AQA C4 2008 January — Question 6 5 marks

Exam BoardAQA
ModuleC4 (Core Mathematics 4)
Year2008
SessionJanuary
Marks5
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TopicImplicit equations and differentiation
TypeFind dy/dx at a point
DifficultyModerate -0.5 This is a straightforward implicit differentiation question requiring application of the product rule and chain rule, followed by substitution of given coordinates. It's slightly easier than average as it's a single-step problem with a clear method and simple algebra, though implicit differentiation itself is a C4 topic requiring some technique.
Spec1.07s Parametric and implicit differentiation
6 A curve has equation \(3 x y - 2 y ^ { 2 } = 4\).
Find the gradient of the curve at the point \(( 2,1 )\).
AnswerMarks Guidance
\(3x\frac{dy}{dx} + 3y - 4y\frac{dy}{dx} = 0\)M1 attempt implicit differentiation
A1product
A1chain
B1constant
\(\frac{dy}{dx} = -\frac{3}{2}\)A1 CSO
Alternative Method:
AnswerMarks Guidance
\(x = \frac{2}{3}y + \frac{4}{3y}\)(M1) solve for \(x =\) expression in \(y\) and differentiate with respect to \(y\)
\(\frac{dx}{dy} = \frac{2}{3} - \frac{4}{3y^2}\)(A1 A1)
\(y = 1\): \(\frac{dy}{dy} = \frac{2}{3} - \frac{4}{3}\)(M1) substitute \(y = 1\)
\(\frac{dx}{dy} = -\frac{3}{2}\)(A1) CSO
Total: 5 marks
| $3x\frac{dy}{dx} + 3y - 4y\frac{dy}{dx} = 0$ | M1 | attempt implicit differentiation |
| | A1 | product |
| | A1 | chain |
| | B1 | constant |
| $\frac{dy}{dx} = -\frac{3}{2}$ | A1 | CSO |

**Alternative Method:**

| $x = \frac{2}{3}y + \frac{4}{3y}$ | (M1) | solve for $x =$ expression in $y$ and differentiate with respect to $y$ |
| $\frac{dx}{dy} = \frac{2}{3} - \frac{4}{3y^2}$ | (A1 A1) | |
| $y = 1$: $\frac{dy}{dy} = \frac{2}{3} - \frac{4}{3}$ | (M1) | substitute $y = 1$ |
| $\frac{dx}{dy} = -\frac{3}{2}$ | (A1) | CSO |

**Total: 5 marks**

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6 A curve has equation $3 x y - 2 y ^ { 2 } = 4$.\\
Find the gradient of the curve at the point $( 2,1 )$.

\hfill \mbox{\textit{AQA C4 2008 Q6 [5]}}
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Q1 5 Q2 10 Q3 6 Q4 9 Q5 10 Q6 5 Q7 14 Q8 5 Q9 11