Question

A pyramid-shaped vat has square cross-section and stands on its tip. The dimensions at the top are 2 m × 2 m, and the depth is 5 m. If water is flowing into the vat at 3 m3 /min, how fast is the water level rising when the depth of water (at the deepest point) is 4 m? Note: the volume of any "conical" shape (including pyramids) is (1/3)(height)(area of base).

Answer 1

the water level increases at a rate of 1.718 m/min when the depth is 4 m

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