Question

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?

Answer 1

Answer:

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.