triangular prism trough with isoceles trapezoid volume

#Triangular prism trough with isoceles trapezoid volume full#

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

#Triangular prism trough with isoceles trapezoid volume full#

Therefore, the rate of change in h at the instant when the trough is ¼ full by volume is -4/3 feet per minute. Plugging in the values, we have -2 = (1/2) * 3 * dh/dt.ĭividing both sides by (3/2), we get -4/3 = dh/dt. 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: 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.ĭividing both sides by (3/2), we find h = 2.5 feet. Since the volume V is related to the height h, we can write the equation V = (1/2) * base area * h. We want to find the value of dh/dt when V = (1/4) * 15 = 3.75 cubic feet.

We are given that dV/dt = -2 (since water is being siphoned out at a rate of 2 cubic feet per minute). Let V represent the volume of water in cubic feet, and h represent the depth of the water in feet. 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. Therefore, the volume of water in the trough when it is full is also 15 cubic feet.ī. Since the trough is 5 feet long, the volume of each cross-section is 3 * 5 = 15 cubic feet. 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. 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. The formula for the volume of a prism is V = base area * height.