6. A trough is in the shape of a triangular prism. It is 5 feet long and its vertical cross sections are isosceles triangles with base 2 feet and height 3 feet. Water is being siphoned out of the trough at the rate of 2 cubic feet per minute. At any time t, let h be the depth and V be the volume of water in the trough.

a. Find the volume of the water in the trough when it is full.
b. What is the rate of change in h at the instant when the trough is ¼ full by volume?

a. To find the volume of water in the trough when it is full, we need to calculate the volume of the triangular prism. The formula for the volume of a prism is V = base area * height.

The base area of each triangular cross-section can be found using the formula for the area of a triangle: A = (1/2) * base * height.

Given that the base of each triangle is 2 feet and the height is 3 feet, we can calculate the base area: A = (1/2) * 2 * 3 = 3 square feet.

Since the trough is 5 feet long, the volume of each cross-section is 3 * 5 = 15 cubic feet.

Therefore, the volume of water in the trough when it is full is also 15 cubic feet.

b. To find the rate of change in h at the instant when the trough is ¼ full by volume, we can use the concept of related rates.

Let V represent the volume of water in cubic feet, and h represent the depth of the water in feet. We are given that dV/dt = -2 (since water is being siphoned out at a rate of 2 cubic feet per minute).

We want to find the value of dh/dt when V = (1/4) * 15 = 3.75 cubic feet.

Since the volume V is related to the height h, we can write the equation V = (1/2) * base area * h.

Using the formula for the base area A = (1/2) * 2 * 3 = 3 square feet, we can substitute into the equation: 3.75 = (1/2) * 3 * h.

Simplifying, we get 3.75 = (3/2) * h.

Dividing both sides by (3/2), we find h = 2.5 feet.

Now that we have both V and h, we can differentiate the equation V = (1/2) * A * h with respect to t (time) using implicit differentiation:

dV/dt = (1/2) * A * dh/dt.

Plugging in the values, we have -2 = (1/2) * 3 * dh/dt.

Simplifying, we find -2 = (3/2) * dh/dt.

Dividing both sides by (3/2), we get -4/3 = dh/dt.

Therefore, the rate of change in h at the instant when the trough is ¼ full by volume is -4/3 feet per minute.

To find the volume of water in the trough when it is full, we need to calculate the volume of the triangular prism.

The volume V of a triangular prism is given by the formula:

V = (1/2) * base * height * length

In this case, the base of the vertical cross-sections is 2 ft, the height is 3 ft, and the length of the trough is 5 ft.

So, substituting these values into the formula, we get:

V = (1/2) * 2 ft * 3 ft * 5 ft
V = 15 ft³

Therefore, the volume of water in the trough when it is full is 15 cubic feet.

Now, let's move on to part (b), where we need to find the rate of change in h (depth) at the instant when the trough is ¼ full by volume.

To solve this problem, we'll use the concept of related rates. We need to find the rate at which the depth is changing with respect to time.

Given that the water is being siphoned out at a rate of 2 cubic feet per minute, we can express the rate of change of volume as:

dV/dt = -2 ft³/min

We want to find dh/dt, the rate at which the depth is changing with respect to time.

To relate the volume V and the depth h, we need to take into account the shape of the trough, which is a triangular prism. The cross-sectional area of the trough at any given depth h can be calculated using similar triangles.

The cross-sectional area A can be expressed as:

A = (1/2) * base * height

In this case, the base is constant at 2 ft and the height can be expressed in terms of h as:

height = 3 - h

Therefore, the area A at any given depth is:

A = (1/2) * 2 ft * (3 - h) ft
A = (3 - h) ft²

Now, we can express the volume V in terms of the depth h as:

V = A * length
V = (3 - h) ft² * 5 ft
V = 15 ft² - 5h ft³

To find the rate of change of h with respect to time, we differentiate both sides of the equation with respect to time (t):

dV/dt = d/dt(15 ft² - 5h ft³)

Since dV/dt is given as -2 ft³/min, we can substitute the values:

-2 ft³/min = d/dt(15 ft² - 5h ft³)

Now, solving for dh/dt gives:

dh/dt = -2 ft³/min / (-5 ft³)
dh/dt = 2/5 ft/min

Therefore, the rate of change in h at the instant when the trough is ¼ full by volume is 2/5 ft/min.

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