Water, at the rate 10 ft^3/min, is pouring into a leaky cistern whose shape is a cone 16' deep and 8' in diameter at the top. At the time the water is 12' deep, the water level is observed to be rising 4''/min. How fast is the water leaking away?
Can anyone please give me some ideas to do it???THANKS A LOT!!!
The above is good, except for the fact that the diameter is given to be 8ft, so the radius is only 4ft.
Well, it seems like you have a leaky situation on your hands! Let's dive into the problem and see if we can solve it with humor.
First, let's find the volume of the cone-shaped cistern. We know the diameter at the top is 8', so the radius is 4'. The height of the cistern is 16', so we have all the measurements we need.
The formula for the volume of a cone is V = (1/3)πr^2h. Plugging in the values, we get V = (1/3)π(4^2)(16) = (1/3)π(64)(16) = (1/3)π(1024) ≈ 1071.24 ft^3.
Now, let's figure out the rate at which the water level is rising. We're told that when the water depth is 12', it's rising at a rate of 4''/min. We need to convert the depth to feet, so 12' = 12 * 12 = 144''. Therefore, the water level is rising at a rate of (4/144) ft/min.
Next, we can use the volume of the cone and the rate at which the water level is rising to find the rate at which water is pouring into the cistern. The formula for the volume of a cone is V = (1/3)πr^2h, but we need to find the derivative of that equation with respect to time.
I'm going to be honest with you, this is where the humor faucet runs dry. We're going to do some math here, involving derivatives and related rates. It's not the funniest topic, but I'll do my best to make it a little more entertaining.
Let's call the rate at which the water is pouring in "dV/dt". We know that dV/dt = 10 ft^3/min, so the water is pouring in at a constant rate. However, the volume of the cone is changing, so we need to relate the rate of change of volume to the rate of change of the water level.
Using the related rates formula, we have dV/dt = (dV/dh) * (dh/dt). We know dV/dt is 10 ft^3/min and dh/dt is (4/144) ft/min. We need to find dV/dh, which is the derivative of the volume equation with respect to height.
dV/dh = (1/3)π * d(r^2)/dh * h + (1/3)π * r^2.
Now we need to find d(r^2)/dh by differentiating the equation r^2 = (h^2)/4. Hang in there, we're almost done!
d(r^2)/dh = (2h)/4 = h/2.
Now we substitute everything back into our original equation:
10 = (1/3)π * (h/2) * h + (1/3)π * r^2.
Plugging in h = 12' and r = 4', we can solve for dV/dh:
10 = (1/3)π * (12/2) * 12 + (1/3)π * (4^2).
Simplifying all that, we can find dV/dh, which represents the rate at which the volume is changing with respect to the height. Once we have that, we can find the rate at which the water is leaking away using the rate of change of the volume and the pouring rate at which water is coming in.
Phew, that was an exercise in math, wasn't it? I hope I didn't drown you in calculations! Keep swimming through those math problems, and remember, even when things get leaky, a little bit of humor can help keep you afloat!
To solve this problem, you can use the formulas for the volume of a cone and the rate of change of volume.
First, let's find the formula for the volume of a cone in terms of its height and radius.
The formula for the volume of a cone is given by:
V = (1/3) * π * r^2 * h
where V is the volume, π is a constant (approximately 3.14159), r is the radius of the cone, and h is the height of the cone.
In this problem, the diameter of the top of the cone is 8', so the radius is half of the diameter, which is 8' / 2 = 4'. The height of the cone is given as 16'.
Now, let's find the rate of change of volume with respect to time.
Since water is pouring into the cistern at a constant rate of 10 ft^3/min, the rate of change of volume, dV/dt, is constant and equal to 10 ft^3/min.
We are also given that, at a certain time, when the water level is 12' deep, the water level is observed to be rising at a rate of 4''/min.
To relate the rate of change of volume to the rate of change of height, we can differentiate the volume formula with respect to time:
dV/dt = (1/3) * π * (2r * dr/dt) * h + (1/3) * π * r^2 * dh/dt
where dr/dt is the rate of change of the radius, and dh/dt is the rate of change of the height.
