# Answer on Differential Equations Question for joe

Question #8553

1. The table below gives the depth of water across a river measured at one metre

intervals between banks.

Distance (m) 0 1 2 3 4

Water depth (m) 0.4 0.5 1.6 0.9 0

Use the Trapezium rule to estimate the cross-sectional area of the river.

A river hydrologist estimates that at the place where this cross sectional data was

measured the average speed of water flow is 0.8m/s. Estimate the volume of water

which passes this section of the river in one minute.

intervals between banks.

Distance (m) 0 1 2 3 4

Water depth (m) 0.4 0.5 1.6 0.9 0

Use the Trapezium rule to estimate the cross-sectional area of the river.

A river hydrologist estimates that at the place where this cross sectional data was

measured the average speed of water flow is 0.8m/s. Estimate the volume of water

which passes this section of the river in one minute.

Expert's answer

1st trapezium points: (0,0), (1,0), (0,0.4), (1,0.5).

1st trapezium area: S1 = (1-0)*min{0.4,0.5} + (1-0)*(0.5-0.4)/2 = 0.45 m²

2nd trapezium points: (1,0), (2,0), (1,0.5), (2,1.6).

2nd trapezium area: S2 = (2-1)*min{0.5,1.6} + (2-1)*(1.6-0.5)/2 = 1.05 m²

3rd trapezium points: (2,0), (3,0), (2,1.6), (3,0.9).

3rd trapezium area: S3 = (3-2)*min{1.6,0.9} + (3-2)*(1.6-0.9)/2 = 1.25 m²

4th trapezium points: (3,0), (4,0), (3,0.9), (4,0).

4th trapezium area: S4 = (4-3)*min{0.9,0} + (4-3)*(0.9-0)/2 = 0.45 m²

Total area: S1+S2+S3+S4 = 0.45 + 1.05 + 1.25 + 0.45 = 3.2 m²

So, the volume of water passes this section of the river in one minute is approximately

V = 3.2m² * 0.8m/s = 2.56 m³/s.

1st trapezium area: S1 = (1-0)*min{0.4,0.5} + (1-0)*(0.5-0.4)/2 = 0.45 m²

2nd trapezium points: (1,0), (2,0), (1,0.5), (2,1.6).

2nd trapezium area: S2 = (2-1)*min{0.5,1.6} + (2-1)*(1.6-0.5)/2 = 1.05 m²

3rd trapezium points: (2,0), (3,0), (2,1.6), (3,0.9).

3rd trapezium area: S3 = (3-2)*min{1.6,0.9} + (3-2)*(1.6-0.9)/2 = 1.25 m²

4th trapezium points: (3,0), (4,0), (3,0.9), (4,0).

4th trapezium area: S4 = (4-3)*min{0.9,0} + (4-3)*(0.9-0)/2 = 0.45 m²

Total area: S1+S2+S3+S4 = 0.45 + 1.05 + 1.25 + 0.45 = 3.2 m²

So, the volume of water passes this section of the river in one minute is approximately

V = 3.2m² * 0.8m/s = 2.56 m³/s.

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