The rate at which the height of sugar inside the cylindrical tank increases can be determined using the formula for the volume of a cylinder:
\(V = \pi r^2 h\)
Given:
Since the radius remains constant, differentiate both sides of the volume equation with respect to time \(t\):
\(\dfrac{dV}{dt} = \pi r^2 \dfrac{dh}{dt}\)
Substitute the known values:
\(100\pi = \pi (10)^2 \dfrac{dh}{dt}\)
\(100\pi = 100\pi \dfrac{dh}{dt}\)
Dividing both sides by \(100\pi\):
\(\dfrac{dh}{dt} = 1 \text{ cm/s}\)
Therefore, the height of the sugar in the tank is increasing at a rate of \(1 \text{ cm/s}\).
Let \(V\) be the volume of the sugar in the cylindrical tank, \(r\) be the radius of the tank, and \(h\) be the height of the sugar in the tank.
The volume of a cylinder is given by \(V = \pi r^2 h\).
Given that the radius \(r = 10\) cm, the volume equation becomes \(V = \pi (10)^2 h = 100\pi h\).
We are given that the tank is being filled with sugar at the rate of \(\frac{dV}{dt} = 100\pi\) cm\(^3\)/s.
We want to find the rate at which the height of the sugar is increasing, which is \(\frac{dh}{dt}\).
Differentiate the volume equation with respect to time \(t\):
\(\frac{dV}{dt} = \frac{d}{dt}(100\pi h)\)
\(\frac{dV}{dt} = 100\pi \frac{dh}{dt}\)
Substitute the given value of \(\frac{dV}{dt}\):
\(100\pi = 100\pi \frac{dh}{dt}\)
Solve for \(\frac{dh}{dt}\):
\(\frac{dh}{dt} = \frac{100\pi}{100\pi}\)
\(\frac{dh}{dt} = 1\) cm/s
Correct Answer: 1 cm/s
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