Year: 2022
Paper: 2
Question Number: 6
Course: LFM Pure
Section: Implicit equations and differentiation
No solution available for this problem.
Candidates appeared to be generally well prepared for most topics within the examination, but there were a few situations in questions where some did not appear to be as proficient in standard techniques as needed. In particular, the method for finding invariant lines required in question 8 and the manipulation of trigonometric functions that were needed in question 10 caused considerable difficulties for some candidates. An additional issue that occurred at numerous points in the paper relates to the direction in which a deduction is required. It is important that candidates make sure that they know which statement is the one that they should start from as they deduce the other and that it is clear in their solution that the logic has gone in the correct direction. Clarity of solution is also an issue that candidates should be aware of, especially in the situations where the result to be reached has been given. It is important to check that there are no special cases that need to be considered separately, and when dividing by a function it is necessary to confirm that the function cannot be equal to 0 (and in the case of inequalities that the function always has the same sign). When drawing diagrams and sketching graphs it is useful if significant points that need to be clear are not drawn over the lines on the page as these can be difficult to interpret during the marking process.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
In this question, you should consider only points lying in the first quadrant, that is with $x > 0$ and $y > 0$.
\begin{questionparts}
\item The equation $x^2 + y^2 = 2ax$ defines a \emph{family} of curves in the first quadrant, one curve for each positive value of $a$. A second family of curves in the first quadrant is defined by the equation $x^2 + y^2 = 2by$, where $b > 0$.
\begin{enumerate}
\item Differentiate the equation $x^2 + y^2 = 2ax$ implicitly with respect to $x$, and hence show that every curve in the first family satisfies the differential equation
\[2xy\frac{\mathrm{d}y}{\mathrm{d}x} = y^2 - x^2.\]
Find similarly a differential equation, independent of $b$, for the second family of curves.
\item Hence, or otherwise, show that, at every point with $y \neq x$ where a curve in the first family meets a curve in the second family, the tangents to the two curves are perpendicular.
A curve in the first family meets a curve in the second family at $(c,\,c)$, where $c > 0$. Find the equations of the tangents to the two curves at this point. Is it true that where a curve in the first family meets a curve in the second family on the line $y = x$, the tangents to the two curves are perpendicular?
\end{enumerate}
\item Given the family of curves in the first quadrant $y = c\ln x$, where $c$ takes any non-zero value, find, by solving an appropriate differential equation, a second family of curves with the property that at every point where a curve in the first family meets a curve in the second family, the tangents to the two curves are perpendicular.
\item A family of curves in the first quadrant is defined by the equation $y^2 = 4k(x + k)$, where $k$ takes any non-zero value.
Show that, at every point where one curve in this family meets a second curve in the family, the tangents to the two curves are perpendicular.
\end{questionparts}
This was one of the better answered questions on the paper and most candidates produced substantial responses to the question. Part (i)(a) was completed well in general, with most candidates achieving good marks for this section. When considering the second family of curves, some candidates forgot to include the dy/dx on the right-hand side following their differentiation with respect to x. In part (i)(b) many candidates failed to justify their division by (x² − y²) when considering the case where x ≠ y. Most candidates were however able to show that the tangents do remain perpendicular even in the case where x = y. In part (ii) many candidates failed to recognise that c needed to be eliminated from the differential equation in order to find the second family of curves. This mistake often led to the candidate only being able to achieve one of the marks available in this section. Many candidates were able to make good progress on part (iii) of the question and many good solutions to this part of the question were seen.