A trough is 15 ft long and 4 ft across the top, as shown in the figure. Its ends are isosceles triangles with height 3 ft. Water runs into the trough at the rate of 2.5 ft^3/min. How fast is the water level rising when it is 2 ft deep?

the width of the water surface when the depth is y is 4y/3

So, the cross-section at depth y has area
1/2 * y * 4y/3 = 2y^2/3

So the volume of water when the depth is y is

v = 10y^2
dv/dt = 20y dy/dt
at y=2,
5/2 = 40 dy/dt
dy/dt = 1/16 ft/min

Well, well, well, looks like we've got ourselves a watery problem to solve! Let's jump right into it, shall we?

First things first, we need to find the volume of the water in the trough. The trough is like a rectangular prism with two isosceles triangle ends, so we'll need to break it down.

The rectangular part of the trough has a length of 15 ft and a width of 4 ft, so its volume is given by V = length x width x height, which is 15 ft x 4 ft x 2 ft (since the water is 2 ft deep). So, the volume of the rectangular part is 120 ft^3.

Now, let's tackle those isosceles triangle ends. Each end has a height of 3 ft and a base of 4 ft (remember, the trough is 4 ft wide). The formula for the area of an isosceles triangle is A = (base x height) / 2. So, the area of one triangle end is (4 ft x 3 ft) / 2, which is 6 ft^2. Since we have two triangle ends, the total area is 2 x 6 ft^2, which is 12 ft^2.

Now, the volume of one triangle end is given by V = base x height x depth, which is 4 ft x 3 ft x 2 ft (since the water is 2 ft deep). So, the volume of one triangle end is 24 ft^3. And since we have two triangle ends, the total volume of the triangles is 2 x 24 ft^3, which is 48 ft^3.

Adding the volume of the rectangular part and the volume of the triangles, we get a total volume of 120 ft^3 + 48 ft^3, which is 168 ft^3.

Now we can finally find the rate at which the water level is rising. We know that water is flowing in at a rate of 2.5 ft^3/min. So, the rate at which the water level is rising is 2.5 ft^3/min / 168 ft^3, which is approximately 0.0149 ft/min.

So, the water level is rising at a rate of about 0.0149 ft/min when it is 2 ft deep. And that's the answer, my friend!

I hope I didn't dive too deep into the math ocean there. Stay afloat and keep smiling!

To find the rate at which the water level is rising when it is 2 ft deep, we need to use similar triangles.

Let's label the height of the trough as y, and the width as x. The volume of the trough can be found using the formula: Volume = (1/2) * width * height * length.

Given that the trough is 15 ft long and 4 ft across the top, we have:
Length = 15 ft
Width (base of the triangle) = 4 ft
Height (height of the triangle) = 3 ft

The volume of the trough is given by:
Volume = (1/2) * 4 ft * 3 ft * 15 ft
Volume = 90 ft^3

Now, we can use the formula for the volume of the trough to find the rate at which the water level is rising.
Volume = x * y * 15 = 90 ft^3

We differentiate both sides of the equation with respect to time t:
d/dt (x * y * 15) = d/dt (90)

Next, we use the product rule of differentiation to differentiate the left side of the equation:
(d/dt (x)) * y * 15 + x * (d/dt (y)) * 15 = 0

We also know that:
(d/dt (x)) = 0 ft/min (since the width of the trough is constant)
(d/dt (y)) = dy/dt ft/min (the rate at which the water level is changing)

Plugging in these values, we get:
0 * y * 15 + x * (dy/dt) * 15 = 0

Simplifying the equation, we have:
15x(dy/dt) = 0

Since x = 4 ft and dy/dt is the rate at which the water level is rising, we can solve for dy/dt:
15 * 4 * (dy/dt) = 0
60(dy/dt) = 0
dy/dt = 0/60
dy/dt = 0 ft/min

Therefore, the water level is not rising when it is 2 ft deep.

To find the speed at which the water level is rising, we need to use related rates and apply the formula for the volume of a trough.

The volume of the trough can be calculated by considering it as a rectangular solid combined with two identical isosceles triangles at each end.

The volume, V, of the trough can be expressed as follows:
V = (length of the trough) * (width of the trough) * (height of the trough)

We are given the following measurements:
Length of the trough (L) = 15 ft
Width of the trough (W) = 4 ft
Height of the trough (H) = 2 ft (since we want to find the rate when it is 2 ft deep)

The rate at which water is flowing into the trough (dV/dt) is given as 2.5 ft³/min.

To find the rate at which the water level is rising, we need to find dH/dt (the rate at which the height is changing) when H = 2 ft.

We can rewrite the formula for volume as:
V = LW*H

Differentiating both sides of the equation with respect to time t:
dV/dt = d(LW*H)/dt
dV/dt = (dL/dt)(W*H) + L(dW/dt)(H) + L(W*dH/dt)

Since the trough's length and width are not changing, dL/dt and dW/dt are both zero.

Substituting the known values, we get:
2.5 ft³/min = 0 + 15(0)(2) + 15(4)(dH/dt)
2.5 ft³/min = 0 + 0 + 60(dH/dt)
2.5 ft³/min = 60(dH/dt)

To solve for dH/dt, divide both sides by 60:
dH/dt = 2.5 ft³/min ÷ 60
dH/dt = 0.0417 ft/min

Therefore, the water level is rising at a rate of 0.0417 ft/min when it is 2 ft deep.