Answer:
the water level increases at a rate of 1.718 m/min when the depth is 4 m
Step-by-step explanation:
the volume of the pyramid is
V = (1/3)(height)(area of base) = 1/3 H*L²
For the diagonal in the pyramid
tg Ф = Side Length/ Height = L / H = x / h
where h= depth of water , x= side of the corresponding cross section
therefore x= L *h/H
the volume of the water is
v= 1/3 h*x² = 1/3 (L/H)² h³
in terms of time
v = Q*t
then
Q*t = 1/3 (L/H)² h³
h³ = 3*(H/L)² *Q *t
h = ∛(3*(H/L)² *Q *t) = ∛(3*(H/L)² *Q) *∛t = k* ∛t , where k=∛(3*(H/L)² *Q)
h = k* ∛t
then the rate of increase in depth is dh/dt
dh/dt = 1/3*k* t^(-2/3)
since
t = (h/k)³
dh/dt = 1/3*k* t^(-2/3) = 1/3*k* (h/k)³ ^(-2/3) = 1/3*k* (h/k)^(-2) = 1/3 k³ / h²
= 1/3 (3*(H/L)²*Q) / h² = (H/L)²*Q /h²
dh/dt= [H/(h*L)]²*Q
replacing values, when h=4m
dh/dt= [H/(h*L)]²*Q = [5m/(4m*2m)]² * (3m³/min)= 1.718 m/min
