Standard +0.3 This is a straightforward partial differentiation problem requiring finding stationary points by setting ∂z/∂x = 0 and ∂z/∂y = 0, then solving the resulting simultaneous equations (3x² - 6y = 0 and 3y² - 6x = 0). While it involves cubic terms, the algebraic manipulation is routine for Further Maths students and requires no novel insight—slightly easier than average.
3 The surface with equation \(z = x ^ { 3 } + y ^ { 3 } - 6 x y\) has two stationary points; one at the origin and the second at the point \(A\). Determine the coordinates of \(A\).
Setting both derivatives to zero ⟹ \(y = \frac{1}{3}x^2\) and \(y^2 = 2x\); Eliminating e.g. y to get \(\frac{1}{3}x^4 = 2x\) ⟹ \(x^3 = 8\) ⟹ \(x = 2\)
M1 M1 A1 E1 [continuing]
2.1 1.1 1.1 1.1
\(x = 0\) implies \(y = z = 0\) which is the origin, so not A
Answer
Marks
Guidance
⟹ \(y = 2\); ⟹ \(z = -8\)
A1 A1 [9]
1.1 1.1
Differentiating $z = x^3 + y^3 - 6xy$ partially w.r.t. x and/or y: $\frac{\partial z}{\partial x} = 3x^2 - 6y$; $\frac{\partial z}{\partial y} = 3y^2 - 6x$ | M1 A1 B1 FT [using symmetry] | 1.1 1.1 1.1 | Must have two terms attempted; 2nd differentiation is either correct or FT from the 1st
Setting both derivatives to zero ⟹ $y = \frac{1}{3}x^2$ and $y^2 = 2x$; Eliminating e.g. y to get $\frac{1}{3}x^4 = 2x$ ⟹ $x^3 = 8$ ⟹ $x = 2$ | M1 M1 A1 E1 [continuing] | 2.1 1.1 1.1 1.1 | Solving e.g. for x
$x = 0$ implies $y = z = 0$ which is the origin, so not A
⟹ $y = 2$; ⟹ $z = -8$ | A1 A1 [9] | 1.1 1.1
3 The surface with equation $z = x ^ { 3 } + y ^ { 3 } - 6 x y$ has two stationary points; one at the origin and the second at the point $A$. Determine the coordinates of $A$.
\hfill \mbox{\textit{OCR Further Additional Pure AS 2017 Q3 [9]}}