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alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

Fxdx is an i nstantaneous element o f area if we try

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Unformatted text preview: ive CHANGE (positive area − area under y’(t) over [t3, t4] y(t3) = DECREASE in height from t3 to t4 y(t4) h3 h4 → 4321 4321 4321 4321 4321 4321 4321 4321 t3 (0,0) t4 y ’(t) = u . s i n θ -- g t vertical speed t4 ∫ t3 t T t4 y ’(t)dt = [ y(t) ] = y (t4) -- y(t3) = DECREASE in height t3 4. (negative f(x)) . (negative direction) = positive CHANGE (negative + area under y’(t) over [t3, t4] y(t3) = INCREASE in height from t4 to t3 y(t4) h3 h4 → 4321 4321 4321 4321 4321 4321 4321 t3 (0,0) t4 t T vertical speed y ’(t) = u . s i n θ -- g t t3 t3 ∫ t4 y ’(t)dt = [ y(t) ] = y (t3) -- y(t4) = INCREASE in height t4 143 hat zer area ero curv imply What does zero ar ea under the cur ve y ’(t) imply ? NO CHANGE in height. In the diagram below let the intervals [ t1, T/2 ] and [ T/2 , t 4 ] be equal in length. Also let ( t1, y ’(t1)) and ( t 4, y ’(t 4)) be equal in length. So from the mensuration mensuration point of view the shaded triangles are equal in area. Let us call this area A. y(t1) zero area under y’(t) over [t1, t4] y(t4) = NO CHANGE in height y’(t1) 654321 654321 654321 654321 654321 654321 654321654321 654321 654321 654321 654321 654321 654321 T/ (0,0) t1 2 y’(t4) t4 t T vertical speed y ’(t) = u . s i n θ -- g t From the Calculation point of view : Calculation Notice that y ’(t) changes SIGN at T/2 . So we may break up the integral into 2 parts. t4 ∫ t1 y ’(t)dt = T/ 2 ∫ t1 y ’(t)dt + t4 ∫ T/ 2 T/ 2 t4 t1 y ’(t)dt T/ 2 = [ y(t)] + [ y(t) ] = { y(T/2) -- y(t1) } + { y(t4) -- y(T/2)} = y(t4) -- y(t1) 144 From the Geometric point of view : Geometric t4 ∫ T/ 2 ∫ t1 ∫ t1 y ’(t)dt + t4 ∫ T/ 2 y ’(t)dt y ’(t)dt = (positive function) . (positive direction) = positive area : + A t4 ∫ T/ 2 t1 y ’(t)dt = T/ 2 y ’(t)dt = (negative function) . (positive direction) = negative area : − A t4 ∫ So : t1 y ’(t)dt = + A − A = 0 From the Analytical point of view : Analytical t4 ∫ t1 y ’(t)dt = y (t4) -- y(t1) = 0 There is NO CHANGE in the height y(t) over interval [ t1, t 4 ] . Now direction integ ov interv Now let us REVERSE the dir ection of integr ation over inter v al [ t1, t 4 ].. From the Calculation point of view : Calculation...
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