| Exam Board | Edexcel |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Parametric arc length calculation |
| Difficulty | Challenging +1.2 This is a standard arc length calculation using the formula for Cartesian curves. While it requires careful algebraic manipulation and integration (likely involving completing a square or recognizing a standard form), the setup is straightforward and the question explicitly guides students to the final form. It's above average difficulty due to the algebraic complexity and integration technique required, but remains a textbook-style FP3 question without requiring novel insight. |
| Spec | 4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{x}{4} - \frac{1}{x}\) | B1 | Correct derivative. Allow any correct equivalent e.g. \(\frac{2x}{8} - \frac{1}{x}\) |
| \(L = \int\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx = \int\sqrt{1+\left(\frac{x}{4}-\frac{1}{x}\right)^2}\,dx\) | M1 | Use of correct formula using their derivative and not the given \(y\) |
| \(= \int\sqrt{\frac{x^2}{16} + \frac{1}{2} + \frac{1}{x^2}}\,dx = \int\sqrt{\left(\frac{x}{4}+\frac{1}{x}\right)^2}\,dx = \int\left(\frac{x}{4}+\frac{1}{x}\right)dx\) | M1 A1 | M1: Squares derivative to obtain \(ax^2 + bx^{-2} + c\) where none of \(a,b,c\) are zero, and adds 1. A1: Correct integrand \(\frac{x}{4}+\frac{1}{x}\) or equivalent e.g. \(\frac{x^2+4}{4x}\) |
| \(= \frac{x^2}{8} + \ln kx\) | A1 | Correct integration |
| \(\left[\frac{x^2}{8}+\ln x\right]_2^3 = \left(\frac{9}{8}+\ln 3\right)-\left(\frac{4}{8}+\ln 2\right)\) | M1 | Substitutes 2 and 3 into expression of form \(px^2 + q\ln x\ (p,q \neq 0)\) and subtracts correct way round. If final single answer given in decimals with no substitution shown, e.g. 1.030… this is M0. |
| \(\frac{5}{8} + \ln\frac{3}{2}\) | A1 | cao and cso (oe e.g. \(0.625 + \ln\frac{3}{2}\)) |
## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{x}{4} - \frac{1}{x}$ | B1 | Correct derivative. Allow any correct equivalent e.g. $\frac{2x}{8} - \frac{1}{x}$ |
| $L = \int\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx = \int\sqrt{1+\left(\frac{x}{4}-\frac{1}{x}\right)^2}\,dx$ | M1 | Use of correct formula using their derivative and not the given $y$ |
| $= \int\sqrt{\frac{x^2}{16} + \frac{1}{2} + \frac{1}{x^2}}\,dx = \int\sqrt{\left(\frac{x}{4}+\frac{1}{x}\right)^2}\,dx = \int\left(\frac{x}{4}+\frac{1}{x}\right)dx$ | M1 A1 | M1: Squares derivative to obtain $ax^2 + bx^{-2} + c$ where none of $a,b,c$ are zero, and adds 1. A1: Correct integrand $\frac{x}{4}+\frac{1}{x}$ or equivalent e.g. $\frac{x^2+4}{4x}$ |
| $= \frac{x^2}{8} + \ln kx$ | A1 | Correct integration |
| $\left[\frac{x^2}{8}+\ln x\right]_2^3 = \left(\frac{9}{8}+\ln 3\right)-\left(\frac{4}{8}+\ln 2\right)$ | M1 | Substitutes 2 and 3 into expression of form $px^2 + q\ln x\ (p,q \neq 0)$ and subtracts correct way round. If final single answer given in decimals with no substitution shown, e.g. 1.030… this is M0. |
| $\frac{5}{8} + \ln\frac{3}{2}$ | A1 | cao and cso (oe e.g. $0.625 + \ln\frac{3}{2}$) |
---\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$y = \frac { x ^ { 2 } } { 8 } - \ln x , \quad 2 \leqslant x \leqslant 3$$
Find the length of the curve $C$ giving your answer in the form $p + \ln q$, where $p$ and $q$ are rational numbers to be found.\\
\hfill \mbox{\textit{Edexcel FP3 2016 Q2 [7]}}