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
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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
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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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- Fall '09
- TAMERDOğAN
- Limit, Δx
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