Mathematics, 14.04.2020 19:55 sammimoose6490

A horizontal trough is 16 m long, and its end are isosceles trapezoids with an altitude of 4 m, a lower base of 4 m, and an upper base of 6 m. If the water level is decreasing at a rate of 25 cm/min when the water is 3 m deep, at what rate is the water being drawn from the trough?

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Answer from: jackchelly

The rate at which the water level (the depth) is decreasing, increases as

more water is drawn from the trough.

The rate at which water is being drawn from the through is 22 m³/min.

Reasons:

The given parameter are;

Length of the through, L = 16 m

Height of the through = 4 m

Lower base of the trapezoidal cross section, a = 4 m

Upper base of the trapezoidal cross section, b = 6 m

Rate at which the water is decreasing at the depth of 3 m, $\dfrac{dh}{dt}$ = 25 cm/min

Required:

Rate at which water is being drawn from the through

Solution:

Let the length of the upper base at height h = 4 + 2·x

Volume of the through = Trapezoidal cross sectional area × Length

$Area \ of \ trapezoid = \dfrac{a + b}{2} \times h$

$Trapezoidal \ cross \ sectional \ area, \ A = \dfrac{4 + (4 + 2 \cdot x)}{2} \times h$ -

$Volume \ of \ water \ in \ the \ through, \ V = \dfrac{4 + (4 + 2 \cdot x)}{2} \times h \times L$

Therefore;

$V = \dfrac{4 + (4 + 2 \cdot x)}{2} \times h \times 16$

By similar triangles, we have;

$\dfrac{1}{x} = \dfrac{4}{h}$

h = 4·x

Which gives;

$V = \dfrac{4 + (4 + 2 \cdot \dfrac{h}{4} )}{2} \times 4 \cdot \dfrac{h}{4} \times 16 = 64\cdot h + 4 \cdot h^2$

$\dfrac{dV}{dh} = \dfrac{dV}{dt} \times \dfrac{dt}{dh}$

$\dfrac{dV}{dh} =64 + 8 \cdot h$

$\dfrac{dh}{dt}$ = 25 cm/min = 0.25 m/min

$\dfrac{dV}{dt} = \dfrac{ \dfrac{dV}{dh} }{ \dfrac{dt}{dh}} = \dfrac{dV}{dh} \times \dfrac{dh}{dt}$

$\dfrac{dV}{dt} = (64 + 8 \times 3 ) \times (0.25 ) = 22$

The rate at which water is being drawn from the through, $\dfrac{dV}{dt}$ = 22 m³/min.

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A horizontal trough is 16 m long, and its end are isosceles trapezoids with an altitude of 4 m, a lo

Answer from: sgslayerkingminecraf

0.28cm/min

Step-by-step explanation:

Given the horizontal trough whose ends are isosceles trapezoid

Volume of the Trough =Base Area X Height

=Area of the Trapezoid X Height of the Trough (H)

The length of the base of the trough is constant but as water leaves the trough, the length of the top of the trough at any height h is 4+2x (See the Diagram)

The Volume of water in the trough at any time

$Volume=\frac{1}{2} (b_{1}+4+2x)h X H$

$Volume=\frac{1}{2} (4+4+2x)h X 16$

=8h(8+2x)

V=64h+16hx

We are not given a value for x, however we can express x in terms of h from Figure 3 using Similar Triangles

x/h=1/4

4x=h

x=h/4

Substituting x=h/4 into the Volume, V

$V=64h+16h(\frac{h}{4})$

$V=64h+4h^2\\\frac{dV}{dt}= 64\frac{dh}{dt}+8h \frac{dh}{dt}$

h=3m,

dV/dt=25cm/min=0.25 m/min

$0.25= (64+8*3) \frac{dh}{dt}\\0.25=88\frac{dh}{dt}\\\frac{dh}{dt}=\frac{0.25}{88}$

=0.002841m/min =0.28cm/min

The rate is the water being drawn from the trough is 0.28cm/min.

Answer from: Quest

step-by-step explanation:

Answer from: Quest

it's a proportional graph.

step-by-step explanation:

8 x 5 = 40 (and you have -8x and -5y)

x= absiss

y= orderly

so you have a proportional graph.

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