{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

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

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: 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...
View Full Document

{[ snackBarMessage ]}