Answer:
a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔
b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]
The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.
c. πa² ≥ k/∝
Step-by-step explanation:
a.
The rate of volume of water in the pond is calculated by
The rate of water entering - The rate of water leaving the pond.
Given
k = Rate of Water flows in
The surface of the pond and that's where evaporation occurs.
The area of a circle is πr² with ∝ as the coefficient of evaporation.
Rate of volume of water in pond with time = k - ∝πr²
dV/dt = k - ∝πr² ----- equation 1
The volume of the conical pond is calculated by πr²L/3
Where L = height of the cone
L = hr/a where h is the height of water in the pond
So, V = πr²(hr/a)/3
V = πr³h/3a ------ Make r the subject of formula
3aV = πr³h
r³ = 3aV/πh
r = ∛(3aV/πh)
Substitute ∛(3aV/πh) for r in equation 1
dV/dt = k - ∝π(∛(3aV/πh))²
dV/dt = k - ∝π((3aV/πh)^⅓)²
dV/dt = K - ∝π(3aV/πh)^⅔
dV/dt = K - ∝π(3a/πh)^⅔V^⅔
b. Equilibrium depth of water
The equilibrium depth of water is when the differential equation is 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 ------ make V the subject of formula
V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides
V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2
V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2
V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2
V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]
V = (hk^3/2)/[(∝^3/2.π^½.(3a))]
The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.
c. Condition that must be satisfied
If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.
i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.
So, we have
k - ∝πa² ≤ 0 ---- subtract k from both w
- ∝πa² ≤ -k divide both sides by - ∝
πa² ≥ k/∝
then the other numbers follow the pattern 
(see :
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