A rectangular trough is 8ft long, 2ft across the top, and 4ft deep. if water flows in at the rate 2 ft^3/min, how fast is the surface rising when the water is 1 ft deep?
CAN ANYONE PLEASE GIVE ME SOME IDEAS TO DO IT???????THANKS A LOT!!!!!
what is the shape of the trough at the bottom?
change in volume of water * area of surface of water * dH
where dH is the change in depth
therefore
flow rate =dV/dt = area of surface of water * dH/dt
or
dH/dt = dV/dt / area of water surface
the depth has noting to do with it. Only the surface area matters.
To find the rate at which the surface is rising when the water is 1 ft deep, we can use related rates.
Let's consider the volume of water in the trough as a function of time, V(t), and the depth of the water as a function of time, h(t).
We are given that the trough is rectangular, with a length of 8 ft, a width of 2 ft at the top, and a depth of 4 ft. Therefore, the volume of water in the trough is given by the formula:
V(t) = length * width * height
V(t) = 8 ft * 2 ft * h(t)
We are also given that water flows into the trough at the rate of 2 ft^3/min. This means that the rate at which the volume of water is changing is:
dV/dt = 2 ft^3/min
We need to find dh/dt, the rate at which the depth of the water is changing when the water is 1 ft deep.
To find dh/dt, we can take the derivative of the volume equation with respect to time:
dV/dt = d(8 ft * 2 ft * h(t))/dt
2 ft^3/min = 16 ft^2 * dh(t)/dt
Now, we substitute the given information that the water depth is 1 ft:
2 ft^3/min = 16 ft^2 * dh(t)/dt
2 ft^3/min = 16 ft^2 * dh/dt (since the depth h(t) is 1 ft)
Now, we can solve for dh/dt, the rate at which the depth of the water is changing:
2 ft^3/min = 16 ft^2 * dh/dt
dh/dt = (2 ft^3/min) / (16 ft^2)
dh/dt = 0.125 ft/min
Therefore, the surface of the water is rising at a rate of 0.125 ft/min when the water is 1 ft deep.
To find the rate at which the water surface is rising, we need to use similar triangles.
Let's first visualize the rectangular trough and note the given measurements:
- Length (l) = 8 ft
- Width (w) = 2 ft
- Depth (d) = 4 ft
Imagine a right-angle triangle forming from the water surface to the top edge of the trough, along with a smaller right-angle triangle formed within this larger triangle.
Let's label the unknown variables:
- Unknown height of the water surface (h)
- Unknown rate at which the water level rises (dh/dt)
To solve for dh/dt, we can use similar triangles and the concept of volumes to relate the change in height (dh) to the change in volume (dV) within a small time interval (dt).
Since the trough has a constant cross-sectional area (A) of 2 ft × 4 ft = 8 ft², the change in volume is directly proportional to the change in height:
dV = A × dh
Now, we need to relate the change in volume to the rate of water flow.
Given that water flows in at a rate of 2 ft³/min, we can express the change in volume as:
dV = 2 ft³/min × dt
By equating the two expressions for dV, we get:
A × dh = 2 ft³/min × dt
Simplifying this equation further, we have:
dh/dt = (2 ft³/min) / A
Plugging in the values for A (8 ft²), we can find the rate at which the surface is rising when the water is 1 ft deep:
dh/dt = (2 ft³/min) / (8 ft²)
= 2/8 ft/min
= 0.25 ft/min
Therefore, the water surface is rising at a rate of 0.25 ft/min when the water is 1 ft deep.