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

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)

Well, this is certainly a conical situation, isn't it? Let's dive right into it!

To express the height h of the water in the tank as a function of time t, we need to use the formula for the volume of a cone, which is V = (1/3) * π * r^2 * h.

In this case, the radius r is constant at R = 19 feet and the rate at which the water is being filled is 9 ft^3/min. So, we can write the equation as:

9 = (1/3) * π * 19^2 * h.

Now, we can solve this equation for h:

h = 9 / [(1/3) * π * 19^2].

Simplifying further, we get:

h = 3 * 9 / (π * 19^2).

And voila! There's your function for the height of the water in the tank as a function of time. Now, grab a towel and be prepared for some wet and wild times!

To express the height h of the water in the tank as a function of time t in minutes, we need to use the formula for the volume of a cone:

V = (1/3) * π * r^2 * h

In this case, the radius r is given as R = 19 feet. We can express the height h as h = f(t), where f(t) is the function that represents the height of the water in terms of time.

To find the rate of change of the volume with respect to time, we differentiate both sides of the equation with respect to t:

dV/dt = d/dt[(1/3) * π * r^2 * h]

The only term that depends on time t is h, so we can treat r, the radius, as a constant.

dV/dt = (1/3) * π * r^2 * (dh/dt)

The rate of change of the volume with respect to time, dV/dt, is given as 9 ft^3/min. We can substitute the values and simplify the equation:

9 = (1/3) * π * (19^2) * (dh/dt)

Now we can solve for dh/dt, the rate at which the height of water is changing with respect to time:

dh/dt = (9 * 3) / (π * 19^2)

Simplifying this further:

dh/dt ≈ 0.0797 ft/min

This means that the height of the water in the tank is increasing at a rate of approximately 0.0797 feet per minute.

To express the height h of the water in the tank as a function of time t, we need to understand the relationship between the volume of water in the tank and the corresponding height.

We know that the volume of a cone can be calculated using the formula:

V = (1/3) * π * r^2 * h

where V is the volume, π is a mathematical constant approximately equal to 3.14159, r is the radius of the base of the cone, and h is the height.

In this case, the radius (r) is given as 19 feet, and we want to express the height (h) as a function of time (t).

Given that the tank is being filled with water at a rate of 9 ft^3/min, we can say that the rate of change of volume with respect to time is constant:

dV/dt = 9 ft^3/min

Now, let's differentiate the volume formula with respect to time:

dV/dt = (d/dt)[(1/3) * π * r^2 * h]

Since the radius (r) is constant, we can bring it outside the derivative:

dV/dt = (1/3) * π * r^2 * (dh/dt)

Remember that dh/dt represents the rate of change of height with respect to time.

We can rearrange the equation to solve for dh/dt:

(1/3) * π * r^2 * (dh/dt) = 9 ft^3/min

(1/3) * (19 ft)^2 * (dh/dt) = 9 ft^3/min

Now, we can solve for dh/dt:

(1/3) * (361 ft^2) * (dh/dt) = 9 ft^3/min

(361/3) * (dh/dt) = 9 ft^3/min

(120.33) * (dh/dt) = 9 ft^3/min

dh/dt = 9 ft^3/min / 120.33

dh/dt = 0.0747 ft/min

Now, we have the rate of change of height with respect to time. We can integrate this equation with respect to time to find the function h(t):

∫(dh) = ∫(0.0747 ft/min) dt

h = 0.0747 ft/min * t + C

where C is the constant of integration. Since we want to express the height at time t, we can set the initial condition h(0) = 0:

0 = 0.0747 ft/min * 0 + C

C = 0

Therefore, the height h of the water in the tank as a function of time t in minutes is:

h(t) = 0.0747 ft/min * t

Hence, the height h of the water in the tank is directly proportional to time t.

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