By E. Poisson
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Extra resources for Advanced Mechanics [Phys 3400 Lecture Notes]
7) into Eq. 6) we must first calculate the derivative of G with respect to y ′ , treating y as a constant parameter. This is ∂G = −yy ′ 1 − y ′2 ∂y ′ −1/2 . We next differentiate this with respect to s. Because ∂G/∂y ′ ≡ Gy′ depends on s through its dependence on both y and y ′ , we must apply the chain rule. This gives d ∂G ds ∂y ′ = = We have and ∂Gy′ dy ∂Gy′ dy ′ + ∂y ds ∂y ′ ds ∂Gy′ ′ ∂Gy′ ′′ y . y + ∂y ∂y ′ ∂Gy′ = −y ′ 1 − y ′2 ∂y ∂Gy′ = −y 1 − y ′2 ∂y ′ −1/2 −3/2 , so that d ∂G = −y ′2 1 − y ′2 ds ∂y ′ The remaining quantity to calculate is −1/2 − yy ′′ 1 − y ′2 ∂G = 1 − y ′2 ∂y 1/2 −3/2 .
What is happening to the major axis of the ellipse? 8 Additional problems 1. An inclined plane makes an angle α with the horizontal. A projectile is launched from point A at the bottom of the inclined plane. Its initial speed is v0 , and its initial velocity vector makes an angle β with the horizontal. The projectile eventually hits the inclined plane at point B. Air resistance is negligible. (a) Calculate the range R of the projectile, the distance between points A and B. Show that it it can be expressed in the form R = R0 sin(β − α) cos β and find an expression for R0 .
We shall examine what happens to A[y] when we displace the path from y(s) = y¯(s) to y(s) = y¯(s) + δy(s). While we shall find that in general, this produces a change δA that is proportional to δy(s), we will instead demand that δA vanish to first order in the displacement; as we shall see, this procedure will permit us to identify the extremum path y¯(s). To carry out this procedure properly it is important to ensure that all the considered paths begin and end at the same two end points. The reference path y¯(s) and the displaced paths y(s) = y¯(s) + δy(s) must all satisfy y(s0 ) = y0 and y(s1 ) = y1 .
Advanced Mechanics [Phys 3400 Lecture Notes] by E. Poisson