Standard +0.3 This is a straightforward implicit differentiation question requiring students to find dy/dx, evaluate it at a given point (first finding the x-coordinate where y=1), then use the negative reciprocal to find the normal equation. While it involves multiple steps (substitution, implicit differentiation, point evaluation, normal gradient, line equation), each step follows standard procedures taught in C4/Further Pure with no novel problem-solving required. Slightly easier than average due to the clean numbers and direct application of techniques.
15 In this question you must show detailed reasoning.
The equation of a curve is
$$\ln y + x ^ { 3 } y = 8$$
Find the equation of the normal to the curve at the point where \(y = 1\), giving your answer in the form \(\mathrm { ax } + \mathrm { by } + \mathrm { c } = 0\), where \(a , b\) and \(c\) are constants to be found.
substitution of their \(x = 2\) and \(y = 1\) to obtain numerical value for \(\frac{dy}{dx}\)
M1*
NB \(-\frac{4}{3}\); dependent on at least two of 3 terms correct on LHS following differentiation; if expression for \(\frac{dy}{dx}\) or evaluation of \(\frac{dy}{dx}\) is incorrect, need to see substitution for award of M1
FT negative reciprocal of their \(-\frac{4}{3}\) and their 2; may see eg \(1 = \frac{3}{4} \times 2 + c\)
\(3x - 4y - 2 = 0\) or \(-3x + 4y + 2 = 0\) oe
A1
must be in required form, but coefficients may be fractions
## Question 15:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = 1$ then $x = 2$ only | B1 | |
| $\frac{1}{y} \times \frac{dy}{dx}$ | B1 | first term correct; allow $y'$ for $\frac{dy}{dx}$ |
| $x^3 \times \frac{dy}{dx} + 3x^2 y$ | M1 | Product Rule; allow one coefficient error or one index error |
| $\frac{1}{y} \times \frac{dy}{dx} + x^3 \times \frac{dy}{dx} + 3x^2 y [= 0]$ | A1 | |
| substitution of their $x = 2$ and $y = 1$ to obtain numerical value for $\frac{dy}{dx}$ | M1* | **NB** $-\frac{4}{3}$; dependent on at least two of 3 terms correct on LHS following differentiation; if expression for $\frac{dy}{dx}$ or evaluation of $\frac{dy}{dx}$ is incorrect, need to see substitution for award of **M1** |
| $y - 1 = \left(their\,\frac{3}{4}\right)(x - their\,2)$ oe | M1dep* | **FT** negative reciprocal of their $-\frac{4}{3}$ and their 2; may see eg $1 = \frac{3}{4} \times 2 + c$ |
| $3x - 4y - 2 = 0$ or $-3x + 4y + 2 = 0$ oe | A1 | must be in required form, but coefficients may be fractions |
15 In this question you must show detailed reasoning.
The equation of a curve is
$$\ln y + x ^ { 3 } y = 8$$
Find the equation of the normal to the curve at the point where $y = 1$, giving your answer in the form $\mathrm { ax } + \mathrm { by } + \mathrm { c } = 0$, where $a , b$ and $c$ are constants to be found.
\hfill \mbox{\textit{OCR MEI Paper 2 2023 Q15 [7]}}