Answer:
a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔
b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]
The small deviations from equilibrium yield similar solutions, indicating stable equilibrium.
c. πa² ≥ k/∝
Step-by-step breakdown:
a.
The volume of water flow in the pond is determined by
the inflow rate minus the evaporation rate.
Given
k = Water inflow rate
The pond's surface, where evaporation happens, influences this equation.
The area of the circle is πr², with ∝ as the evaporation coefficient.
The volume's rate with respect to time = k - ∝πr²
dV/dt = k - ∝πr² ----- equation 1
The conical pond's volume is calculated using πr²L/3
Where L = the height of the cone
L = hr/a where h is the height of the water
Thus, V = πr²(hr/a)/3
V = πr³h/3a ------ Rearranging for r gives
3aV = πr³h
r³ = 3aV/πh
r = ∛(3aV/πh)
Substituting ∛(3aV/πh) for r in equation 1 yields
dV/dt = k - ∝π(∛(3aV/πh))²
dV/dt = k - ∝π((3aV/πh)^⅓)²
dV/dt = K - ∝π(3a/πh)^⅔V^⅔
dV/dt = K - ∝π(3a/πh)^⅔V^⅔
b. Water's equilibrium depth is when the differential equation equals 0
i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0
k - ∝π(3a/πh)^⅔V^⅔ = 0
∝π(3a/πh)^⅔V^⅔ = k ------ solving for V results in
V^⅔ = k/∝π(3a/πh)^⅔ -------- extracting 3/2 root from both sides yields
V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2
Thus, V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅓)]^3/2
Final expression is V = (hk^3/2)/[(∝^3/2.π^½.(3a))]
Following small deviations from equilibrium yields comparable results, hence equilibrium is stable.
c. Condition for avoiding overflow
If we continue adding water when the inflow rate is zero, the pond will overflow.
This implies dV/dt = k - ∝πr², where r = a and the rate is now ≤ 0.
Thus, we arrive at
k - ∝πa² ≤ 0 ---- rearranging gives
- ∝πa² ≤ -k dividing each side by - ∝ yields
πa² ≥ k/∝
