set8 - Lecture 21 Derivatives and rates of change...

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Unformatted text preview: Lecture 21 Derivatives and rates of change Applications to classical mechanics (contd) We proceed with our analysis of the projectile problem, in which a ball is launched vertically upward from the ground with velocity v > 0. It is sketched again below. x ( t ) f =- mg v initial velocity x ground level In the last lecture, we showed that Newtons Second Law translated into the following equation of motion for x ( t ), the position of the mass at time t : a = d 2 x dt 2 = g. (1) We may reduce the order of the derivative by expressing the above equation in terms of the velocity v ( t ) = x ( t ), i.e., dv dt = g. (2) As discussed previously, this is a first-order DE in the function v ( t ). We are looking for a function v ( t ), the derivative of which, with respect to t, is g . By inspection, we know that the function v ( t ) = gt satisfies (2). But you also know that we can add a constant to this function, i.e., v ( t ) = gt + C 1 . (3) When we take derivatives, we see that this function satisfies (2). But is it possible that there are other functions that satisfy (2)? We need to discuss the idea of antiderivatives . 150 Antiderivatives Suppose that we have a function f ( x ). (Well let x denote the independent variable in this section.) Definition: A function F such that F ( x ) = f ( x ) for all x (in its domain of definition) is called an antiderivative of f . Examples: 1. f ( x ) = x . An antiderivative of x is 1 2 x 2 . But so is 1 2 x 2 + 10. Or 1 2 x 2 + C , where C is any constant. 2. f ( x ) = sin x . An antiderivative of sin x is cos x . But so is cos x + C where, once again, C is any constant. From the above examples, we see that if we have an antiderivative F of a function f , then we can add an arbitrary constant C to it and still have an antiderivative of f . But are there any other antiderivatives? The following result answers this question. Theorem: Suppose that F 1 and F 2 are antiderivatives of a function f . Then F 1 ( x ) = F 2 ( x ) + C (4) for some constant C . Proof: From the assumptions, for all x in the domains of definition, F 1 ( x ) = f ( x ) F 2 ( x ) = f ( x ) . (5) Now subtract the second equation from the first: F 1 ( x ) F 2 ( x ) = 0 . (6) 151 But this implies that d dx [ F 1 ( x ) F 2 ( x )] = 0 . (7) Since this is true for all x , it follows that the LHS is a constant, i.e., F 1 ( x ) F 2 ( x ) = C, (8) for some constant C . (This statement seems to be correct well prove it rigorously in a few lectures with the help of the Mean Value Theorem.) Therefore, F 1 ( x ) = F 2 ( x ) + C, (9) and the theorem is proved. A consequence of this theorem is that once we know a single antiderivative F of a function f , we can generate all others by simply adding a constant to F ( x )....
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This note was uploaded on 09/22/2011 for the course MATH 137 taught by Professor Speziale during the Fall '08 term at Waterloo.

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set8 - Lecture 21 Derivatives and rates of change...

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