8. The ends of a horizontal water trough 10 ft . long are isosceles trapezoids with lower base 3 ft ., upper base 5 ft ., and altitude 2 ft . If the water level is rising at a rate of \( .25 \mathrm{in} / \mathrm{min} \) when the depth of the water is 1 ft , water entering the trough?

To solve this problem, we need to find the rate at which water is entering the trough when the water level is rising at a rate of \(0.25 \, \text{in/min}\) and the depth of the water is \(1 \, \text{ft}\). ### Step 1: Understand the Geometry of the Trough The ends of the trough are isosceles trapezoids. The dimensions given are: - Lower base \(b_1 = 3 \, \text{ft}\) - Upper base \(b_2 = 5 \, \text{ft}\) - Altitude (height) \(h = 2 \, \text{ft}\) ### Step 2: Calculate the Area of the Trapezoid The area \(A\) of a trapezoid can be calculated using the formula: \[ A = \frac{1}{2} (b_1 + b_2) h \] Substituting the values: \[ A = \frac{1}{2} (3 + 5) \cdot 2 = \frac{1}{2} \cdot 8 \cdot 2 = 8 \, \text{ft}^2 \] ### Step 3: Determine the Volume of Water in the Trough The volume \(V\) of water in the trough when the depth of the water is \(d\) can be expressed as: \[ V = A \cdot L \] where \(L\) is the length of the trough. Given \(L = 10 \, \text{ft}\): \[ V = A \cdot 10 = 8 \cdot 10 = 80 \, \text{ft}^3 \] ### Step 4: Relate the Volume to the Depth of Water The area of the trapezoid changes as the water level rises. We need to find the area of the trapezoid when the water depth is \(d = 1 \, \text{ft}\). The area of the trapezoid at a depth \(d\) can be calculated using similar triangles. The height of the trapezoid is \(2 \, \text{ft}\), and the bases are \(3 \, \text{ft}\) and \(5 \, \text{ft}\). The width of the trapezoid at height \(d\) can be found using the ratio of the heights. The width of the trapezoid at height \(d\) is: \[ b_d = b_1 + \left( \frac{b_2 - b_1}{h} \cdot d \right) = 3 + \left( \frac{5 - 3}{2} \cdot d \right) = 3 + d \] ### Step 5: Calculate the Area at \(d = 1 \, \text{ft}\) Substituting \(d = 1\): \[ b_d = 3 + 1 = 4 \, \text{ft} \] Now, the area \(A_d\) at \(d = 1 \, \text{ft}\) is: \[ A_d = \frac{1}{2} (b_1 + b_d) d = \frac{1}{2} (3 + 4) \cdot 1 = \frac{1}{2} \cdot 7 = 3.5 \, \text{ft}^2 \] ### Step 6: Calculate the Volume of Water at \(d = 1 \, \text{ft}\) The volume \(V_d\) of water when the depth is \(1 \, \text{ft}\): \[ V_d = A_d \cdot L = 3.5 \cdot 10 = 35 \, \text{ft}^3 \] ### Step 7: Find the Rate of Change of Volume The rate of change of volume with respect to time can be found using the relationship: \[ \frac{dV}{dt} = A_d \cdot \frac{dh}{dt} \] where \(\frac{dh}{dt} = 0.25 \, \text{in/min} = \frac{0.25}{12} \, \text{ft/min}\). Now substituting the values: \[ \frac{dV}{dt} = 3.5 \cdot \frac{0.25}{12} \] ### Step 8: Calculate the Rate of Water Entering the Trough Now we can calculate: \[ \frac{dV}{dt} = 3.5 \cdot \frac{0.25}{12} = \frac{3.5 \cdot 0.25}{12} = \frac{0.875}{12} \approx 0.0729 \, \text{ft}^3/\text{min} \] ### Conclusion The rate at which water is entering the trough when the depth of the water is \(1 \, \text{ft}\) is approximately \(0.0729 \, \text{ft}^3/\text{min}\).