The trough is in the shape of a triangular prism. It is 5 ft long and its vertical cross sections are isosceles triangles with base 2ft and height 3ft. 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 volumeof water in the trough.
To find the depth and volume of water in the trough at any time t, we can follow these steps:
1. Given that the trough is in the shape of a triangular prism, we know that its cross sections are isosceles triangles with a base of 2 ft and a height of 3 ft.
2. The length of the trough is 5 ft, so the volume V of the trough can be calculated by multiplying the area A of the triangular cross-section by the length of the trough. The formula for the area of an isosceles triangle is (base * height) / 2. So, the volume formula becomes V = A * 5.
3. We are told that water is being siphoned out of the trough at a rate of 2 cubic feet per minute. This means that the volume of water V is decreasing at a rate of -2 ft^3/min. This can be written as dV/dt = -2.
4. Now, let's find the depth h of the trough at any time t. Since the cross section of the trough is an isosceles triangle, we know that the depth h is equal to the height of this triangle. So, h = 3 ft.
5. We need to find the rate dh/dt at which the depth is changing with respect to time. To do this, differentiate both sides of the equation h = 3 with respect to t. We get dh/dt = 0 since the height of the triangle is constant.
6. Finally, we know that dh/dt = dV/dt / A, where A is the area of the cross section. Substituting the known values, we get 0 = -2 / A. Rearranging the equation, we find A = -2 / 0 = undefined.
Therefore, there seems to be an error in the given information. Please double-check the given data for accuracy.