A conical water tank with vertex down has a radius of 12 feet at the top and is 28 feet high. If water flows into the tank at a rate of 30 ft^3/min, how fast is the depth of the water increasing when the water is 16 feet deep?
Make a sketch of a cross-section of the cone containing some water.
Let the radius of the circular water level be r ft, let the height of the water be h ft.
by similar triangles,
r/h = 12/28
28r = 12h
r = 3h/7
V = (1/3)πr^2 h
= (1/3)π(9h^2/49)h = (3/49)π h^3
dV/dt = (9/49)π h^2 dh/dt
fill in our data ...
30 = (9/49)π (16^2) dh/dt
30 =(2304/49 π) dh/dt
leaving it up to you to solve for dh/dt in ft/min
let me know what your answer is, will be here for another 20 min
I got .2031
correct , that is what I had
To find the rate at which the depth of the water is increasing, we need to differentiate the equation that relates the volume of water in the tank with respect to time.
The volume of a cone can be calculated using the formula V = (1/3)πr^2h, where V is the volume, r is the radius of the base of the cone, and h is the height of the cone.
In this case, the radius of the top of the cone is given as 12 feet. Since the cone is symmetrical, the radius at any given height (h) can be calculated using the similar triangles property. We can use the formula r = (R/H) * h, where R is the radius at the top of the cone, H is the total height of the cone, and h is the height from the vertex.
Now, we can differentiate the volume equation with respect to time (t) to find the rate at which the volume of water is changing:
dV/dt = (1/3)π[(2rh)(dh/dt) + (r^2)(dh/dt)]
Simplifying and substituting the given values:
dV/dt = (1/3)π[(2(12)(h)(dh/dt)) + ((h^2/R^2)(dh/dt))]
The rate at which the depth of the water is increasing, dh/dt, is given as 30 ft^3/min. We need to find the value of dh/dt when the water is 16 feet deep, h = 16.
Substituting the given values into the equation:
30 = (1/3)π[(2(12)(16)(dh/dt)) + ((16^2/12^2)(dh/dt))]
Now, solve for dh/dt to find the rate at which the depth of the water is increasing.