OCR C4 2006 June — Question 1 4 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Year2006
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind dy/dx at a point
DifficultyModerate -0.3 This is a straightforward implicit differentiation question requiring application of the product rule and chain rule, then substitution of given coordinates. It's slightly easier than average as it's a standard technique with clear steps and no conceptual surprises, typical of a first question on a C4 paper.
Spec1.07s Parametric and implicit differentiation

Find the gradient of the curve \(4x^2 + 2xy + y^2 = 12\) at the point \((1, 2)\). [4]

AnswerMarks Guidance
\(\frac{d}{dx}(xy) = x\frac{dy}{dx} + y\)B1 s.o.i. e.g. \(2x\frac{dy}{dx} + y\)
\(\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}\)B1
Substitute (1,2) into their differentiated equation and attempt to solve for \(\frac{dy}{dx}\)M1 dep at Or attempt to solve their diff equation for \(\frac{d\phi}{dx}\)
[Allow subst of (2,1)] least 1 x B1 and then substitute (1,2)
\(\frac{dy}{dx} = -2\)A1 4
$\frac{d}{dx}(xy) = x\frac{dy}{dx} + y$ | B1 | s.o.i. e.g. $2x\frac{dy}{dx} + y$

$\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$ | B1 |

Substitute (1,2) into their differentiated equation and attempt to solve for $\frac{dy}{dx}$ | M1 dep at | Or attempt to solve their diff equation for $\frac{d\phi}{dx}$

[Allow subst of (2,1)] least 1 x B1 and then substitute (1,2) | | 

$\frac{dy}{dx} = -2$ | A1 | 4

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Find the gradient of the curve $4x^2 + 2xy + y^2 = 12$ at the point $(1, 2)$. [4]

\hfill \mbox{\textit{OCR C4 2006 Q1 [4]}}