Question

A coffee pot in the form of a circular cylinder of radius 2 in. is being filled with water flowing at a constant rate. If the water level is rising at the rate of 0.9 in./s, what is the rate (in in3/s) at which water is flowing into the coffee pot? (Round your answer to one decimal place.)

Answer 1

Answer:

11.3 in^3/s

Step-by-step explanation:

Water level rising rate = 0.9 in/s

The cross-sectional area of the coffee pot is given by;

$A= \pi r^2\\A = \pi 2^2\\A= 12.5664\ in^2$

The rate at which water is flowing into the coffee pot is given by the product of the cross-sectional area by the rate at which the level rises:

$R_{v} = 12.5664\ in^2 * 0.9\ in/s \\R_{v} = 11.3 \ in^3/s$

Water flows into the coffee pot at 11.3 in^3/s

Answer 2

Using implicit differentiation, it is found that water is flowing into the coffee pot at a rate of 11.3 cubic inches per second.

The volume of a cylinder of radius r and height h is given by:

$V = \pi r^2h$

It's rate the water is flowingis given by the implicit differentiation of V as function of t, thus:

$\frac{dV}{dt} = 2\pi r\frac{dr}{dt} + \pi r^2\frac{dh}{dt}$

In this problem:

• Radius of 2 in, thus $r = 2$
• Radius does not change, thus $\frac{dr}{dt} = 0$.
• Water level rising at a rate of 0.9 in/s, thus $\frac{dh}{dt} = 0.9$.

Then:

$\frac{dV}{dt} = \pi r^2\frac{dh}{dt}$

$\frac{dV}{dt} = \pi(2)^2(0.9)$

$\frac{dV}{dt} = 11.3$

Water is flowing into the coffee pot at a rate of 11.3 cubic inches per second.

A similar problem is given at https://brainly.com/question/1602395