A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k

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.

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/∝