CAIE P1 2016 June — Question 4 6 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2016
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConnected Rates of Change
TypeCurve motion: find dy/dt
DifficultyModerate -0.3 Part (i) is a straightforward application of the chain rule (dy/dt = dy/dx × dx/dt) with given values substituted at x=0. Part (ii) requires integration with a substitution for the power term, which is standard P1 technique. Both parts are routine applications of core calculus methods with no novel problem-solving required, making this slightly easier than average.
Spec1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates

4 A curve is such that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 - 8 ( 3 x + 4 ) ^ { - \frac { 1 } { 2 } }\).
  1. A point \(P\) moves along the curve in such a way that the \(x\)-coordinate is increasing at a constant rate of 0.3 units per second. Find the rate of change of the \(y\)-coordinate as \(P\) crosses the \(y\)-axis. The curve intersects the \(y\)-axis where \(y = \frac { 4 } { 3 }\).
  2. Find the equation of the curve.

Question 4(i):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\frac{dy}{dx} = 2 - 8(3x+4)^{-\frac{1}{2}}\)
\((x = 0 \rightarrow \frac{dy}{dx} = -2)\)
\(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} \rightarrow -0.6\)M1A1 Ignore notation. Must be \(\frac{dy}{dx} \times 0.3\)
Question 4(ii):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(y = \{2x\}\left\{-\frac{8\sqrt{3x+4}}{\frac{1}{2}} \div 3\right\}\) \((+c)\)B1 B1 No need for \(+c\)
\(x = 0, y = \frac{4}{3} \rightarrow c = 12\)M1 A1 Uses \(x\), \(y\) values after \(\int\) with \(c\)
## Question 4(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = 2 - 8(3x+4)^{-\frac{1}{2}}$ | | |
| $(x = 0 \rightarrow \frac{dy}{dx} = -2)$ | | |
| $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} \rightarrow -0.6$ | **M1A1** | Ignore notation. Must be $\frac{dy}{dx} \times 0.3$ |

## Question 4(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \{2x\}\left\{-\frac{8\sqrt{3x+4}}{\frac{1}{2}} \div 3\right\}$ $(+c)$ | **B1 B1** | No need for $+c$ |
| $x = 0, y = \frac{4}{3} \rightarrow c = 12$ | **M1 A1** | Uses $x$, $y$ values after $\int$ with $c$ |
4 A curve is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 - 8 ( 3 x + 4 ) ^ { - \frac { 1 } { 2 } }$.\\
(i) A point $P$ moves along the curve in such a way that the $x$-coordinate is increasing at a constant rate of 0.3 units per second. Find the rate of change of the $y$-coordinate as $P$ crosses the $y$-axis.

The curve intersects the $y$-axis where $y = \frac { 4 } { 3 }$.\\
(ii) Find the equation of the curve.

\hfill \mbox{\textit{CAIE P1 2016 Q4 [6]}}