| Exam Board | CAIE |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2012 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find vertical tangent points |
| Difficulty | Standard +0.3 This is a straightforward implicit differentiation question with standard techniques. Part (i) requires routine application of chain rule and product rule to differentiate ln(xy) and y³, then rearranging for dy/dx. Part (ii) requires recognizing that a vertical tangent occurs when dx/dy = 0 (denominator = 0), leading to 3y³ - 1 = 0, then substituting back to find x. While it requires understanding of vertical tangents, the execution is mechanical with no novel insight needed, making it slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | ||
| EITHER: State or imply \(\frac{1}{x} + \frac{1}{y}\frac{dy}{dx}\) as derivative of \(\ln xy\), or equivalent | B1 | |
| State or imply \(3y^2\frac{dy}{dx}\) as derivative of \(y^3\), or equivalent | B1 | |
| Equate derivative of LHS to zero and solve for \(\frac{dy}{dx}\) | M1 | |
| Obtain the given answer | A1 | |
| OR | ||
| Obtain \(xy = \exp(1+y^3)\) and state or imply \(y + x\frac{dy}{dx}\) as derivative of \(xy\) | B1 | |
| State or imply \(3y^2\frac{dy}{dx}\exp(1+y^3)\) as derivative of \((1+y^3)\) | B1 | |
| Equate derivatives and solve for \(\frac{dy}{dx}\) | M1 | |
| Obtain the given answer | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) | ||
| Equate denominator to zero and solve for \(y\) | M1* | |
| Obtain \(y = 0.693\) only | A1 | |
| Substitute found value in the equation and solve for \(x\) | M1(dep*) | |
| Obtain \(x = 5.47\) only | A1 | [4] |
**(i)** | |
**EITHER:** State or imply $\frac{1}{x} + \frac{1}{y}\frac{dy}{dx}$ as derivative of $\ln xy$, or equivalent | B1 |
State or imply $3y^2\frac{dy}{dx}$ as derivative of $y^3$, or equivalent | B1 |
Equate derivative of LHS to zero and solve for $\frac{dy}{dx}$ | M1 |
Obtain the given answer | A1 |
**OR** | |
Obtain $xy = \exp(1+y^3)$ and state or imply $y + x\frac{dy}{dx}$ as derivative of $xy$ | B1 |
State or imply $3y^2\frac{dy}{dx}\exp(1+y^3)$ as derivative of $(1+y^3)$ | B1 |
Equate derivatives and solve for $\frac{dy}{dx}$ | M1 |
Obtain the given answer | A1 | [4]
[The M1 is dependent on at least one of the B marks being earned]
**(ii)** | |
Equate denominator to zero and solve for $y$ | M1* |
Obtain $y = 0.693$ only | A1 |
Substitute found value in the equation and solve for $x$ | M1(dep*) |
Obtain $x = 5.47$ only | A1 | [4]7 The equation of a curve is $\ln ( x y ) - y ^ { 3 } = 1$.\\
(i) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y } { x \left( 3 y ^ { 3 } - 1 \right) }$.\\
(ii) Find the coordinates of the point where the tangent to the curve is parallel to the $y$-axis, giving each coordinate correct to 3 significant figures.
\hfill \mbox{\textit{CAIE P3 2012 Q7 [8]}}