alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

# Sin0 gt t0 instantaneous ver ii over t1 t2 the

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online