Explanation Given information: Refer to the given figure in the textbook for the ellipse with a parabolic trajectory. Calculation: Consider that the expression for the maximum height. y = ( v 0 sin α ) 2 2 g (1) The range at which the maximum height in Equation (1) occurs as, x = v 0 2 2 g sin 2 α = v 0 2 2 g 2 sin α cos α { ∵ sin 2 α = 2 sin α cos α } x = v 0 2 sin α cos α g (2) Consider that equations (1) and (2) describe parametrically the points on a curve in the xy -plane associated with the maximum heights on parabolic trajectories in terms of the parameter α that is launch angle. Square on both sides on the Equation (2). x 2 = ( v 0 2 sin α cos α g ) 2 = v 0 4 sin 2 α cos 2 α g 2 = v 0 4 sin 2 α ( 1 − sin 2 α ) g 2 = v 0 4 sin 2 α g 2 − v 0 4 sin 4 α g 2 Re-arrange the equation. x 2 = v 0 2 g [ v 0 2 sin 2 α g ] − v 0 4 sin 4 α g 2 = v 0 2 g [ 2 v 0 2 sin 2 α 2 g ] − 4 v 0 4 sin 4 α 4 g 2 = 2 v 0 2 g [ v 0 2 sin 2 α 2 g ] − 4 [ v 0 2 sin 2 α 2 g ] 2 = 2 v 0 2 g

Thomas' Calculus Format: Unbound (saleable) With Access Card

14th Edition

ISBN: 9780134768762

Author: Hass, Joel R.^heil, Christopher D.^weir, Maurice D.

Publisher: Prentice Hall

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Chapter 13.2, Problem 34E

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Verify that the points in the plane that provide the maximum heights of these parabolic trajectories all lie on the given ellipse.

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Chapter 13.2, Problem 33E

Chapter 13.2, Problem 35E

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Thomas' Calculus Format: Unbound (saleable) With Access Card

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