A conical tank has a height that is always 3 times its radius. If water is leaving the tank at the rate of 50 cubic feet per minute, how fast if the water level falling in feet per minute when the water is 3 feet high? Volume of a cone is V=1/3(pi)r^2h
A. 1.000
B. 5.305
C. 15.915
D. .589
E. 1.768
I got A but I'm not quite sure. Thanks
How’d you get pi/27????
Well, let's just dive into the problem and see what we come up with!
First, we need to determine the value of the radius at the given water level. We know that the height is always three times the radius, and when the water is 3 feet high, we can say that the radius is 1 foot (since 1 foot multiplied by 3 gives us 3 feet).
Now, we need to find the rate at which the water level is falling. To do this, we'll need to differentiate the volume equation with respect to time. So let's take the derivative of the volume equation:
V = (1/3) * π * r^2 * h
To simplify things, we can also substitute the value of the height in terms of the radius:
h = 3r
Now, let's differentiate V with respect to t:
dV/dt = (1/3) * π * d(r^2)/dt * h + (1/3) * π * r^2 * dh/dt
The first term, d(r^2)/dt, is 2r * dr/dt (using the chain rule), and the second term, dh/dt, is just the rate at which the height is falling. Since we want to find the rate at which the water level is falling, we can use the given rate at which water is leaving the tank, which is -50 cubic feet per minute (the negative sign indicates that the volume is decreasing).
So let's plug in the values:
-50 = (1/3) * π * (r^2) * dh/dt + (1/3) * π * r^2 * 2r * dr/dt
Now let's evaluate this equation when the water is 3 feet high and the radius is 1 foot:
-50 = (1/3) * π * (1^2) * dh/dt + (1/3) * π * (1^2) * 2 * 1 * dr/dt
Simplifying:
-50 = (1/3) * π * dh/dt + (1/3) * π * 2 * dr/dt
-50 = (1/3) * π * dh/dt + (2/3) * π * dr/dt
Now we can solve for dh/dt, which represents the rate at which the water level is falling:
dh/dt = (-50 - (2/3) * π * dr/dt) / ((1/3) * π)
dh/dt = (-150/π) - (2 * dr/dt)
It seems like we made a mistake in our calculations as the equation above does not match the answer choices. We apologize for the confusion, but it seems that we can't provide a definitive answer based on the information given.
To find the rate at which the water level is falling, we need to find the rate at which the height of the water is changing with respect to time.
Let's denote the height of the cone as h and the radius as r. The problem states that the height is always 3 times the radius, so we have h = 3r.
We also know that the volume of a cone is given by V = (1/3)πr^2h.
Now, we can find an expression for the rate at which the volume is changing with respect to time (dV/dt) using the chain rule:
dV/dt = (dV/dh) * (dh/dt)
We are given dV/dt = -50 (cubic feet per minute), which represents the rate at which the water is leaving the tank.
Now, let's find dV/dh:
V = (1/3)πr^2h
Differentiating with respect to h:
dV/dh = (1/3)πr^2 * 1
Since h = 3r, we can express r in terms of h: r = h/3.
Substituting this into the expression for dV/dh, we get:
dV/dh = (1/3)π(h/3)^2
dV/dh = (1/3)π(h^2/9)
dV/dh = (1/27)πh^2
Now, let's find dh/dt:
Differentiating h = 3r = 3(h/3) with respect to t:
dh/dt = 3 * (dh/dt)
dh/dt = 3 * (dr/dt)
We need to find dr/dt. We know that the water is leaving the tank at 50 cubic feet per minute, which means that the rate at which the radius is changing with respect to time (dr/dt) is equal to -50 / (dV/dr).
We can find dV/dr by differentiating the expression for the volume of the cone with respect to r:
V = (1/3)πr^2h
Differentiating with respect to r:
dV/dr = (1/3)π * 2r * h
Substituting h = 3r:
dV/dr = (1/3)π * 2r * 3r
dV/dr = 2πr^2
Now we can find dr/dt:
dr/dt = -50 / (2πr^2)
dr/dt = -25 / (πr^2)
Finally, we can substitute these values into the expression for dV/dt:
-50 = (1/27)πh^2 * (-25 / (πr^2))
-50 = (1/27) * (h^2 / r^2) * (-25)
We are interested in finding the value of dh/dt when h = 3. Substituting this into the equation above, we get:
-50 = (1/27) * (3^2 / r^2) * (-25)
-50 = -(1/27) * (9 / r^2) * 25
To find dh/dt, we need to isolate it on one side of the equation:
dh/dt = -50 / (-(1/27) * (9 / r^2) * 25)
dh/dt = 1350 / (9 / r^2)
dh/dt = 150r^2
We know that h = 3r, so substituting this into the expression for dh/dt, we get:
dh/dt = 150(3r)^2
dh/dt = 1350r^2
When the water level is 3 feet high, h = 3, so r = h/3 = 3/3 = 1. Plug this value into the expression for dh/dt:
dh/dt = 1350(1)^2
dh/dt = 1350
Therefore, the water level is falling at a rate of 1350 feet per minute. Therefore, the correct answer is E. 1.768.
To find the rate at which the water level is falling, we need to find the derivative of the volume with respect to time.
Let's first write an equation relating the height and radius of the conical tank. We are given that the height is always 3 times the radius. So, we can write this relationship as h = 3r.
The volume of a cone is given by the formula V = (1/3)πr^2h.
To get rid of the radius in the volume equation, we can substitute the value of h in terms of r:
V = (1/3)πr^2(3r)
= πr^3
Now, we have the volume equation solely in terms of r.
Taking the derivative of V with respect to time (t) gives us:
dV/dt = d(πr^3)/dt
= 3πr^2(dr/dt)
We are given that the water is leaving the tank at a rate of 50 cubic feet per minute, which we can denote as dV/dt = -50 (negative sign indicates that the volume is decreasing).
We need to find dr/dt, the rate at which the radius is changing, when the water level is 3 feet high (h = 3).
Now, let's substitute the given information and solve for dr/dt:
-50 = 3πr^2(dr/dt)
dr/dt = -50/(3πr^2)
To find dr/dt when the water level is 3 feet high (h = 3), we need to find the corresponding value of r.
Using the equation h = 3r:
3 = 3r
r = 1
Now, substitute r = 1 into the equation for dr/dt:
dr/dt = -50/(3π(1)^2)
dr/dt = -50/(3π)
Calculating this value, you will get:
dr/dt ≈ -5.305
This represents the rate at which the radius is changing. However, we are interested in the rate at which the water level is falling. Since the height is always 3 times the radius, the rate at which the water level is falling is three times the rate at which the radius is changing.
So, the rate at which the water level is falling, dh/dt, is:
dh/dt = 3(dr/dt)
dh/dt = 3(-5.305)
dh/dt = -15.915
Therefore, the water level is falling at a rate of approximately 15.915 feet per minute when the water is 3 feet high.
The correct answer is C. 15.915.
since h = 3r,
v = 1/3 pi r^2 h = pi/27 h^3
dv/dt = pi/9 h^2 dh/dt
-50 = pi/9 * 9 dh/dt
dh/dt = -50/pi
C is correct