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Calc Help! Please explain this I'm so lost!?
A trough is 9 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of y=x^{10} from x=-1 to x=1 . The trough is full of water. Find the amount of work in foot-pounds required to empty the trough by pumping the water over the top. Note: The weight of water is 62 pounds per cubic foot.
2) A tank in the shape of an inverted right circular cone has height 11 meters and radius 10 meters. It is filled with 7 meters of hot chocolate.
Find the work required to empty the tank by pumping the hot chocolate over the top of the tank. Note: the density of hot chocolate is \delta = 1550 kg/m^3
3)A trough is 8 meters long, 3 meters wide, and 3 meters deep. The vertical cross-section of the trough parallel to an end is shaped like an isoceles triangle (with height 3 meters, and base, on top, of length 3 meters). The trough is full of water (density 1000 kg/m^3 ). Find the amount of work in joules required to empty the trough by pumping the water over the top. (Note: Use g=9.8 m/s^2 as the acceleration due to gravity.)
3 Answers
- 1 decade agoFavorite Answer
I give you no guarantees on these answers, but if you have an answer sheet to check them with then maybe you will find the process useful.
1)
Consider lifting each layer out of the trough. Each layer has dimensions of 9 * 2x, but when integrating each layer the variable of integration is y, so express x in terms of y.
y = x^10
x = y^(1/10)
Each layer:
9 * 2 * y^(1/10)
18 * y^(1/10)
The volume of each layer is
18 * y^(1/10) * dy
The mass is volume times density.
18 * y^(1/10) * dy * 62
1116 * y^(1/10) * dy
The distance that the mass must be lifted is 1 - y because the trough is 1 foot high.
Work = Force * Distance
1116 * y^(1/10) * dy * ( 1 - y )
The limits of integration are still 0 and 1 because the equation states
y = x^10
y = 0^10 = 0
y = 1^10 = 1
1116 * integrate( y^(1/10) * ( 1 - y ) * dy, 0, 1 )
1116 * integrate( [ y^(1/10) - y^(11/10) ] * dy, 0, 1 )
1116 * [ y^(11/10) / ( 11/10 ) - y^(21/10) / ( 21/10 ) ] | 1, 0
1116 * [ 10/11 * y^(11/10) - 10/21 * y^(21/10) ] | 1, 0
1116 * ( [ 10/11 * (1)^(11/10) - 10/21 * (1)^(21/10) ] - [ 10/11 * (0)^(11/10) - 10/21 * (0)^(21/10) ] )
1116 * [ 10/11 * (1)^(11/10) - 10/21 * (1)^(21/10) ]
1116 * [ 10/11 - 10/21 ]
483 foot pounds
2)
The diagonal of the inverted right circular cone is a line with points ( 0, 11 ) and ( 5, 0 ).
y = -11/5x + 11
x = -5/11 * ( y - 11 )
x is the radius of the circular layer at any elevation.
The area of the layer is
PI * ( -5/11 * ( y - 11 ) )^2
The volume of the layer is
PI * ( -5/11 * ( y - 11 ) )^2 * dy
The mass of the layer is
PI * ( -5/11 * ( y - 11 ) )^2 * dy * 1550
The work done to move the layer is
PI * ( -5/11 * ( y - 11 ) )^2 * dy * 1550 * y, if we set up our integral to use 4 through 11. 4 comes from 11-7.
PI * ( -5/11 )^2 * ( y - 11 )^2 * dy * 1550 * y
320.248 * PI * y * ( y - 11 )^2 * dy
320.248 * PI * integrate( y * ( y - 11 )^2 * dy, 4, 11 )
320.248 * PI * integrate( ( y^3 - 22y^2 + 121y ) * dy, 4, 11 )
320.248 * PI * [ 1/4 * y^4 - 22/3 * y^3 + 121/2 * y^2 ] | 11, 4
320.248 * PI * ( [ 1/4 * (11)^4 - 22/3 * (11)^3 + 121/2 * (11)^2 ] - [ 1/4 * (4)^4 - 22/3 * (4)^3 + 121/2 * (4)^2 ] )
661419 kilogram * meters
3)
The diagonal of the trough has points ( 0, 0 ) and ( 3, 3 ).
y = x
x = y
y is half the length of the rectangular layer.
The area of the layer is
2y * 8
The volume of the layer is
2y * 8 * dy
The mass of the layer is
2y * 8 * dy * 1000
The work done to move the layer is
2y * 8 * dy * 1000 * ( 3 - y ), The factor must be 3-y because the layer that has the longest distance has the smallest volume if you draw a diagram.
16000 * y * ( 3 - y ) * dy
16000 * integrate( y * ( 3 - y ) * dy, 0, 3 )
16000 * integrate( ( 3y - y^2 ) * dy, 0, 3 )
16000 * [ 3/2 * y^2 - 1/3 * y^3 ] | 3, 0
16000 * [ 3/2 * (3)^2 - 1/3 * (3)^3 ]
72000 kilogram * meters
Force = mass * acceleration
Force = 72000 * 9.8 = 705600 newtons
705600 newton * meters
705600 joules
- mccuistionLv 44 years ago
a. take g(x) and plug interior the cost of f(x) for each x. so it is like a million/x^3 b. the alternative of a. so (a million/x)^3 c. (x^3)^3 sorry, i won't be able to truly clarify this. seem up composite applications
- ?Lv 44 years ago
a. take g(x) and plug in the cost of f(x) for each x. so that's like a million/x^3 b. the different of a. so (a million/x)^3 c. (x^3)^3 sorry, i won't be able to truly clarify this. study composite purposes
