Water is flowing into a vertical cylindrical tank at the rate of 24 cu. ft. per minute. If the radius of the tank is 4 feet, how fast is the surface rising?
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volume=PI*r^2*h
dV/dt=pi(2hr dr/dt+r^2 dh/dt)
wouldn't dr/dt=0?
Solution pls
To find the rate at which the surface of the water is rising, we need to use the formula for the volume of a cylinder: V = πr^2h, where V is the volume, r is the radius, and h is the height.
In this case, the volume is increasing at a rate of 24 cubic feet per minute. The radius is given as 4 feet. We need to find the rate at which the height is changing (dh/dt).
Differentiating both sides of the volume equation with respect to time (t), we get:
dV/dt = π(2r)(dh/dt)
Substituting the given values:
24 = π(2)(4)(dh/dt)
Simplifying this equation:
24 = 8π(dh/dt)
Now, we can solve for dh/dt by dividing both sides by 8π:
dh/dt = 24/(8π)
= 3/(π)
Therefore, the surface of the water is rising at a rate of 3/π feet per minute.
To find how fast the surface of the water is rising, we can use the formula for the volume of a cylinder:
V = πr^2h
Where V is the volume, π is a constant (approximately 3.14159), r is the radius of the cylinder, and h is the height of the water in the cylinder.
Differentiating both sides of the equation with respect to time (t), we get:
dV/dt = d(πr^2h)/dt
Now, we need to find how fast the height of the water is changing (dh/dt). We know that the water is flowing into the tank at a rate of 24 cu. ft. per minute. Therefore,
dV/dt = 24 ft^3/min
Now, let's differentiate the formula for the volume:
d(πr^2h)/dt = 24
Using the product rule, we differentiate each term:
(πr^2)(dh/dt) + (2πrh)(dr/dt) = 24
Since we're interested in finding dh/dt (the rate at which the height is changing), we can solve for it by rearranging the equation:
(πr^2)(dh/dt) = 24 - (2πrh)(dr/dt)
dh/dt = (24 - (2πrh)(dr/dt)) / (πr^2)
Now, we can substitute the given values into the equation. The radius of the tank is 4 feet, so r = 4. We can assume that the rate at which the radius is changing (dr/dt) is zero, because the problem does not specify otherwise. Therefore, the equation simplifies to:
dh/dt = (24 - 0) / (π(4)^2)
= 24 / (16π)
= 3 / (2π) ft/min
Thus, the surface of the water is rising at a rate of 3 / (2π) feet per minute.