By Snieder R.

Unlike conventional textbooks for college students and pros within the actual sciences, this ebook offers its fabric within the type of difficulties. the second one variation comprises new chapters on dimensional research, variational calculus, and the asymptotic evaluate of integrals. The publication can be utilized through undergraduates and lower-level graduate scholars. it might probably function a stand-alone textual content, or as a resource of difficulties and examples to enrich different textbooks. First variation Hb (2001): 0-521-78241-4 First variation Pb (2001): 0-521-78751-3

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**Example text**

Since we know that M−1 = MT , it follows that (MT )−1 = (M−1 )−1 = M. Problem e Use these results to show that the spherical coordinate components of u are related to the Cartesian coordinates by the following transformation rule: ⎛ ⎞ ⎛ ⎞⎛ ⎞ ur ux sin θ cos ϕ sin θ sin ϕ cos θ ⎝ u θ ⎠ = ⎝ cos θ cos ϕ cos θ sin ϕ − sin θ ⎠ ⎝ u y ⎠ . 3 Acceleration in spherical coordinates You may wonder whether we really need all these transformation rules between a Cartesian coordinate system and a system of spherical coordinates.

25). 25). Problem d We are now in the position to compute the acceleration in spherical coordinates. 23) to eliminate the time-derivatives of the basis vectors. Use this to show that the acceleration a is given by: a = v˙ r − θ˙ vθ − sin θ ϕv ˙ ϕ rˆ + v˙ θ + θ˙ vr − cos θ ϕv ˙ ϕ θˆ ˆ. 3 Acceleration in spherical coordinates 39 z-axis r(t) } r(t + ∆ t) ∆r= dr ∆ t = vr ∆ t dt y-axis x-axis Fig. 2 Deﬁnition of the geometric variables used to derive the radial component of the velocity. Problem e This expression is not quite satisfactory because it contains both the components of the velocity as well as the time-derivatives θ˙ and ϕ˙ of the angles.

1 Properties of the gradient vector 47 Fig. 1 Deﬁnition of the points A, B, and C. 3) ∂f ⎪ ⎪ fC − f B = (x + δx, y) δ y. 2) and derive that to leading order in δx and δ y the result can be written as: ∂f ∂f (x, y) δx + (x, y) δ y. 4) δf = ∂x ∂y Note that ∂ f /∂ y is evaluated at (x + δx, y). However, using a Taylor expansion of ∂ f /∂ y we ﬁnd that ∂f ∂f ∂2 f (x + δx, y) δ y = (x, y) δ y + (x, y) δxδ y. 5) The last term is of second order in δx and δ y and can be ignored when these quantities are small.