A trough is 15 ft long and 4 ft across the top. Its ends are isosceles triangles with height 3 ft. Water runs into the trough at the rate of 2.5 ft3/min. How fast is the water level rising when it is 0.89 ft deep? Give your answer correct to 3 decimal places.
when the water has depth y, the surface has with 4/3 y. That means the volume is
v = y(4/3 y)/2 * 15 = 10y^2
dv/dt = 20y dy/dt
Now just plug in your numbers to find dy/dt
To answer this question, we can use related rates, which involves finding the rate of change of one variable with respect to another variable. In this case, we want to find how fast the water level is rising (rate of change of depth) with respect to time.
Let's assign variables to the given quantities:
- Length of the trough: L = 15 ft
- Width of the trough: W = 4 ft
- Height of the triangular ends: h = 3 ft
- Rate of water flow: V = 2.5 ft^3/min
- Depth of the water: d (which varies with time)
- Rate of change of depth with respect to time: dd/dt (what we want to find)
We can use the formula for the volume of the trough to relate the variables:
Volume = Length × Width × Depth
V = L × W × d
Now, differentiate both sides of the equation with respect to time (t):
dV/dt = d(L × W × d)/dt
To evaluate d(L × W × d)/dt, we need to use the product rule:
d(L × W × d)/dt = L × W × (dd/dt) + dL/dt × W × d + L × dW/dt × d
However, we can simplify this equation because the trough's length (L) and width (W) are constant. Therefore, their rates of change with respect to time (dL/dt and dW/dt) are zero in this particular scenario. This simplifies the equation to:
dV/dt = L × W × (dd/dt)
Now, we can substitute the given values into the equation:
2.5 = 15 × 4 × (dd/dt)
Solve for dd/dt:
dd/dt = 2.5 / (15 × 4)
Simplifying:
dd/dt = 2.5 / 60
Calculating:
dd/dt = 0.0417 ft/min
Therefore, the water level is rising at a rate of 0.0417 ft/min when the depth is 0.89 ft.