Standard +0.3 This is a straightforward partial differentiation exercise requiring computation of two second partial derivatives and verification of a given identity. While it involves multiple steps, each differentiation is routine application of product and chain rules with no conceptual difficulty or problem-solving insight required. Being from Further Maths adds some difficulty, but the mechanical nature keeps it slightly easier than average.
3 Given \(z = x \sin y + y \cos x\), show that \(\frac { \partial ^ { 2 } z } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } z } { \partial y ^ { 2 } } + z = 0\).
Question 3:
3 | (cid:119)z (cid:119)z
(cid:32)sin y (cid:16) ycosx (cid:32) xcosy (cid:14)cosx
(cid:119)x (cid:119)y
p
(cid:119)2z (cid:119)2z S
= – y cos x = – x sin y
(cid:119)x2 (cid:119)y2
(cid:119)2z (cid:119)2z
(cid:14) (cid:14)z
(cid:119)x2 (cid:119)y2
(cid:32)(cid:16)ycosx(cid:16)xsiny(cid:14)(xsiny(cid:14) ycosx)(cid:32)0 | M1
e
A1
A1 FT
A1 FT
E1
[5] | i
c
2.1
1.1
1.1
1.1
2.1 | Attempt at 1st and 2nd partial
derivatives of z with respect to x and y;
at least one correct
FT for each correct second partial
derivative
AG Shown clearly
3 Given $z = x \sin y + y \cos x$, show that $\frac { \partial ^ { 2 } z } { \partial x ^ { 2 } } + \frac { \partial ^ { 2 } z } { \partial y ^ { 2 } } + z = 0$.
\hfill \mbox{\textit{OCR Further Additional Pure Q3 [5]}}