A tank in the form of an inverted cone having an altitude of 2 meters and a base radius of 50 cm. Water if flowing into the tank at the rate of 10 cube centimeters/sec. How fast is the water level rising when the water level is 80 cm deep?
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To find how fast the water level is rising, we need to use related rates. Let's start by finding the volume of the water in the tank at any given time.
We know that the tank has the shape of an inverted cone. The volume of a cone is given by the formula:
Volume = (1/3) * π * r^2 * h,
where r is the radius of the base of the cone and h is the height (or altitude) of the cone.
In this case, we are given that the base radius (r) is 50 cm, and the altitude (h) is 2 meters (or 200 cm), so we can substitute these values into the formula:
Volume = (1/3) * π * (50 cm)^2 * (200 cm) = (1/3) * π * 250000 cm^3.
Now, we need to find the rate at which the volume is changing.
We are given that water is flowing into the tank at a rate of 10 cubic centimeters per second (cm^3/sec). This rate refers to how fast the volume is increasing.
Let's denote the height of the water in the tank as H (in cm). We are asked to find how fast the water level is rising, which refers to how fast dH/dt (the rate of change of height with respect to time) is changing.
To find this rate, we need to differentiate the volume formula with respect to time (t):
dV/dt = (d/dt) [(1/3) * π * r^2 * H].
Differentiating with respect to time (d/dt), we get:
dV/dt = (1/3) * π * 2r * dr/dt * H + (1/3) * π * r^2 * dH/dt.
Note that dr/dt is the rate at which the radius of the cone is changing, which is not given in the problem. However, since the problem only asks for the rate of change of the water level (dH/dt) and not the rate of change of the radius, we can assume that the radius remains constant. Therefore, dr/dt can be considered 0.
This simplifies the equation to:
dV/dt = (1/3) * π * r^2 * dH/dt.
Now, we can substitute the given values into the formula:
10 cm^3/sec = (1/3) * π * (50 cm)^2 * dH/dt.
Simplifying further:
10 cm^3/sec = (1/3) * π * 2500 cm^2 * dH/dt.
To solve for dH/dt, we can rearrange the equation:
dH/dt = (10 cm^3/sec) / [(1/3) * π * 2500 cm^2],
dH/dt = 3.8 * 10^-5 cm/sec.
Therefore, when the water level is 80 cm deep, the water level is rising at approximately 3.8 * 10^-5 cm/sec.