| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Normal line motion |
| Difficulty | Standard +0.3 This is a straightforward connected rates of change question with three standard parts: finding a normal equation (routine differentiation and negative reciprocal), applying dy/dt = (dy/dx)(dx/dt) on the curve, and applying the same chain rule on the normal line. All techniques are direct applications of standard methods with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = x^{-\frac{1}{2}} \rightarrow m = \frac{1}{2}\) at \(x = 4\) | B1 | |
| Equation of normal is \(y - 3 = -2(x - 4)\) | M1 | Through \((4, 3)\) with gradient \(-\frac{1}{\text{their } m}\). Dependent on differentiation used. |
| \(y = -2x + 11\) | A1 | Only acceptable answer. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dt} = \text{their } \frac{1}{2} \times 3\) | M1 | Correct use of the chain rule with a numerical gradient. |
| \(\frac{3}{2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Required gradient \(\left[= \frac{dy}{dx}\right] = -2\) | B1FT | SOI. FT from their part (a) if a normal gradient has been found from \(m_1 m_2 = -1\) and differentiation. |
| \(\frac{dx}{dt} = \frac{1}{\text{their normal gradient}} \times -5\) | M1 | Correct use of chain rule. Allow method mark also for \(+5\), must be numerical. Their normal gradient must come from \(m_1 m_2 = -1\) and differentiation in part(a) unless 'restarted' here. |
| \(\frac{5}{2}\) | A1 |
## Question 9(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = x^{-\frac{1}{2}} \rightarrow m = \frac{1}{2}$ at $x = 4$ | B1 | |
| Equation of normal is $y - 3 = -2(x - 4)$ | M1 | Through $(4, 3)$ with gradient $-\frac{1}{\text{their } m}$. Dependent on differentiation used. |
| $y = -2x + 11$ | A1 | Only acceptable answer. |
## Question 9(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dt} = \text{their } \frac{1}{2} \times 3$ | M1 | Correct use of the chain rule with a numerical gradient. |
| $\frac{3}{2}$ | A1 | |
## Question 9(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Required gradient $\left[= \frac{dy}{dx}\right] = -2$ | B1FT | SOI. FT from their part (a) if a normal gradient has been found from $m_1 m_2 = -1$ and differentiation. |
| $\frac{dx}{dt} = \frac{1}{\text{their normal gradient}} \times -5$ | M1 | Correct use of chain rule. Allow method mark also for $+5$, must be numerical. Their normal gradient must come from $m_1 m_2 = -1$ and differentiation in part(a) unless 'restarted' here. |
| $\frac{5}{2}$ | A1 | |9 A curve has equation $y = 2 x ^ { \frac { 1 } { 2 } } - 1$.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the normal to the curve at the point $A ( 4,3 )$, giving your answer in the form $y = m x + c$.\\
A point is moving along the curve $y = 2 x ^ { \frac { 1 } { 2 } } - 1$ in such a way that at $A$ the rate of increase of the $x$-coordinate is $3 \mathrm {~cm} \mathrm {~s} ^ { - 1 }$.
\item Find the rate of increase of the $y$-coordinate at $A$.\\
At $A$ the moving point suddenly changes direction and speed, and moves down the normal in such a way that the rate of decrease of the $y$-coordinate is constant at $5 \mathrm {~cm} \mathrm {~s} ^ { - 1 }$.
\item As the point moves down the normal, find the rate of change of its $x$-coordinate.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q9 [8]}}