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alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

# D d we may drop the subscript of particular angle 1

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Unformatted text preview: ar instant in time we get a numerical value. In Calculus we also learn how to interpret this value. We will be able to say things such as the height is increasing, decreasing, is at a maximum or a minimum and so on. If the duration of flight is T, then from the physics of the situation we know the maximum height is reached when t = T/ 2. We also know that at the maximum maximum height the vertical speed should be zero. Substituting t = T/ 2 in the vertical speed equation u . s i n θ -- g t = 0, we get T = 2u. sin θ / g . u. 74 ertical Example 4: What is the instantaneous ver tical speed of the bouncing ball instantaneous ver (i) over (t0, t1) ? (ii) over (t1, t2) ? (iii) at t1 ? Over [t0, t 1] the height y(t) is : y0(t) = u0.sinθ0.(t − t0) − 1 g (t − t0)2 . -2 Over [t1, t2] the height y(t) is : y1(t) = u1. sinθ1.(t − t1) − 1 g (t − t1)2 . -2 ver y(t) = v er tical position n t0 t1 t2 t3 t ver y ’(t) = ver tical speed ertical (i) Over (t0, t1) the instantaneous ver tical speed y0’(t) = u0.sinθ0 − g(t − t0). instantaneous ver (ii) Over (t1, t2) the instantaneous ver tical speed y1’(t) = u1.sinθ1 − g(t − t1). ertical instantaneous ver ertical (iii) At t1 we have two instantaneous ver tical speeds : the speed on impact at t1 instantaneous ver while descending and the speed after impact at t1 while rising . We know from descending rising the previous example the duration of flight over [t0, t1] is T0 = 2u0. sinθ/ g. To find the speed on impact at t1 while descending we substitute t0 = 0 and t = T0 descending in y0’(t) = u0.sinθ0 − g(t − t0) to get y0’(t− ) = − u0.sinθ0 . 1 After impact at t1 we know the speed on rising is y1’(t+ ) = +u1.sinθ1. Alternatively, rising 1 we may substitute t = t1 in y1’(t) = u1.sinθ1 − g(t − t1) to get the same result. 75 5: Example 5 Let us differentiate f(x) = sin (x) differentiate VERAGE step: 1. AVERAGE ste p: ∆ f(x) ∆x = f(x + ∆ x) − f(x) (x + ∆ x) − x = f(x + ∆ x) − f(x) ∆x = sin(x + ∆ x) − sin(x) ∆x = {sin(x) . cos(∆ x) + cos (x)...
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