alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

# If fx is the expression of change in position then

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Unformatted text preview: ATIVES. If F(x) is one of them, then any other one may be given by the expression F(x) + C, where C is constant. From a Geometric point of view we have a family of curves. Geometric ∫ R ATE OF CHANGE f(x) = CHANGE F(x) + C ( a constant ) The constant C is called the CONSTANT OF INTEGRATION. C could be an arbitrary constant. But in general it depends on the situation. ∫ f n (x)dx = f n -- 1(x) + C n -- 1 In general: Let us see how we determine the constant of integration C n -- 1 by looking at three examples. Example1: We know y”(t), the function that expresses the INSTANTANEOUS RATE OF CHANGE in ver tical speed (i.e. acceleration). We can integrate y”(t) and get y’ (t) the function that expresses the CHANGE in ver tical speed. We may think of y’ (t), the function that expresses the CHANGE in ver tical speed, as the function that expresses the INSTANTANEOUS RATE OF CHANGE in height. We can integrate y’ (t) and get y(t) the function that expresses the CHANGE in height. We know : y ’’(t) = -- g [ meters / sec 2 ] y ’(t) = ∫ y ’’(t) d t = ∫ ( -- g ) d t = -- g t + C 1 [ meters/ sec ]. The constant C 1 = u . sin θ i s the initial ver tical velocity. So y ’(t) = u.sinθ -- g t [ meters/ sec ]. . Now: y (t) = ∫ y ’(t) d t = ∫ ( u.sinθ -- g t ) d t = u. si n θ . t -- 1 g t 2 + C 0 [ meters ]. -. 2 The constant C0 is the initial height. 163 How do you determine the constant of integration C 0 ? height y 5(t) = u . s i n θ . t -- 1 g t 2 + 5 -2 +5 T/ 2 t Case 2: 5 meters above ground level T height y 0(t) = u . s i n θ . t -- 1 g t 2 -2 (0,0) T/ 2 t T +5 meters y ’(t) = u . s i n θ -- g t Case 1: ground level y ’(t) = u . s i n θ -- g t In both cases y ’(t) is the same. So : y (t) = ∫ y ’(t) d t = ∫ ( u.sinθ -- g t ) d t -So: y (t) = u. si n θ . t -- 1 g t 2 is also the same. . 2 We need to know something more about the initial conditions. For example : Case 1: C 0 = 0 a t ground level Case 2: C 0 = + 5 m eters above ground level. 164 ti y(ti ) = ∫ y’(t). dt give...
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## This note was uploaded on 11/29/2012 for the course PHYSICS 105 taught by Professor Tamerdoğan during the Fall '09 term at Middle East Technical University.

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