OCR MEI C3 2015 June — Question 5 6 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Year2015
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicImplicit equations and differentiation
TypeVerify point and find gradient
DifficultyModerate -0.3 This is a straightforward implicit differentiation question with two routine parts: substituting coordinates to verify a point (trivial algebra) and finding dy/dx at that point using standard implicit differentiation rules. The differentiation requires the product rule for the 2x ln y term but evaluating at (1,1) simplifies nicely since ln(1)=0. Slightly easier than average due to the convenient point chosen and being a standard textbook exercise.
Spec1.07s Parametric and implicit differentiation
5 A curve has implicit equation \(y ^ { 2 } + 2 x \ln y = x ^ { 2 }\).
Verify that the point \(( 1,1 )\) lies on the curve, and find the gradient of the curve at this point.
AnswerMarks Guidance
\(y^2 + 2x\ln y = x^2\)B1 clear evidence of verification needed
\(1^2 + 2 \times 1 \times \ln 1 = 1^2\) so \((1, 1)\) lies on the curve.M1 must be correct
\(2y\frac{dy}{dx} + 2\ln y + 2x \cdot \frac{1}{y}\frac{dy}{dx} = 2x\)M1 \(\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}\); \(\frac{d}{dx}(2x\ln y) = 2\ln y + 2y\frac{dy}{dx}\) must be correct
\(\left[\Rightarrow \frac{dy}{dx} = \frac{2x - 2\ln y}{2y + 2x/y}\right]\)M1 substituting both \(x = 1\) and \(y = 1\) into their \(\frac{dy}{dx}\) or their equation in \(x, y\) and \(\frac{dy}{dx}\)
when \(x = 1, y = 1\), \(\frac{dy}{dx} = \frac{2 - 2\ln 1}{2 + 2} = \frac{1}{2}\)A1 cao not from wrong working
Total: [6] at least "\(1 + 0 = 1\)"; must be correct; condone \(\frac{dy}{dx} = \ldots\) unless pursued; \(2\frac{dy}{dx} + 2\ln 1 + 2\frac{dy}{dx} = 2\) 
$y^2 + 2x\ln y = x^2$ | B1 | clear evidence of verification needed
$1^2 + 2 \times 1 \times \ln 1 = 1^2$ so $(1, 1)$ lies on the curve. | M1 | must be correct
$2y\frac{dy}{dx} + 2\ln y + 2x \cdot \frac{1}{y}\frac{dy}{dx} = 2x$ | M1 | $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$; $\frac{d}{dx}(2x\ln y) = 2\ln y + 2y\frac{dy}{dx}$ must be correct
$\left[\Rightarrow \frac{dy}{dx} = \frac{2x - 2\ln y}{2y + 2x/y}\right]$ | M1 | substituting both $x = 1$ and $y = 1$ into their $\frac{dy}{dx}$ or their equation in $x, y$ and $\frac{dy}{dx}$
when $x = 1, y = 1$, $\frac{dy}{dx} = \frac{2 - 2\ln 1}{2 + 2} = \frac{1}{2}$ | A1 cao | not from wrong working

**Total: [6]** | | at least "$1 + 0 = 1$"; must be correct; condone $\frac{dy}{dx} = \ldots$ unless pursued; $2\frac{dy}{dx} + 2\ln 1 + 2\frac{dy}{dx} = 2$

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5 A curve has implicit equation $y ^ { 2 } + 2 x \ln y = x ^ { 2 }$.\\
Verify that the point $( 1,1 )$ lies on the curve, and find the gradient of the curve at this point.

\hfill \mbox{\textit{OCR MEI C3 2015 Q5 [6]}}
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