modelling of a parabolic tank

Question

A water tank is to be placed on the moon as part of an on-going space mission. It will be 10 metres long, with cross-section the shape of a downward-opening parabola 4 metres high and 4 metres wide at the base. i. How deep is the water in the tank when it is three-quarters full (by volume)?

Answer 1

If we can find the cross-sectional area, A, of the tank in terms of the depth, D, then, since the cross-section is implied to be constant along the length, L, we can calculate the volume, V=AL then backsolve for D when V=0.75V_max.

If we place a coordinate system (x,y) with the ordinate axis pointing down at the parabola's vertex, then when the tank is full x=2 and y=4.

The equation of the parabola is f(x)=cx² where c is a constant. We know that f(2)=4, so 4=c(2)² so c=1 and f(x)=x².

To find the total area of the parabola when the tank is full, A_f,

A_f = ∫f(x)dx = ∫x²dx = (1/3)x³ evaluated with from -2 to 2:

A_f = (1/3)[2³-(-2)³) = 8/3 m = 5.33 m²

So when the tank is 3/4 full, A = 0.75(5.33) = 4 m²

Now we seek the value of x coinciding with y= D when the tank is 3/4 full.

If b = the value of x on either side of the ordinate when the tank is 3/4 full, then

from the integration we know A = (1/3)x³ evaluated from -b to +b so

A = (1/3)[b³ -(-b)³]=(1/3)(2b³] = (2/3)b³

b³ = 3(A)/2 = 3(4)/2 = 6 m

b = 1.82 m

Finally, since y = x²,

D = (1.82 m)² = 3.3 m

Solution: Depth is 3.3 m (a big tank!).