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

By Mejlbro L.

Similar analysis books

Analisi matematica

Nel quantity vengono trattati in modo rigoroso gli argomenti che fanno parte tradizionalmente dei corsi di Analisi matematica I: numeri reali, numeri complessi, limiti, continuità, calcolo differenziale in una variabile e calcolo integrale secondo Riemann in una variabile. Le nozioni di limite e continuità sono ambientate negli spazi metrici, di cui viene presentata una trattazione elementare ma precisa.

Multicriteria and Multiobjective Models for Risk, Reliability and Maintenance Decision Analysis

This ebook integrates a number of standards innovations and strategies for difficulties in the chance, Reliability and upkeep (RRM) context. The options and foundations relating to RRM are thought of for this integration with multicriteria methods. within the e-book, a basic framework for construction determination versions is gifted and this can be illustrated in a variety of chapters by way of discussing many alternative determination versions relating to the RRM context.

Extremal Lengths and Closed Extensions of Partial Differential Operators

Test of print of Fuglede's paper on "small" households of measures. A strengthening of Riesz's theorem on subsequence is received for convergence within the suggest. This result's utilized to calculus of homologies and classes of differential types.

Additional resources for Calculus 2c-2, Examples of Description of Surfaces Partial Derivatives, Gradient, Directional Derivative and Taylor's Formula

Sample text

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.