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Unformatted text preview: height hi
i =1
n ≈ ∑ ( 7 − hi ) ⋅ δ ⋅ π 24 h
⋅ 22 ∆
123 14 3
i =1
force 7 ( ft ⋅ lbs
ft 3 ⋅ ft 3 = ft ⋅ lbs ) distance 5 n ≈ ∑ 200π ( 7 − hi ) ∆h hi i =1 work = 200π ∫ 5
0 ( 7 − h ) dh = 4500π ft ⋅ lbs n 2.9 work ≈ ∑ work to raise the chain, with length yi + 3 ft, vertically ∆y ft
i =1
n ≈ ∑ δ ( yi + 3) ⋅ ∆y
1 24 {
43
i =1
force (weight) work = 5∫ 2.10 17
0 ( lbs ⋅ ft ⋅ ft = ft ⋅ lbs )
ft distance ( y + 3) dy = 997.5 ft ⋅ lbs ∆h
∆F
∆F
ft
lbs
bs
= 1 sec and
= −20 sec . So,
= −20 lft .
∆t
∆t
∆h
n work ≈ ∑ work to raise the bucket, at height hi ft, vertically ∆h ft
i =1
n ≈ ∑ (1000 − 20hi ) ⋅ ∆h
14 244 distance
4
3{
i =1 ( ( lbs − lbs
ft ⋅ ft ) ⋅ ft = ft ⋅ lbs ) force (weight) work = ∫ 2.11 (a) (b) 30
0 (1000 − 20h ) dh = 21, 000 ft ⋅ lbs The center of the ellipse is 3 units to the right of the origin.
So, we will find the equation of the ellipse with center (0,0) and then add 3
to th...
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 Fall '08
 POOLE
 Calculus

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