Water is running into an open conical tank at the rate of 9 cubic feet per minute. The tank is standing inverted, and has a height of 10 feet and a base diameter of 10 feet. At what rate is the radius of the water in the tank increasing when the radius is 2 feet?
draw a diagram. We have at a depth of h feet, a water surface with a diameter of h feet. So, h=2r
v = 1/3 π r^2 h = 2π/3 r^3
dv/dt = 2πr^2 dr/dt
so we have
9 = (2π)(4) dr/dt
Now you have dr/dt, the rate that the water is rising.
I mean, the rate at which the radius is increasing.
Oh, the ol' inverted conical tank problem! Time to put on my "math clown" nose. So, let's break it down. We're looking for the rate at which the radius of the water in the tank is increasing, given certain conditions.
We know the rate at which water is flowing in (9 cubic feet per minute), and we need to find the rate at which the radius is increasing. We also know the height of the tank (10 feet) and the base diameter (10 feet).
To solve this, we'll use a combination of geometry and calculus. The volume of a cone is given by the formula V = (1/3)πr²h, where V is volume, r is the radius of the base, and h is the height.
Differentiating both sides of this equation with respect to time (t) gives us dV/dt = (2/3)πrh(dr/dt) + (1/3)πr²(dh/dt). Let's break it down step by step.
We are given dV/dt (the rate of water flowing in), which is 9 cubic feet per minute. We are also given h (height) as 10 feet. We need to solve for dr/dt (rate at which the radius is increasing) when r (radius) is 2 feet.
Plugging in the known values, we have 9 = (2/3)π(2)(10)(dr/dt) + (1/3)π(2²)(dh/dt).
Now, we need to solve for dr/dt. Manipulating the equation further, we have 9 = (4/3)π(20)(dr/dt) + (4/3)π(dh/dt).
Simplifying, we have 9 = (80/3)π(dr/dt) + (4/3)π(dh/dt).
To isolate dr/dt, we subtract (4/3)π(dh/dt) from both sides, giving us 9 - (4/3)π(dh/dt) = (80/3)π(dr/dt).
Finally, we divide both sides by (80/3)π to solve for dr/dt, yielding (dr/dt) = [9 - (4/3)π(dh/dt)] / [(80/3)π].
Put those numbers in, do the math, and you'll get your answer. Just remember, if you ever need a laugh while crunching numbers, just summon the Clown Bot!
We can start by finding the formula for the volume of a cone. The volume of a cone is given by the formula:
V = (1/3)πr^2h
Where V is the volume, r is the radius, and h is the height.
In this problem, the height of the tank remains constant at 10 feet, so the volume of water in the tank can be represented as a function of the radius, V(r), by substituting the given height into the volume formula:
V(r) = (1/3)πr^2(10)
Now, we need to determine the rate at which the volume is changing with respect to time, dV/dt. Given that the water is entering the tank at a constant rate of 9 cubic feet per minute, we have:
dV/dt = 9
To find the rate at which the radius is changing when the radius is 2 feet, we need to find dr/dt. Applying the Chain Rule to the formula for the volume, we have:
dV/dt = dV/dr * dr/dt
From the volume formula, we can differentiate both sides with respect to r:
dV/dr = (1/3)π * 2r * 10
= (20/3)πr
Now we can substitute the known values into the equation:
9 = (20/3)π * 2 * dr/dt
Simplifying, we have:
9 = (40/3)π * dr/dt
Dividing both sides by (40/3)π, we get:
9 / ((40/3)π) = dr/dt
To calculate the value of dr/dt, we evaluate the expression on the left side of the equation:
dr/dt = 9 / ((40/3)π)
= 27π / (40π)
= 27/40
Therefore, when the radius is 2 feet, the rate at which the radius of the water in the tank is increasing is 27/40 feet per minute.
To find the rate at which the radius of the water in the tank is increasing, we need to use related rates.
Let's start by visualizing the problem. We have a cone-shaped tank that is filling with water at a constant rate, and we want to find how quickly the radius of the water is changing when the radius is 2 feet.
Step 1: Set up the equation
The volume of a cone can be calculated using the formula V = (1/3)πr^2h, where V represents the volume, r is the radius, and h is the height.
In this case, the volume of water is increasing at a rate of 9 cubic feet per minute, so dV/dt = 9.
Step 2: Express the variables in terms of time
We are interested in finding how quickly the radius is changing with respect to time, so we need to express r in terms of t (time).
Let's assume that the radius of the water at any given time is represented by r. We want to find dr/dt, the rate at which r is changing with respect to time.
Step 3: Express the height in terms of the radius
We are given that the tank has a height of 10 feet and a base diameter of 10 feet. Since the base diameter is equal to twice the radius, we can conclude that the height is equal to twice the radius. Therefore, h = 2r.
Step 4: Find an equation relating V, r, and t
Substitute the expression for h into the equation for volume V, resulting in V = (1/3)πr^2(2r) = (2/3)πr^3.
Step 5: Differentiate the equation with respect to time
Differentiate both sides of the equation with respect to time (t) using the chain rule:
dV/dt = d((2/3)πr^3)/dt
Simplifying and substituting dV/dt = 9, we get:
9 = (2/3)πr^2(dr/dt)
Step 6: Solve for dr/dt
Rearrange the equation to solve for dr/dt:
(2/3)πr^2(dr/dt) = 9
Multiply both sides by 3/(2πr^2):
dr/dt = 27/(2πr^2)
Step 7: Substitute the radius
We are interested in finding the rate at which the radius is changing when r = 2. Substitute r = 2 into the equation to get:
dr/dt = 27/(2π(2^2))
Simplifying further:
dr/dt = 27/(8π)
So, when the radius is 2 feet, the rate at which the radius of the water in the tank is increasing is 27/(8π) feet per minute.