a conical tank is 15 feet deep and has an open top whose radius is 15 feet. Assume that starting at t = 0 water is added to the tank at a rate of pi ft^3/hr, and water evaporates from the tank at a rate proportional to the suface area with the constant of proportionality being 0.01. The tank is assumed to be empty at time t = 0. Let V and h represent the volume and depth of water in the tank.
solve the equation dh/dt = (1-0.01h^2)/h^2
Use separation of variables.
Integrate
dt = h^2/(1-0.01h^2) dh
with h = 0 at t = 0
To solve the given differential equation dh/dt = (1-0.01h^2)/h^2, we will use the method of separation of variables.
Step 1: Rearrange the equation:
dh/h^2 = (1 - 0.01h^2) dt
Step 2: Separate the variables by multiplying both sides by h^2:
dh = (1 - 0.01h^2) dt/h^2
Step 3: Integrate both sides with respect to their respective variables:
∫(1/h^2) dh = ∫(1 - 0.01h^2) dt/h^2
Step 4: Evaluate the integrals:
-1/h = ∫(1 - 0.01h^2)/h^2 dt
To integrate the right side, we can use u-substitution. Let u = h^2, then du = 2h dh.
After the substitution, the equation becomes:
-1/h = ∫(1 - 0.01u)/u^2 (1/2h) du
Simplifying further:
-1/h = (1/2) ∫((1 - 0.01u)/u^2) du
Step 5: Evaluate the integral:
-1/h = (1/2) [∫(1/u^2) du - ∫(0.01u/u^2) du]
The first integral is the integral of the power rule, while the second integral is the integral of a constant multiplied by a function.
-1/h = (1/2) [-u^(-1) - 0.01∫(u/u^2) du]
-1/h = (1/2) [-u^(-1) - 0.01∫(1/u) du]
-1/h = (1/2) [-u^(-1) - 0.01ln|u| + C1]
Where C1 is the constant of integration.
Step 6: Substitute back u = h^2:
-1/h = (1/2) [-h^(-2) - 0.01ln|h^2| + C1]
Step 7: Multiply both sides by 2h to isolate h on the left side:
-2 = -h^(-1) - 0.01ln|h^2| + C1h
Step 8: Rearrange the equation:
0.01ln|h^2| = -2h + C1h - 2
Step 9: Combine like terms and solve for the constant C1:
0.01ln|h^2| = (C1 - 2)h - 2
We'll assume C1 - 2 = C2, where C2 is another constant:
0.01ln|h^2| = C2h - 2
Step 10: Exponentiate both sides:
e^(0.01ln|h^2|) = e^(C2h - 2)
h^2 = e^(C2h - 2)
Taking the square root of both sides:
h = ±√(e^(C2h - 2))
Step 11: Solve for h using the initial condition provided:
Since the tank is empty at t = 0, h = 0 at t = 0.
Therefore, we substitute h = 0 and t = 0 into the equation:
0 = ±√(e^(C2*0 - 2))
We can see that one of the solutions will be h = 0.
Step 12: Find the other solution by solving the equation for h when h ≠ 0:
√(e^(C2h - 2)) = ±h
e^(C2h - 2) = h^2
Taking the natural logarithm of both sides:
C2h - 2 = ln(h^2)
C2h = ln(h^2) + 2
h = (ln(h^2) + 2) / C2
Note: The constant C2 cannot be determined without further information or constraints.
Therefore, the general solution for the differential equation is h(t) = (ln(h^2) + 2) / C2, where h represents the depth of water in the tank and t represents time.