alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

# The first par t of the study of calculus deals with

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Unformatted text preview: 25 m/sec for t ≥ 10 seconds is constant. Limit v(t) − v(10) v(t) − v(10) ≠ right derivative t → 10+ left derivative Limit0− right t→1 t − 10 t − 10 we say : v(t) is not differentiable at t = 10 secs. differentiable Even though v(t) is not differentiable at t = 10 secs, we can still compute the left differentiable left derivative and right derivative and interpret them according to the situation. deriv tiv right deriv tiv Exer ercise1: Exercise1: Given f ’(x) = AVERAGE RATE OF CHANGE of f(x), what is f(x) ? a) ex b) |x| c) constant d) x e) ax + b 90 Example 4: Is the function y(t), that describes the height of a bouncing ball, bouncing differentiable at instants t0, t1, t2, t3, . . . ? y(t) SLOPE of tangent at t1 from the left is negative left negative h e i g h t t0 SLOPE of tangent at t1 from the right is postive right postive n t1 t2 t3 t ertical At t1 we have two instantaneous ver tical speeds : the speed on impact at t1 while instantaneous ver descending and the speed after impact at t1 while rising. We know the duration of rising flight T = 2u0. sinθ/ g (see page 75). To find the speed on impact at t1 while descending we substitute T = 0 and t = T0 in y0’(t) = u0. sinθ0 − g(t − t0) to get y0’(t− ) = −u0. sinθ0 . 1 After impact at t1 we know this 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. Geometrically Geometrically, we can see that we have two tangents at t1. left derivative y0’(t− ) = −u0.sinθ0 ≠ 1 y1’(t+ ) = +u1.sinθ1 right derivative right 1 we say : y(t) is not differentiable at t = t1 secs. differentiable Even though y(t) is not differentiable at t = t0, t1, t2, t3, . . . , we can still differentiable compute the left derivative and right derivative and interpret them according to left deriv tiv right deriv tiv the situation. 91 Example 5 : is f(x) = + √x differentiable at x = 0 ? differentiable f(x) f(x) = +√ x (0, 0) x We know that f(x) =...
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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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