CAIE P1 2016 June — Question 7 7 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2016
SessionJune
Marks7
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TopicVariable acceleration (1D)
TypeRelated rates with point moving along a curve or two moving objects
DifficultyStandard +0.8 This is a related rates problem requiring implicit differentiation with respect to time, understanding that dy/dt is constant, and solving a resulting equation involving dx/dt. It goes beyond routine differentiation by requiring students to connect rates of change through the chain rule (dy/dt = dy/dx × dx/dt) and set up an equation where dx/dt = (1/2)dy/dt, which requires conceptual understanding rather than just mechanical application.
Spec1.07i Differentiate x^n: for rational n and sums1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates
7 The point \(P ( x , y )\) is moving along the curve \(y = x ^ { 2 } - \frac { 10 } { 3 } x ^ { \frac { 3 } { 2 } } + 5 x\) in such a way that the rate of change of \(y\) is constant. Find the values of \(x\) at the points at which the rate of change of \(x\) is equal to half the rate of change of \(y\).
Question 7:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{dy}{dx}=2x-5x^{1/2}+5\)B1
\(\frac{dy}{dx}=2\)B1
\(2x-5x^{1/2}+5=2\)M1 Equate their \(dy/dx\) to *their* 2 or \(\frac{1}{2}\)
\(2x-5x^{1/2}+3(=0)\) or equivalent 3-term quadraticA1
Attempt to solve for \(x^{1/2}\) e.g. \((2x^{1/2}-3)(x^{1/2}-1)=0\)DM1 Dep. on 3-term quadratic
\(x^{1/2}=3/2\) and \(1\)A1 ALT: \(5x^{\frac{1}{2}}=2x+3\rightarrow 25x=(2x+3)^2\)
\(x=9/4\) and \(1\)A1 [7] \(4x^2-13x+9(=0)\), \(x=9/4\) and \(1\) 
## Question 7:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx}=2x-5x^{1/2}+5$ | B1 | |
| $\frac{dy}{dx}=2$ | B1 | |
| $2x-5x^{1/2}+5=2$ | M1 | Equate their $dy/dx$ to *their* 2 or $\frac{1}{2}$ |
| $2x-5x^{1/2}+3(=0)$ or equivalent 3-term quadratic | A1 | |
| Attempt to solve for $x^{1/2}$ e.g. $(2x^{1/2}-3)(x^{1/2}-1)=0$ | DM1 | Dep. on 3-term quadratic |
| $x^{1/2}=3/2$ and $1$ | A1 | ALT: $5x^{\frac{1}{2}}=2x+3\rightarrow 25x=(2x+3)^2$ |
| $x=9/4$ and $1$ | A1 [7] | $4x^2-13x+9(=0)$, $x=9/4$ and $1$ |

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7 The point $P ( x , y )$ is moving along the curve $y = x ^ { 2 } - \frac { 10 } { 3 } x ^ { \frac { 3 } { 2 } } + 5 x$ in such a way that the rate of change of $y$ is constant. Find the values of $x$ at the points at which the rate of change of $x$ is equal to half the rate of change of $y$.

\hfill \mbox{\textit{CAIE P1 2016 Q7 [7]}}
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