Since the radius is constant in this problem (since the top of the cone isn't changing shape), dr/dt is equal to 0, and we can rewrite the formula as:
dV/dt = (1/3) * π * r^2 * dh/dt
Now, we can substitute the values we know into the formula:
10 ft^3/min = (1/3) * π * (4')^2 * (4''/min)
Simplifying the equation, we have:
10 = (1/3) * π * 16 * (4/12)
Now, solve for π:
10 = (1/3) * π * (16/3)
Multiplying both sides by 3, we get:
30 = π * (16/3)
Dividing both sides by (16/3), we get:
30 / (16/3) = π
Simplifying, we have:
π ≈ 5.68
Therefore, the value of π is approximately 5.68.
Now, to find the rate at which water is leaking away, we can use the formula for the volume of the cone and differentiate it with respect to time:
V = (1/3) * π * r^2 * h
Differentiating this formula to find dV/dt, we have:
dV/dt = (1/3) * π * r^2 * dh/dt
We already know that dh/dt is -4''/min (negative because the height of the water is decreasing). So, we can solve for dV/dt:
dV/dt = (1/3) * π * (4')^2 * (-4''/min)
Simplifying, we have:
dV/dt = (1/3) * π * 16 * (-4/12) ft^3/min
Now, substitute the value of π we found earlier (approximately 5.68):
dV/dt = (1/3) * 5.68 * 16 * (-4/12) ft^3/min
Simplifying, we get:
dV/dt ≈ -30 ft^3/min
Therefore, the rate at which the water is leaking away from the cistern is approximately -30 ft^3/min. Note that the negative sign indicates that the water is being lost.
To solve this problem, we will use related rates. We need to find how fast the water is leaking away when the water level is 12' deep.
Let's break down the problem into different components:
1. Volume of the cone:
The volume of a cone can be calculated using the formula V = (1/3) * π * r^2 * h, where V is the volume, r is the radius, and h is the height.
2. Time derivatives:
We are given the rate at which water is pouring into the cistern, which is 10 ft^3/min. We want to find how fast the water level is rising, which is given as 4''/min. We need to find the rate at which the water is leaking out.
Now, let's do the calculations step by step:
1. Find the radius of the cone:
Given the diameter at the top is 8', the radius (r) will be half of that, which is 4'.
2. Find the rate at which the height of the water is changing:
Given that the water level is rising at a rate of 4''/min, we need to convert this to feet by dividing by 12 (since there are 12 inches in a foot). So, the height rate (dh/dt) is 4/12 = 1/3 ft/min.
3. Find the volume of the cone at the time the water level is 12' deep:
Using the volume formula, V = (1/3) * π * r^2 * h, we can substitute the given values, where r = 4' and h = 12'. Hence, V = (1/3) * π * (4)^2 * 12 = 64π ft^3.
4. Find the rate at which the volume is changing with respect to time:
We know that the water is pouring into the cistern at a rate of 10 ft^3/min, so dV/dt (rate at which the volume is changing) is 10 ft^3/min.
5. Apply the chain rule to find the rate at which the radius is changing:
Using the volume formula V = (1/3) * π * r^2 * h, we can differentiate both sides with respect to time t to find dV/dt. Since the only variable changing with time is the radius (r), we can rewrite the formula as V = (1/3) * π * (4)^2 * h * r^2. Now, differentiating both sides with respect to t, we get dV/dt = (1/3) * π * (4)^2 * h * 2r * dr/dt.
6. Substitute the known values:
Using the values we found earlier, dV/dt = 10 ft^3/min and dr/dt = ? (which we are trying to find), we can substitute those values into the equation. So, we have 10 = (1/3) * π * (4)^2 * (12) * (4) * dr/dt.
7. Solve for dr/dt:
Rearranging the equation, we find dr/dt = 10 / [(1/3) * π * (4)^2 * (12) * (4)].
Now, evaluate the expression and simplify to find the value of dr/dt, which will tell us how fast the water is leaking away.
at some time of t minutes, let the height of the water be h ft, and the radius of its surface be r feet.
Let the rate of leakage be x ft^3/min
Volume = (1/3)pi(r^2)h
by similar triangles r/h = 8/16
so r = h/2
Volume = (1/3)pi(h^2/4)(h)
= (1/12)pi(h^3)
then d(Volume)/dt = (1/4)pi(h^2)dh/dt
10+x = (1/4)pi(144)(1/3)
x = 12pi - 10 ft^3/min