alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

Sin0 gt t0 instantaneous ver ii over t1 t2 the

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Unformatted text preview: 1 b ( x 2 -- x 1 ) L imit x 2 -- x 1 = x 2 → x1 b = b . The instantaneous rate of change of y is CONSTANT. chang hange instantaneous ra dy d ( bx ) = dx = b dx The general linear function is : y = mx + C where m is the slope . slope So the DERIVATIVE of a linear function is its slope . slope Exercise: What is the slope of f (x) = x + a ? Draw the graph. (Ref. page 26 Example 2 and the graph on page 180) 72 Example 3: Let us find the general expression for the INSTANTANEOUS RATE OF CHANGE in height of the ball from the function of its height. differentia eneral We need to differentiate y(t) = u . s i n θ . t -- 1 g t 2 at general instant t . dif entiate -2 VERAGE step: 1. AVERAGE ste p: ∆y ∆t = y(t + ∆ t) − y(t) (t + ∆ t) − t {u . s i n θ . (t + ∆ t) -- 1 g (t + ∆ t)2 } -- {u . s i n θ . t -- 1 g t2} --2 2 = ∆t 2 {u . s i n θ . (t + ∆ t) -- 1 g (t + 2 t ∆ t + ∆ t 2)} -- {u . s i n θ . t -- 1 g t2} --2 2 = ∆t u . s i n θ . ∆ t -- g t ∆ t -- 1 g ∆ t 2 -2 = ∆t u . s i n θ . δ t -- g t δ t -- 1 g δ t 2 -δy 2 2. TENDS TO step: = δt δt Here we may simplify since δt ≠ 0. δy = u . s i n θ -- g t -- 1 g δ t -δt 2 step: 3. LIMIT step: Finally, we take the LIMIT as δt → 0 to get the INSTANTANEOUS VERTICAL SPEED at general instant t . general dy LIMIT δy LIMIT 1 = δt→ 0 δt = δt→ 0 {u . s i n θ -- g t -- -- g δ t } 2 dt y ’(t) = u . s i n θ -- g t (refer Preface iii & iv) 73 We may compare this with the AVERAGE SPEED. If we measure the height at any chosen instant t , then in terms of the height function y(t) = u . s i n θ . t -- 1 g t 2 -2 the AVERAGE SPEED over time t = 0 to t is : y (t ) distance = u . s i n θ -- 1 g t -time t= 2 height y(t) = u . s i n θ . t -- 1 g t 2 -2 (0,0) T T/ 2 t y ’(t) = u . s i n θ -- g t y(t) = u . s i n θ . t -- 1 g t 2 is a polynomial of the for m a 0 + a 1 t + a 2 t 2 -2 1 with coefficients a 0 = 0 , a 1 = u . s i n θ a nd a 2 = -- -- g . We may directly 2 apply our formula for differentiating polynomials to get : dy d 2 Vertical speed y ’(t) = -= (u . s i n θ . t -- 1 g t ) = u . s i n θ -- g t 2 dt dt When we evaluate this expression for the INSTANTANEOUS RATE OF CHANGE at a par ticul...
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