Water is being pumped into an inverted conical tank at a constant rate. The tank has height 9 meters and the diameter at the top is 4 meters. If the water level is rising at a rate of 28 centimeters per minute when the height of the water is 3.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

I got it down to pi/4 * (4/9)^2 *(350)^2 * 28 = 532131.4956 but it's not right. Any suggestions?

let the height of the water be h cm

let the radius of the water level be r cm
by similar triangles, r/h = 2/9 , r = 2h/9

V = (1/3)pi(r^2)h
= (1/3)pi(4/81)(h^2)h = (4pi/243)h^3
dV/dt = (4pi/81)(h^2)dh/dt
= (4pi/81)(350)(28) = 1520.375 cm^3/min

check my calculations

I did it again!

forgot to square the 350 in the last line

(4pi/81)(350)^2(28) = 532131.4964 cm^3/min

which makes it exactly the same as your answer!

MMMHHHH ?

What the heck? haha.

I was going crazy could I could've swore I did my calculations right. Oh well, thanks for the help man. I appreciate it.

To find the rate at which water is being pumped into the tank in cubic centimeters per minute, we need to use the concept of similar triangles and related rates.

Let's start by considering the two similar triangles that are formed by the water level in the tank.

First, we have a triangle formed by the water, with a height of 3.5 meters and the corresponding base at the top of the water, which is the diameter of the tank, 4 meters.

Second, we have a smaller triangle formed by the water, with a height of x meters (unknown) and the corresponding base at the top of the water, which is the radius of the tank, 2 meters.

Since these two triangles are similar, we can set up a proportion to relate the heights of the triangles:

3.5 meters / 4 meters = x meters / 2 meters

Simplifying this proportion, we get:

x = (3.5 meters / 4 meters) * 2 meters
x = 1.75 meters

Therefore, the height of the water in the small triangle is 1.75 meters.

Now, let's focus on the rates:

We are given that the water level is rising at a rate of 28 centimeters per minute when the height of the water is 3.5 meters. So, we have:

(dh/dt) = 28 cm/min
h = 3.5 meters

We want to find the rate at which water is being pumped into the tank (dV/dt) in cubic centimeters per minute.

To relate the rates of change, we can use the formula for the volume of a cone:

V = (1/3) * π * r^2 * h

Taking the derivative with respect to time (t), we get:

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

We know that the diameter at the top of the tank is 4 meters, so the radius (r) is 2 meters.

Substituting the known values into the equation:

(dV/dt) = (1/3) * π * (2 meters) * (dr/dt) * (1.75 meters) + (1/3) * π * (2 meters)^2 * (28 cm/min)

To convert the rate of change of radius from meters per minute to centimeters per minute, we need to multiply by 100:

(dr/dt) = (28 cm/min) * (100 cm/m) = 2800 cm/min

Substituting this value into the equation:

(dV/dt) = (1/3) * π * (2 meters) * (2800 cm/min) * (1.75 meters) + (1/3) * π * (2 meters)^2 * (28 cm/min)

Evaluating this expression will give you the rate at which water is being pumped into the tank in cubic centimeters per minute.