Water is being pumped into an inverted right circular conical tank at the rate 30 ft^3/min. the tank, which stands vertex down and base up, has a height of 12 ft and a base diameter of 16 ft. how fast is the water level rising when the water is 9 ft deep?

Question

Water is being pumped into an inverted right circular conical tank at the rate 30 ft^3/min. the tank, which stands vertex down and base up, has a height of 12 ft and a base diameter of 16 ft. how fast is the water level rising when the water is 9 ft deep?

Answer 1

at depth h, h = 3/2 r

v = 1/3 pi r^2 h
v = 1/3 pi r^2 (3r/2)
v = pi/2 r^3

dv/dt = 3pi/2 r^2 dr/dt

30 = 3pi/2 * 36 dr/dt
dr/dt = 5/(9pi)

Answer 2

thanks steve

Answer 3

To find the rate at which the water level is rising, we need to use related rates. Let's call the height of the water in the conical tank "h" (in feet) and the radius of the water level "r" (in feet).

We are given that the water is being pumped into the tank at a rate of 30 ft^3/min, which means the volume is increasing at a constant rate of 30 ft^3/min.

The volume V of a cone is given by the formula: V = (1/3) * π * r^2 * h.

We have the dimensions of the conical tank: height (h) = 12 ft and base diameter (d) = 16 ft. From the base diameter, we can find the radius using r = d/2 = 16/2 = 8 ft.

Now, let's differentiate the volume equation with respect to time (t) to find an equation relating the rates of change:

dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt).

Now, we need to find expressions for dr/dt and dh/dt.

Because the water level is rising, we are interested in finding dh/dt.

We know that when the water is 9 ft deep, h = 9 ft. We are required to find how fast the height of the water is changing, so we need to find dh/dt.

To find dh/dt, we can use similar triangles. The ratio of the change in height (dh) to the change in radius (dr) is constant throughout the conical tank:

dh/dr = h/r.

Rearranging the equation, we get dh = (h/r) * dr.

Now, we can substitute dh and h into the volume equation and solve for dh/dt:

dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt),

30 ft^3/min = (1/3) * π * (2 * 8 * dr/dt * 9 + 8^2 * (9/8) * dr/dt),

30 = (1/3) * π * (18dr/dt + 9dr/dt),

90 = π * 27dr/dt,

dr/dt = 90 / (π * 27) = 10 / (3π) ft/min.

So, the water level is rising at a rate of (10 / (3π)) ft/min when the water is 9 ft deep.