Question

A conical tank has height 8 meters and radius 4 meters at the base. Water flows at a rate of 4m^3/min. How fast is the water level rising when the water level is 2 meters?

Answer 1

To find how fast the water level is rising in a conical tank when the water level reaches 2 meters, we use the cone's volume formula, substitute for the tank's proportions, differentiate with respect to time, then solve for the rate of change of the water level with the given values.

Step-by-step explanation:

The question involves applying principles of calculus, specifically related to rates of change, to determine how fast the water level is rising in a conical tank. The conical tank has a height of 8 meters and a base radius of 4 meters. With water flowing in at a rate of 4 m³/min, we are asked to find the rate at which the water level rises when it is at 2 meters.

First, we use the formula for the volume of a cone, V = 1/3πr²h. Because the radius and height of the cone are proportional (r/h is constant), we can express the radius r as a function of the height h. We then differentiate both sides of the volume equation with respect to time t to find dV/dt (volume flow rate) and dh/dt (rate at which water level rises).

By plugging in the given volume flow rate (4 m³/min) and solving for dh/dt when the water height is 2 meters, we can find the rate of change of the water level.

Answer 2

The water level in the conical tank is rising at a rate of approximately 0.25 meters per minute when the water level is at 2 meters.

Step-by-step explanation:

To determine the rate at which the water level is rising in the conical tank, we can apply related rates involving the volume of a cone formula
`(\(V = (1)/(3)\pi r^2 h\))`and the chain rule of differentiation.

Given the dimensions of the conical tank (height \(h = 8\) meters and r
`adius \(r = 4\) meters at the base) and the rate of water flow (\(4 \, \text{m}^3/\text{min}\)), we need to find the rate at which the water level (\(\frac{{dh}}{{dt}}\))`is rising when the water level is at \(h = 2\) meters.

First, we derive the volume of the water in the tank using the cone volume formula and differentiate it with respect to time:

`\[V = (1)/(3)\pi r^2 h\]\(\frac{{dV}}{{dt}} = (1)/(3)\pi \left(2rh \frac{{dr}}{{dt}} + r^2 \frac{{dh}}{{dt}}\right)\)`

We are given \(V = 4 \, \text{m}^3/\text{min}\) and \(h = 2\) meters. We substitute these values along with \(r = 4\) meters and \(h = 2\) meters into the volume derivative equation to solve for \(\frac{{dh}}{{dt}}\).

Solving for \(\frac{{dh}}{{dt}}\) yields the rate at which the water level is rising when the water level is at 2 meters, which is approximately \(0.25\) meters per minute.

Understanding related rates in calculus and their application in real-world scenarios like fluid dynamics aids in solving problems involving changing quantities in various geometrical shapes.