OCR Further Additional Pure 2023 June — Question 6 11 marks

Exam BoardOCR
ModuleFurther Additional Pure (Further Additional Pure)
Year2023
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVector Product and Surfaces
TypeClassifying stationary points on surfaces
DifficultyChallenging +1.8 This is a Further Maths question on multivariable calculus requiring computation of partial derivatives, the Hessian determinant, and solving a transcendental equation. While conceptually advanced (Further Maths content), the execution is methodical: compute ∂z/∂x, ∂z/∂y, find second partials for the Hessian, then apply the second derivative test. Part (c) requires setting up stationary point conditions and algebraic manipulation. The techniques are standard for this syllabus level, making it moderately challenging but not requiring exceptional insight.
Spec8.05d Partial differentiation: first and second order, mixed derivatives8.05f Nature of stationary points: classify using Hessian matrix
6 The surface \(S\) has equation \(z = x \sin y + \frac { y } { x }\) for \(x > 0\) and \(0 < y < \pi\).
  1. Determine, as a function of \(x\) and \(y\), the determinant of \(\mathbf { H }\), the Hessian matrix of \(S\).
  2. Given that \(S\) has just one stationary point, \(P\), use the answer to part (a) to deduce the nature of \(P\).
  3. The coordinates of \(P\) are \(( \alpha , \beta , \gamma )\). Show that \(\beta\) satisfies the equation \(\beta + \tan \beta = 0\).
Question 6:
AnswerMarks Guidance
6(a)  z y z 1
= s in y − = xcosy+
 x x 2 y x
 2 z 2 y  2 z
= and = − x s in y
 x 2 x 3  y 2
2z 2z 1
= =cosy−
xy yx x2
2 y 1
c o s y −
x 3 x 2
H = =
1
c o s y − − x s in y
x 2
2ysin y  1  2
− −cosy− 
AnswerMarks
x2  x2 B1
B1
B1
B1
M1
A1
AnswerMarks
[6]1.1
1.1
1.1
1.1
1.1
AnswerMarks
1.1Both correct
Only one needs to be seen
H attempted with their correct partial derivatives (as
functions) in the appropriate places
Any correct (unsimplified) form www
AnswerMarks Guidance
(b)P is a saddle-point
SinceH <0 (for x, y in the given domain).
Both y and sin y are positive (in (0, )) and the square is
AnswerMarks
non-negativeB1
B1
AnswerMarks Guidance
[2]2.2a
2.4Convincing argument that H
(c)z y  z 1
=sin y− =0 and = x c o s y + = 0
x x2  y x
1 s in y
 = = − c o s y
x 2 y
AnswerMarks
 + t a n = 0  M1*
M1dep
A1
AnswerMarks
[3]1.1
1.1
AnswerMarks
1.1Setting both their 1st partial derivatives to zero
Valid method for eliminating the x’s and use of y =
 at some stage
AG www
Question 6:
6 | (a) |  z y z 1
= s in y − = xcosy+
 x x 2 y x
 2 z 2 y  2 z
= and = − x s in y
 x 2 x 3  y 2
2z 2z 1
= =cosy−
xy yx x2
2 y 1
c o s y −
x 3 x 2
H = =
1
c o s y − − x s in y
x 2
2ysin y  1  2
− −cosy− 
x2  x2  | B1
B1
B1
B1
M1
A1
[6] | 1.1
1.1
1.1
1.1
1.1
1.1 | Both correct
Only one needs to be seen
H attempted with their correct partial derivatives (as
functions) in the appropriate places
Any correct (unsimplified) form www
(b) | P is a saddle-point
Since |H|<0 (for x, y in the given domain).
Both y and sin y are positive (in (0, )) and the square is
non-negative | B1
B1
[2] | 2.2a
2.4 | Convincing argument that |H| is always negative.
(c) | z y  z 1
=sin y− =0 and = x c o s y + = 0
x x2  y x
1 s in y
 = = − c o s y
x 2 y
 + t a n = 0   | M1*
M1dep
A1
[3] | 1.1
1.1
1.1 | Setting both their 1st partial derivatives to zero
Valid method for eliminating the x’s and use of y =
 at some stage
AG www
6 The surface $S$ has equation $z = x \sin y + \frac { y } { x }$ for $x > 0$ and $0 < y < \pi$.
\begin{enumerate}[label=(\alph*)]
\item Determine, as a function of $x$ and $y$, the determinant of $\mathbf { H }$, the Hessian matrix of $S$.
\item Given that $S$ has just one stationary point, $P$, use the answer to part (a) to deduce the nature of $P$.
\item The coordinates of $P$ are $( \alpha , \beta , \gamma )$.

Show that $\beta$ satisfies the equation $\beta + \tan \beta = 0$.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Additional Pure 2023 Q6 [11]}}
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