A rectangular trough is 8 feet long, 2 feet aross the top and 4 feet deep, if water flows in At a rate of 2 cubic feet per minute. How fast is the surface rises when the water is 1 feet deep.?
At a given time of t minutes, let the height of the water be h ft, h < 4
V = (8)(2)(h) = 16h feet^3
dV/dt = 16 dh/dt
2 = 16dh/dt
dh/dt = 1/8 ft/min
notice that the fact that the container is 4 ft high, did not enter the picture
YUTNIINAYO WRONG T ANSWER DAPAT T ANSWER YOU KET NO TUMAKI KAYO ADA MASARAKAM A ANSWER U ISO NOKWA DJAY
To solve this problem, we can use related rates. Let's start by identifying the given information:
Length of the trough (l) = 8 feet
Width of the top of the trough (w) = 2 feet
Depth of the trough (h) = 4 feet
Rate of water flow (dV/dt) = 2 cubic feet per minute
We are asked to find the rate at which the water surface rises when the water is 1 foot deep (dh/dt when h = 1).
Since the trough is rectangular, we can use the formula for the volume of a rectangular prism to relate the dimensions:
Volume of the trough (V) = l * w * h
Differentiating both sides of the equation with respect to time (t) using the chain rule, we get:
dV/dt = (dl/dt) * w * h + l * (dw/dt) * h + l * w * (dh/dt)
Since dl/dt and dw/dt are not given, we can assume that the dimensions of the trough are fixed and not changing over time. Therefore, we can simplify the equation to:
dV/dt = l * w * (dh/dt)
Now we can substitute the given values into the equation:
2 = 8 * 2 * (dh/dt)
Simplifying further:
2 = 16 * (dh/dt)
Now, divide both sides of the equation by 16:
2/16 = dh/dt
1/8 = dh/dt
So, the rate at which the water surface rises when the water is 1 foot deep is 1/8 feet per minute.
To find out how fast the surface of the water rises when it is 1 foot deep, we can use related rates.
Let's define some variables:
- Let V be the volume of water in the trough at any given time (in cubic feet).
- Let h be the depth of the water (in feet).
- Let A be the surface area of the water (in square feet) as a function of h.
Given:
- The trough is rectangular and 8 feet long, 2 feet across the top, and 4 feet deep. This means the bottom of the trough also has dimensions 8 feet long and 2 feet wide.
- Water flows into the trough at a rate of 2 cubic feet per minute.
To determine the equation relating V and h, we need to find the equation for the surface area of the water, A, as a function of h.
Since the trough is rectangular, the surface area A is the product of the length and width. The width would be constant at 2 feet, but the length will change as the water level rises.
To determine the length of the water surface, we need to use similar triangles. The length l of the surface will be proportional to the depth h.
Using the similar triangles, we can write the equation:
l / h = (8 feet) / (4 feet)
l = 2h
The surface area A is the product of the length l and the width w:
A = l * w
A = (2h) * 2
A = 4h
Now, we can find the volume V as a function of h. Since the trough has a constant width of 2 feet, the volume of water V is simply the cross-sectional area A multiplied by the change in depth dh:
V = A * dh
V = (4h) * dh
V = 4h * dh
Now, let's differentiate the equation with respect to time t (since we're interested in how fast the surface rises with respect to time):
dV/dt = 4h * dh/dt
Given that the rate at which the water flows into the trough is 2 cubic feet per minute, we have:
dV/dt = 2 cubic feet per minute
We want to find how fast the surface rises when the water is 1 foot deep, so let's substitute h = 1 foot into the equation:
2 = 4(1) * dh/dt
Simplifying the equation:
2 = 4dh/dt
dh/dt = 2/4
dh/dt = 0.5 feet per minute
Therefore, the surface of the water rises at a rate of 0.5 feet per minute when the water is 1 foot deep.
Well, well, well, it looks like we have a watery situation on our hands! Let's see if we can splash some humor into this math problem!
So, we have a rectangular trough that's 8 feet long, 2 feet across the top, and 4 feet deep. Now, water is flowing into this trough at a rate of 2 cubic feet per minute. And our mission is to find out how fast the surface rises when the water is 1 foot deep. Let's dive in, shall we?
To get to the bottom of this, we'll need to use a little bit of math. One thing we know is that the cross-sectional area of the trough is constant. It's 2 square feet because the top is 2 feet across (get it? "across"—that's a pun!).
Now, the rate at which the water level rises depends on the flow rate. According to the problem, the water is flowing into the trough at a rate of 2 cubic feet per minute. But since we know the cross-sectional area is 2 square feet, we can figure out how fast the surface rises!
If the trough is 4 feet deep and the water is rising at a rate of 2 cubic feet per minute, that means the water is rising 0.5 feet per minute (2 divided by 4 gives us 0.5).
Therefore, when the water is 1 foot deep, it will be rising at a rate of 0.5 feet per minute. So, the surface is rising at a speed of 0.5 feet per minute when the water is 1 foot deep!
I hope my watery humor didn't make you too seasick!