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C3 question.

Hi, this question is a bit annoying, just need some help since I'm going wrong at some point.

(Q) A river is 30 metres wide in a certain region and its depth, 'd' metres, at a point 'x' metres from one side is given by the formula d = 1/60 √(x(30-x)(30+x))

(a) Use Simpson's rule with 6 intervals to estimate the cross sectional area of the river in this region.

I started by multiplying 'd' by 30 and then using the resultant formula to input values of 'x' with the width of the interval being 5. The correct answer is meant to be 38.3 m².

Thanks a lot for your time
from john
Reply 1
Avatar for gyrase
gyrase
3
Neo1
Hi, this question is a bit annoying, just need some help since I'm going wrong at some point.

(Q) A river is 30 metres wide in a certain region and its depth, 'd' metres, at a point 'x' metres from one side is given by the formula d = 1/60 √(x(30-x)(30+x))

(a) Use Simpson's rule with 6 intervals to estimate the cross sectional area of the river in this region.

I started by multiplying 'd' by 30 and then using the resultant formula to input values of 'x' with the width of the interval being 5. The correct answer is meant to be 38.3 m².
6 intervals = 30/6 = 5
d = 1/60 √(x(30-x)(30+x))
= √[(900x-x^3)/60^2] = √[x/4 - x^3/60^2]

d(0) = 0
d(5/2) = √[5/8 - 5^3/(60^2.2^3)] = √(715/1152)
d(5) = √[5/4 - 5^3/60^2] = √(175/144) = √175/12
d(15/2) = √[15/8 - 15^3/(60^2.2^3)] = 15/√128
d(10) = √[10/4 - 10^3/60^2] = √20/3
d(25/2) = √[25/8 - 25^3/(60^2.2^3)] = √(2975/1152)
d(15) = √[15/4 - 15^3/60^2] = √45/4
d(35/2) = √[35/8 - 35^3/(60^2.2^3)] = √(3325/1152)
d(20) = √[20/4 - 20^3/60^2] = 5/3
d(45/2) = √[45/8 - 45^3/(60^2.2^3)] = √(315/128)
d(25) = √[25/4 - 25^3/60^2] = √275/12
d(55/2) = √[55/8 - 55^3/(60^2.2^3)] = √(1265/1152)
d(30) = 0

M(6) = 5[d(5/2)+d(15/2)+d(25/2)+d(35/2)+d(45/2)+d(55/2)]
...... = 5[√(715/1152)+15/√128+√(2975/1152)+√(3325/1152)+√(315/128)+√(1265/1152)]
..... = 40.18096189 m^2
T(6) = 5/2[d(0)+2d(5)+2d(10)+2d(15)+2d(20)+2d(25)+d(30)]
...... = 5[(1/2)(d(0)+d(30)+d(5)+d(10)+d(15)+d(20)+d(25)]
...... = 5[√175/12+√20/3+√45/4+5/3+√275/12] = 36.59376505 m^2

S(6) = (1/3)(2M(6)+T(6)) = 38.99m^2

[Which is close enough, I might have made calculation mistakes ...]
Reply 2
Avatar for Neo1
Neo1
OP
4
yeah I managed to get 38.3. I used simpson's rule for 6 intervals only, thanks for the help

The second part of this question is a problem too, I guess you have to use the formula ſ pi (fx)² and then what?

Thanks a lot
Hi,
i dont understand why youre x values are going up in 2.5 when the height is 5
Original post by faisa12345
Hi,
i dont understand why youre x values are going up in 2.5 when the height is 5


The chances of someone remembering a question they answered 10 years ago is slim.

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