| Exam Board | AQA |
|---|---|
| Module | Further Paper 1 (Further Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Polar coordinates |
| Type | Polar curve with exponential function |
| Difficulty | Challenging +1.8 This is a substantial Further Maths polar coordinates question requiring derivation of the Cartesian gradient formula from polar form (part a) and then proving an angle is constant using logarithmic differentiation and trigonometric manipulation (part b). While the techniques are standard for Further Maths (polar calculus, implicit differentiation), the multi-step proof structure and algebraic manipulation required place it above average difficulty, though not at the extreme end since the path is relatively guided. |
| Spec | 1.07s Parametric and implicit differentiation4.09b Sketch polar curves: r = f(theta) |
| Answer | Marks |
|---|---|
| 15(a) | Correctly converts |
| Answer | Marks | Guidance |
|---|---|---|
| coordinates (PI) | AO3.1a | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| dθ dθ | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| dθ dθ | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| dx | AO2.1 | R1 |
| Answer | Marks |
|---|---|
| 15(b) | Differentiates r |
| Answer | Marks | Guidance |
|---|---|---|
| standard result. | AO1.2 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| gradient from (a) | AO3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| formula. | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| tan | AO2.2a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone | AO2.2a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| independent of θ | AO2.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| required result. | AO2.1 | R1 |
| Total | 11 |
Question 15:
--- 15(a) ---
15(a) | Correctly converts
polar to Cartesian
coordinates (PI) | AO3.1a | B1 | x=rcosθ and y =rsinθ
dy dr
= sinθ+rcosθ
dθ dθ
dx dr
= cosθ−rsinθ
dθ dθ
dr
sinθ+rcosθ
dy dθ
Gradient = =
dx dr
cosθ−rsinθ
dθ
Finds correct
expression for at least
dy dx
one of and
dθ dθ | AO1.1a | M1
dy dx
Divides their by
dθ dθ | AO1.1a | M1
Completes proof,
including statement
dy
that gradient =
dx | AO2.1 | R1
--- 15(b) ---
15(b) | Differentiates r
correctly using
standard result. | AO1.2 | B1 | dr
=( ) (cotb)θ
cotb e
dθ
dy ( cotb ) e (cotb)θ sinθ+e (cotb)θ cosθ
=
dx ( cotb ) e (cotb)θ cosθ−e (cotb)θ sinθ
cotbsinθ+cosθ
=
cotbcosθ−sinθ
cosb
sinθ+cosθ
sinb
=
cosb
cosθ−sinθ
sinb
cosbsinθ+sinbcosθ
=
cosbcosθ−sinbsinθ
(θ+b )
sin
= =tan (θ+b )
(θ+b )
cos
STP=θ+b
angle
angle OPT =b (exterior angle of a
∴
triangle)
∴
So the angle between the line OP and
the tangent TPQ does not depend onθ.
Substitutes r and their
dr
into expression for
dθ
gradient from (a) | AO3.1a | M1
Rearranges into a form
that can be recognised
as a compound angle
formula. | AO1.1a | M1
Deduces that gradient
of TPQ is equal to
(θ+b )
tan | AO2.2a | M1
Deduces that angle
STP =θ+b
−(θ+b )
Condone | AO2.2a | M1
Uses a geometric
argument to explain
why OPT is
independent of θ | AO2.4 | M1
Completes a rigorous
argument to show the
required result. | AO2.1 | R1
Total | 11The diagram shows part of a spiral curve.
The point $P$ has polar coordinates $(r, \theta)$ where $0 \leq \theta \leq \frac{\pi}{2}$
The points $T$ and $S$ lie on the initial line and $O$ is the pole.
$TPQ$ is the tangent to the curve at $P$.
\includegraphics{figure_15}
\begin{enumerate}[label=(\alph*)]
\item Show that the gradient of $TPQ$ is equal to
$$\frac{\frac{dr}{d\theta} \sin \theta + r \cos \theta}{\frac{dr}{d\theta} \cos \theta - r \sin \theta}$$
[4 marks]
\item The curve has polar equation
$$r = e^{(\cot b)\theta}$$
where $b$ is a constant such that $0 < b < \frac{\pi}{2}$
Use the result of part (a) to show that the angle between the line $OP$ and the tangent $TPQ$ does not depend on $\theta$.
[7 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 1 2019 Q15 [11]}}