Edexcel C34 2015 January — Question 6 10 marks

Exam BoardEdexcel
ModuleC34 (Core Mathematics 3 & 4)
Year2015
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConnected Rates of Change
TypeCube or cuboid: related rates
DifficultyStandard +0.3 Part (i) requires chain rule differentiation of a composite trigonometric function and algebraic manipulation to reach the specified form—moderately routine for C3/C4 students. Part (ii) is a standard connected rates problem (dV/dt and dx/dt for a cube) requiring only V=x³ and chain rule—textbook exercise level. Overall slightly easier than average A-level question.
Spec1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.07s Parametric and implicit differentiation
6. (i) Given \(x = \tan ^ { 2 } 4 y , 0 < y < \frac { \pi } { 8 }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) as a function of \(x\). Write your answer in the form \(\frac { 1 } { A \left( x ^ { p } + x ^ { q } \right) }\), where \(A , p\) and \(q\) are constants to
be found.
(ii) The volume \(V\) of a cube is increasing at a constant rate of \(2 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\). Find the rate at which the length of the edge of the cube is increasing when the volume of the cube is \(64 \mathrm {~cm} ^ { 3 }\).
Question 6:
Part (i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(x = \tan^2 4y \Rightarrow \frac{dx}{dy} = 8\tan 4y \sec^2 4y\)M1A1 Differentiates \(\tan^2 4y\) to get form \(C\tan 4y\sec^2 4y\); fully correct
\(\frac{dy}{dx} = \frac{1}{8\tan 4y \sec^2 4y} = \frac{1}{8\tan 4y(1+\tan^2 4y)} = \frac{1}{8\sqrt{x}(1+x)} = \frac{1}{8(x^{0.5}+x^{1.5})}\)M1, M1A1 Uses \(\frac{dy}{dx} = 1\big/\frac{dx}{dy}\); uses \(\sec^2 4y = 1+\tan^2 4y\) where \(x=\tan^2 4y\); correct final answer
Accept \(\frac{1}{8(x^{0.5}+x^{1.5})}\) or \(\frac{1}{8\left(x^{\frac{1}{2}}+x^{\frac{3}{2}}\right)}\) or \(A=8, p=0.5, q=1.5\) Candidates need not explicitly state values of \(A\), \(p\), \(q\)
Part (ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{dV}{dt} = 2\)B1 May be awarded if embedded within chain rule
\(V = x^3 \Rightarrow \frac{dV}{dx} = 3x^2\)B1 May be awarded if embedded within chain rule; accept any variable in place of \(x\)
Uses \(\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt}\)M1 Correct chain rule with their values
\(\left.\frac{dx}{dt}\right_{x=4} = \frac{2}{3x^2} = \frac{1}{24}\ (\text{cm s}^{-1})\) M1A1 
## Question 6:

### Part (i):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = \tan^2 4y \Rightarrow \frac{dx}{dy} = 8\tan 4y \sec^2 4y$ | M1A1 | Differentiates $\tan^2 4y$ to get form $C\tan 4y\sec^2 4y$; fully correct |
| $\frac{dy}{dx} = \frac{1}{8\tan 4y \sec^2 4y} = \frac{1}{8\tan 4y(1+\tan^2 4y)} = \frac{1}{8\sqrt{x}(1+x)} = \frac{1}{8(x^{0.5}+x^{1.5})}$ | M1, M1A1 | Uses $\frac{dy}{dx} = 1\big/\frac{dx}{dy}$; uses $\sec^2 4y = 1+\tan^2 4y$ where $x=\tan^2 4y$; correct final answer |
| Accept $\frac{1}{8(x^{0.5}+x^{1.5})}$ or $\frac{1}{8\left(x^{\frac{1}{2}}+x^{\frac{3}{2}}\right)}$ or $A=8, p=0.5, q=1.5$ | | Candidates need not explicitly state values of $A$, $p$, $q$ |

### Part (ii):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dV}{dt} = 2$ | B1 | May be awarded if embedded within chain rule |
| $V = x^3 \Rightarrow \frac{dV}{dx} = 3x^2$ | B1 | May be awarded if embedded within chain rule; accept any variable in place of $x$ |
| Uses $\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt}$ | M1 | Correct chain rule with their values |
| $\left.\frac{dx}{dt}\right|_{x=4} = \frac{2}{3x^2} = \frac{1}{24}\ (\text{cm s}^{-1})$ | M1A1 | Substitutes $x=4$ to find numerical value; $\frac{1}{24}$ (accept awrt $0.0417$) |

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6. (i) Given $x = \tan ^ { 2 } 4 y , 0 < y < \frac { \pi } { 8 }$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ as a function of $x$.

Write your answer in the form $\frac { 1 } { A \left( x ^ { p } + x ^ { q } \right) }$, where $A , p$ and $q$ are constants to\\
be found.\\
(ii) The volume $V$ of a cube is increasing at a constant rate of $2 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$. Find the rate at which the length of the edge of the cube is increasing when the volume of the cube is $64 \mathrm {~cm} ^ { 3 }$.

\hfill \mbox{\textit{Edexcel C34 2015 Q6 [10]}}
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