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

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Lecture 21 Derivatives and rates of change Applications to classical mechanics (cont’d) We proceed with our analysis of the projectile problem, in which a ball is launched vertically upward from the ground with velocity v 0 > 0. It is sketched again below. x ( t ) f = - mg v 0 initial velocity x ground level 0 In the last lecture, we showed that Newton’s 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

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Antiderivatives Suppose that we have a function f ( x ). (We’ll 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 – we’ll 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 ). We now return to our classical mechanics problem. Solving the DE dv dt = g (10) amounts to finding an antiderivative of the function g . An obvious antiderivative of g is gt . From the previous theorem, it follows that all antiderivatives of gt , hence all solutions of the above DE, have the form v ( t ) = gt + C 1 , (11) where C 1 is an arbitary constant. We can remove the arbitrariness of the constant C 1 by imposing

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

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