OCR C4 2010 January — Question 7 8 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Year2010
SessionJanuary
Marks8
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Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeFind normal equation at point
DifficultyStandard +0.3 This is a straightforward implicit differentiation question requiring students to differentiate both sides with respect to x, substitute the given point to find dy/dx, then find the normal gradient and write the equation. While it involves multiple steps (implicit differentiation, product rule, algebraic manipulation, and finding the normal), these are all standard C4 techniques with no novel problem-solving required, making it slightly easier than average.
Spec1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation
7 Find the equation of the normal to the curve \(x ^ { 3 } + 2 x ^ { 2 } y = y ^ { 3 } + 15\) at the point \(( 2,1 )\), giving your answer in the form \(a x + b y + c = 0\), where \(a , b\) and \(c\) are integers.
Question 7:
AnswerMarks Guidance
Treat \(\frac{d}{dx}(x^2 y)\) as a productM1
\(\frac{d}{dx}(y^3) = 3y^2\frac{dy}{dx}\)B1
\(3x^2 + 2x^2\frac{dy}{dx} + 4xy = 3y^2\frac{dy}{dx}\)A1 Ignore \(\frac{dy}{dx} =\) if not used
Subst \((2,1)\) and solve for \(\frac{dy}{dx}\) or vice-versaM1
\(\frac{dy}{dx} = -4\)A1 WWW
grad normal \(= -\frac{1}{\text{their } \frac{dy}{dx}}\)\(\sqrt{}\)A1 stated or used
Find eqn of line, through \((2,1)\), with either gradientM1 using their \(\frac{dy}{dx}\) or \(-\frac{1}{\text{their }\frac{dy}{dx}}\)
\(x - 4y + 2 = 0\)A1 AEF with integral coefficients 
# Question 7:

| Treat $\frac{d}{dx}(x^2 y)$ as a product | M1 | |
|---|---|---|
| $\frac{d}{dx}(y^3) = 3y^2\frac{dy}{dx}$ | B1 | |
| $3x^2 + 2x^2\frac{dy}{dx} + 4xy = 3y^2\frac{dy}{dx}$ | A1 | Ignore $\frac{dy}{dx} =$ if not used |
| Subst $(2,1)$ and solve for $\frac{dy}{dx}$ or vice-versa | M1 | |
| $\frac{dy}{dx} = -4$ | A1 | WWW |
| grad normal $= -\frac{1}{\text{their } \frac{dy}{dx}}$ | $\sqrt{}$A1 | stated or used |
| Find eqn of line, through $(2,1)$, with either gradient | M1 | using their $\frac{dy}{dx}$ or $-\frac{1}{\text{their }\frac{dy}{dx}}$ |
| $x - 4y + 2 = 0$ | A1 | AEF with integral coefficients |

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7 Find the equation of the normal to the curve $x ^ { 3 } + 2 x ^ { 2 } y = y ^ { 3 } + 15$ at the point $( 2,1 )$, giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.

\hfill \mbox{\textit{OCR C4 2010 Q7 [8]}}
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