| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2003 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find tangent equation at point |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question requiring standard techniques: differentiate √x + √y = √a to find dy/dx, then find point P by solving √x + √x = √a, and finally substitute into point-slope form. While it involves multiple steps, each is routine and the question follows a standard template for implicit differentiation problems, making it slightly easier than average. |
| Spec | 1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| EITHER: Obtain terms \(\frac{1}{2\sqrt{x}}\) and \(\frac{1}{2\sqrt{y}}\frac{dy}{dx}\), or equivalent | B1+B1 | |
| Obtain answer in any correct form, e.g. \(\frac{dy}{dx} = -\sqrt{\frac{y}{x}}\) | B1 | |
| OR: Using chain or product rule, differentiate \((\sqrt{a} - \sqrt{x})^2\) | M1 | |
| Obtain derivative in any correct form | A1 | |
| Express \(\frac{dy}{dx}\) in terms of \(x\) and \(y\) only in any correct form | A1 | |
| OR: Expand \((\sqrt{a} - \sqrt{x})^2\), differentiate and obtain term \(-2\cdot\frac{\sqrt{a}}{2\sqrt{x}}\), or equivalent | B1 | |
| Obtain term 1 by differentiating an expansion of the form \(a + x \pm 2\sqrt{a}\sqrt{x}\) | B1 | |
| Express \(\frac{dy}{dx}\) in terms of \(x\) and \(y\) only in any correct form | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State or imply coordinates of \(P\) are \((\frac{1}{4}a, \frac{1}{4}a)\) | B1 | |
| Form equation of the tangent at \(P\) | M1 | |
| Obtain 3 term answer \(x + y = \frac{1}{2}a\) correctly, or equivalent | A1 |
## Question 4(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| **EITHER:** Obtain terms $\frac{1}{2\sqrt{x}}$ and $\frac{1}{2\sqrt{y}}\frac{dy}{dx}$, or equivalent | B1+B1 | |
| Obtain answer in any correct form, e.g. $\frac{dy}{dx} = -\sqrt{\frac{y}{x}}$ | B1 | |
| **OR:** Using chain or product rule, differentiate $(\sqrt{a} - \sqrt{x})^2$ | M1 | |
| Obtain derivative in any correct form | A1 | |
| Express $\frac{dy}{dx}$ in terms of $x$ and $y$ only in any correct form | A1 | |
| **OR:** Expand $(\sqrt{a} - \sqrt{x})^2$, differentiate and obtain term $-2\cdot\frac{\sqrt{a}}{2\sqrt{x}}$, or equivalent | B1 | |
| Obtain term 1 by differentiating an expansion of the form $a + x \pm 2\sqrt{a}\sqrt{x}$ | B1 | |
| Express $\frac{dy}{dx}$ in terms of $x$ and $y$ only in any correct form | B1 | |
**Total: [3]**
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## Question 4(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply coordinates of $P$ are $(\frac{1}{4}a, \frac{1}{4}a)$ | B1 | |
| Form equation of the tangent at $P$ | M1 | |
| Obtain 3 term answer $x + y = \frac{1}{2}a$ correctly, or equivalent | A1 | |
**Total: [3]**
---4 The equation of a curve is
$$\sqrt { } x + \sqrt { } y = \sqrt { } a$$
where $a$ is a positive constant.\\
(i) Express $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in terms of $x$ and $y$.\\
(ii) The straight line with equation $y = x$ intersects the curve at the point $P$. Find the equation of the tangent to the curve at $P$.
\hfill \mbox{\textit{CAIE P3 2003 Q4 [6]}}