| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2015 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Curve motion: find dy/dt |
| Difficulty | Standard +0.3 This is a straightforward connected rates of change question requiring chain rule (dy/dt = dy/dx × dx/dt), basic integration to find the curve equation, and standard tangent/normal geometry. All parts use routine A-level techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.07m Tangents and normals: gradient and equations1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| At \(x=4\), \(\frac{dy}{dx}=2\) | B1 | |
| \(\frac{dy}{dt}=\frac{dy}{dx}\times\frac{dx}{dt}=2\times3=6\) | M1A1 [3] | Use of Chain rule |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((y)=x+4x^{\frac{1}{2}}(+c)\) | B1 | |
| Sub \(x=4\), \(y=6\rightarrow6=4+(4\times4^{\frac{1}{2}})+c\) | M1 | Must include \(c\) |
| \(c=-6\rightarrow(y=x+4x^{\frac{1}{2}}-6)\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Eqn of tangent is \(y-6=2(x-4)\) or \((6-0)/(4-x)=2\) | M1 A1 | Correct eqn thru \((4,6)\) & with \(m=\) *their* 2 |
| \(B=(1,0)\) (Allow \(x=1\)) | M1 | [Expect eqn of normal: \(y=-\frac{1}{2}x+8\)] |
| Gradient of normal \(=-\frac{1}{2}\) | A1 | |
| \(C=(16,0)\) (Allow \(x=16\)) | A1 | |
| Area of triangle \(=\frac{1}{2}\times15\times6=45\) | [5] | Or \(AB=\sqrt{45}\), \(AC=\sqrt{180}\rightarrow\) Area \(=45.0\) |
# Question 9:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| At $x=4$, $\frac{dy}{dx}=2$ | B1 | |
| $\frac{dy}{dt}=\frac{dy}{dx}\times\frac{dx}{dt}=2\times3=6$ | M1A1 [3] | Use of Chain rule |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(y)=x+4x^{\frac{1}{2}}(+c)$ | B1 | |
| Sub $x=4$, $y=6\rightarrow6=4+(4\times4^{\frac{1}{2}})+c$ | M1 | Must include $c$ |
| $c=-6\rightarrow(y=x+4x^{\frac{1}{2}}-6)$ | A1 [3] | |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Eqn of tangent is $y-6=2(x-4)$ or $(6-0)/(4-x)=2$ | M1 A1 | Correct eqn thru $(4,6)$ & with $m=$ *their* 2 |
| $B=(1,0)$ (Allow $x=1$) | M1 | [Expect eqn of normal: $y=-\frac{1}{2}x+8$] |
| Gradient of normal $=-\frac{1}{2}$ | A1 | |
| $C=(16,0)$ (Allow $x=16$) | A1 | |
| Area of triangle $=\frac{1}{2}\times15\times6=45$ | [5] | Or $AB=\sqrt{45}$, $AC=\sqrt{180}\rightarrow$ Area $=45.0$ |
---9 A curve passes through the point $A ( 4,6 )$ and is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = 1 + 2 x ^ { - \frac { 1 } { 2 } }$. A point $P$ is moving along the curve in such a way that the $x$-coordinate of $P$ is increasing at a constant rate of 3 units per minute.\\
(i) Find the rate at which the $y$-coordinate of $P$ is increasing when $P$ is at $A$.\\
(ii) Find the equation of the curve.\\
(iii) The tangent to the curve at $A$ crosses the $x$-axis at $B$ and the normal to the curve at $A$ crosses the $x$-axis at $C$. Find the area of triangle $A B C$.
\hfill \mbox{\textit{CAIE P1 2015 Q9 [11]}}