CAIE FP1 2012 June — Question 3 8 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2012
SessionJune
Marks8
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TopicImplicit equations and differentiation
TypeFind second derivative d²y/dx²
DifficultyStandard +0.3 This is a straightforward implicit differentiation question requiring two applications of the chain rule and product rule. While finding the second derivative adds a step beyond basic implicit differentiation, the algebra is manageable and the method is standard for Further Pure 1. The verification of dy/dx = -3/4 provides scaffolding, making this slightly easier than average.
Spec1.07s Parametric and implicit differentiation
3 The curve \(C\) has equation $$x y + ( x + y ) ^ { 3 } = 1$$ Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 3 } { 4 }\) at the point \(A ( 1,0 )\) on \(C\). Find the value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) at \(A\).
Question 3:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(y + x\frac{dy}{dx} + 3(x+y)^2\left(1+\frac{dy}{dx}\right) = 0\)B1B1
\(0 + y' + 3 + 3y' = 0\)—
\(\Rightarrow y' = -\frac{3}{4}\) (AG)B1 Part Mark: 3
\(\frac{dy}{dx}+\frac{dy}{dx}+x\frac{d^2y}{dx^2}+6(x+y)\left(1+\frac{dy}{dx}\right)^2+3(x+y)^2\frac{d^2y}{dx^2}=0\)B1, B1B1
\(-\frac{3}{4}-\frac{3}{4}+y''+6\times\frac{1}{16}+3y''=0\)M1
\(\Rightarrow y'' = \frac{9}{32}\)A1 Part Mark: 5; Total: [8]
*N.B. Mark similarly if expression expanded before differentiating.*
## Question 3:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $y + x\frac{dy}{dx} + 3(x+y)^2\left(1+\frac{dy}{dx}\right) = 0$ | B1B1 | |
| $0 + y' + 3 + 3y' = 0$ | — | |
| $\Rightarrow y' = -\frac{3}{4}$ (AG) | B1 | Part Mark: 3 |
| $\frac{dy}{dx}+\frac{dy}{dx}+x\frac{d^2y}{dx^2}+6(x+y)\left(1+\frac{dy}{dx}\right)^2+3(x+y)^2\frac{d^2y}{dx^2}=0$ | B1, B1B1 | |
| $-\frac{3}{4}-\frac{3}{4}+y''+6\times\frac{1}{16}+3y''=0$ | M1 | |
| $\Rightarrow y'' = \frac{9}{32}$ | A1 | Part Mark: 5; Total: **[8]** |

*N.B. Mark similarly if expression expanded before differentiating.*

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3 The curve $C$ has equation

$$x y + ( x + y ) ^ { 3 } = 1$$

Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = - \frac { 3 } { 4 }$ at the point $A ( 1,0 )$ on $C$.

Find the value of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ at $A$.

\hfill \mbox{\textit{CAIE FP1 2012 Q3 [8]}}
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