Question 12.11.4: Suppose that the two equations (z + 2w)^5 + xy² = 2z − yw (1......
Suppose that the two equations
(z+2w)^{5}+x y^{2}=2z-y w(1+z^{2})^{3}-z^{2}w=8x+y^{5}w^{2}
define z and w as differentiable functions z = ϕ(x, y) and w = ψ(x, y) of x and y in a neighbourhood around (x, y, z, w) = (1, 1, 1, 0).
(a) Compute ∂z/∂x, ∂z/∂y, ∂w/∂x, and ∂w/∂y at (1, 1, 1, 0).
(b) Use the results in (a) to find an approximate value of φ(1 + 0.1, 1 + 0.2).
(a) Equating the differentials of each side of the two equations, treated as functions of (x, y), we obtain
5(z+2w)^{4}(\mathrm{d}z+2\,\mathrm{d}w)+y^{2}\,\mathrm{d}x+2x y\,\mathrm{d}y=2\,\mathrm{d}z-w\,\mathrm{d}y-y\,\mathrm{d}w3(1+z^{2})^{2}2z\,\mathrm{d}z-2z w\,\mathrm{d}z-z^{2}\,\mathrm{d}w=8\,\mathrm{d}x+5y^{4}w^{2}\,\mathrm{d}y+2y^{5}w\,\mathrm{d}w
At the particular point (x, y, z, w) = (1, 1, 1, 0) this system reduces to:
3\,{\mathrm{d}}z+11\,{\mathrm{d}}w=-{\mathrm{d}}x-2\,{\mathrm{d}}y;\quad24\,{\mathrm{d}}z-{\mathrm{d}}w=8\,{\mathrm{d}}xSolving these two equations simultaneously for dz and dw in terms of dx and dy yields
\mathrm{d}z={\frac{29}{89}}\,\mathrm{d}x-{\frac{2}{267}}\,\mathrm{d}y\,\,\,\mathrm{and}\ \ \mathrm{d}w=-{\frac{16}{89}}\,\mathrm{d}x-{\frac{16}{89}}\,\mathrm{d}yHence, ∂z/∂x = 29/89, ∂z/∂y = −2/267, ∂w/∂x = −16/89, and ∂w/∂y = −16/89.
(b) If x = 1 is increased by dx = 0.1 and y = 1 is increased by dy = 0.2, the associated change in z = ϕ(x, y) is approximately dz = (29/89) · 0.1 − (2/267) · 0.2 ≈ 0.03. Hence φ(1 + 0.1, 1 + 0.2) ≈ φ(1, 1) + dz ≈ 1 + 0.03 = 1.03.