}} + \underline{\underline{\underline{\left(\dfrac{10}{x - 3}\right)}}}\) or their \(y = \underline{x + 5} + \underline{\underline{\left(\dfrac{10}{x - 3}\right)}}\) by \(\pm k(x - 3)\). Note that all terms in their equation must be multiplied by \(\pm k(x - 3)\).
Note: Condone “invisible” brackets for dM1.
A1: Correct algebra with no incorrect working leading to \(y = \dfrac{x^2 + 2x - 5}{x - 3}\) or \(a = 2\) and \(b = -5\)
Note: Some examples for the award of dM1 in (b):
dM0 for \(y = x - 3 + 8 + \dfrac{10}{x - 3} \to y = \dfrac{(x - 3)(x - 3) + 8 + 10}{x - 3}\). Should be \(\ldots + 8(x - 3) + \ldots\)
dM0 for \(y = x - 3 + \dfrac{10}{x - 3} \to y = \dfrac{(x - 3)(x - 3) + 10}{x - 3}\). The “8” part has been omitted.
dM0 for \(y = x + 5 + \dfrac{10}{x - 3} \to y = \dfrac{x(x - 3) + 5 + 10}{x - 3}\). Should be \(\ldots + 5(x - 3) + \ldots\)
dM0 for \(y = x + 5 + \dfrac{10}{x - 3} \to y(x - 3) = x(x - 3) + 5(x - 3) + 10(x - 3)\). Should be just 10.
Note: \(y = x + 5 + \dfrac{10}{x - 3} \to y = \dfrac{x^2 + 2x - 5}{x - 3}\) with no intermediate working is dM1A1.
Alternative Method 1 of Equating Coefficients
| Scheme | Marks |
|---|---|
| \(y = \dfrac{x^2 + ax + b}{x - 3} \Rightarrow y(x - 3) = x^2 + ax + b\) \(y(x - 3) = (4t + 3)^2 + 2(4t + 3) - 5 = 16t^2 + 32t + 10\) \(x^2 + ax + b = (4t + 3)^2 + a(4t + 3) + b\) | |
| \((4t + 3)^2 + a(4t + 3) + b = 16t^2 + 32t + 10\) Correct method of obtaining an equation in only \(t\), \(a\) and \(b\) | M1 |
| \(t: \quad 24 + 4a = 32 \Rightarrow a = 2\) constant: \(\quad 9 + 3a + b = 10 \Rightarrow b = -5\) Equates their coefficients in \(t\) and finds both \(\underline{a = \ldots}\) and \(\underline{b = \ldots}\) \(a = 2\) and \(b = -5\) | dM1 A1 |
| (3) |
Alternative Method 2 of Equating Coefficients
| Scheme | Marks |
|---|---|
| \(\left\{t = \dfrac{x - 3}{4} \Rightarrow\right\}\ y = 4\left(\dfrac{x - 3}{4}\right) + 8 + \dfrac{5}{2\left(\frac{x - 3}{4}\right)}\) Eliminates \(t\) to achieve an equation in only \(x\) and \(y\) | M1 |
| \(y = x - 3 + 8 + \dfrac{10}{x - 3} \Rightarrow y = x + 5 + \dfrac{10}{(x - 3)}\) \(\underline{\underline{y(x - 3)}} = (x + 5)(x - 3) + 10 \Rightarrow x^2 + ax + b = \underline{\underline{(x + 5)(x - 3) + 10}}\) | dM1 |
| \(\Rightarrow y = \dfrac{x^2 + 2x - 5}{x - 3}\) or equating coefficients to give \(a = 2\) and \(b = -5\) Correct algebra leading to \(y = \dfrac{x^2 + 2x - 5}{x - 3}\) or \(a = 2\) and \(b = -5\) | A1 cso |
| (3) |