Calculus 2c-2, Examples of Description of Surfaces Partial by Mejlbro L.

By Mejlbro L.

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Calculate f (0; e)e or |f (0; e|e for every unit vector e. com 47 Calculus 2c-2 Directional derivative 2 y –2 1 0 –1 2 1 x –1 –2 Figure 22: The sketched diameter is f (0, 0). I We can obviously assume that (x0 , y0 ) = (0, 0). Then let ∂f ∂f (0), (0) ∂x ∂y f (0) = := (a, b). Any unit vector can be written in the form e(ϕ) = (cos ϕ, sin ϕ), ϕ ∈ [0, 2π[, so (2) f (0; e(ϕ))e(ϕ) = (a cos ϕ + b sin ϕ)(cos ϕ, sin ϕ) = (x(ϕ), y(ϕ)), where x(ϕ) = a cos2 ϕ + b sin ϕ cos ϕ = a 1 {a cos 2ϕ + b sin 2ϕ} + 2 2 and y(ϕ) = a sin ϕ cos ϕ + b sin2 ϕ = b 1 {a sin 2ϕ − b cos 2ϕ} + .

Y 2 {1 + cos2 (xy)}2 Finally, ∂2f ∂2f = ∂x∂y ∂y∂x = − = − sin(2xy) 2xy cos(2xy) − 1 + cos2 (xy) 1 + cos2 (xy) x sin(2xy) {−2 cos(xy) sin(xy) · y} + {1 + cos2 (xy)}2 xy sin2 (2xy) sin(2xy) + 2xy cos(2xy) − . com 57 Calculus 2c-2 Partial derivatives of higher order whence ∂2f ∂x2 ∂2f ∂x∂y = (1 + y)2 f (x, y) = (1 + y)2 exp(x + xy − 2y), = ∂2f = 1 · f (x, y) + (1 + y)(x − 2)f (x, y) ∂y∂x = (x + xy − 2y − 1) exp(x + xy − 2y), ∂2f ∂y 2 = (x − 2)2 f (x, y) = (x − 2)2 exp(x + xy − 2y). 6) When f (x, y) = Arctan(x − y), we get ∂f 1 = , ∂x 1 + (x − y)2 1 ∂f =− , ∂y 1 + (x − y)2 Please click the advert what‘s missing in this equation?

V Test. com 32 Calculus 2c-2 The chain rule and 2u(v − u) ∂F = . ∂v (u + v)3 The results agree. ♦ 2) Consider f (x, y) = Arctan y , x X(u, v) = (u2 − uv + v 2 , 2uv), (u, v) = (0, 0). We first check that the composite function is defined (and of class C ∞ , where it is defined). Here we shall just check that x = 0 for (u, v) = (0, 0). Now x(u, v) = u2 − uv + v 2 = u− 1 v 2 2 + 3 = 0 for (u, v) = (0, 0). 4 Therefore, f (X(u, v)) is defined and of class C ∞ for (u, v) = (0, 0). In the calculations we shall need x2 + y 2 expressed by u and v.

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