assuming the tank is pointed-end down, the water forms a cone, so
v = π/3 r^2 h
Now, using similar triangles, (r/h) = 9/16, so
v = (π/3)(9/16 h)^2 h = 27π/256 h^3
h = 3/8 ∛(2πv)
since this is precal, not calculus, I'm not sure what tools you have available to convert that to a function of time
A conical tank of radius R=19 feet and height of H=16 feet is being filled with water at a rate of 9ft 3 /min .
(a) Express the height h of the water in the tank, in feet, as a function of time t in minutes
2 answers
Is the point at the bottom or the top?
whichever way you draw it
Volume of a cone = (1/3)pi R^2 x
where r is the radius of the surface and x is distance from the tip
r/x = 9.5/ 16 = .594
so
r = .594 x
v = (1/3) pi (.594x)^2 x
v = .369 x^3
dv/dx = 1.11 x^2
dv/dt = 1.11 x^2 dx/dt
but dv/dt = 9 ft^3/min
x^2 dx/dt = 9/1.11 = 8.11
x^2 dx = 8.11 dt
(1/3) x^3 = 8.11 t
x^3 = 24.3 t
x = (24.3 t)^(1/3)
whichever way you draw it
Volume of a cone = (1/3)pi R^2 x
where r is the radius of the surface and x is distance from the tip
r/x = 9.5/ 16 = .594
so
r = .594 x
v = (1/3) pi (.594x)^2 x
v = .369 x^3
dv/dx = 1.11 x^2
dv/dt = 1.11 x^2 dx/dt
but dv/dt = 9 ft^3/min
x^2 dx/dt = 9/1.11 = 8.11
x^2 dx = 8.11 dt
(1/3) x^3 = 8.11 t
x^3 = 24.3 t
x = (24.3 t)^(1/3)
2 answers