Sand is poured at the rate of 10 m cube /min so as to form a conical pile whose altitude is always equal to the radius of its base.
dV/dt = 10 m^3/min
V = (1/3) pi r^2 h = (1/3) pi r^3
dV/dr = pi r^2
dV/dr* dr/dt = pi r^2 dr/dt
so
dV/dt = 10 = pi r^2 dr/dt
dr/dt = dh/dt = 10 /pi r^2
or whatever it is you are supposed to solve for
After you finish typing the rest of your problem, take a look at the series of
"Similar Questions" at the bottom of this page. There is a nice variety of them
going back quite a while. One of them should match your problem, all you have to
do is change the numbers.
To solve this problem, we need to determine the rate of change of various quantities and find the relationship between them. Let's consider the following variables:
V = Volume of the conical pile (in cubic meters)
r = Radius of the base of the conical pile (in meters)
h = Altitude (or height) of the conical pile (in meters)
Given information:
The rate of pouring sand is 10 m^3/min.
Now, let's find the relationship between the variables using the formula for the volume of a cone:
V = (1/3) * π * r^2 * h
Since the altitude is always equal to the radius, we have h = r. Therefore, we can rewrite the volume equation as:
V = (1/3) * π * r^3
Now we need to find the rate at which the height of the pile is changing with respect to time. Let's differentiate both sides of the equation with respect to time (t):
dV/dt = (1/3) * π * (3r^2) * (dr/dt)
Since we know dV/dt (rate of pouring sand) is 10 m^3/min, we can substitute the values into the equation:
10 = (1/3) * π * (3r^2) * (dr/dt)
Now, let's solve for (dr/dt), which represents the rate of change of the radius with respect to time:
(dr/dt) = (10 / (π * r^2))
So, the rate at which the radius of the conical pile is changing with respect to time is (10 / (π * r^2)).
Note: If you need any numerical calculations, please provide the value of the radius (r) at a specific time.
To solve this problem, we need to find the rate of change of the height (altitude) of the pile with respect to time when the altitude is equal to the radius of the base.
Let's first define the variables:
- h: height (altitude) of the conical pile (in meters)
- r: radius of the conical base (also in meters)
- V: volume of the conical pile (in cubic meters)
- t: time (in minutes)
We are given that sand is poured into the cone at a constant rate of 10 m^3/min. This means that the volume of the cone is increasing at a constant rate of 10 m^3/min.
The volume of a cone can be calculated using the formula:
V = (1/3) * π * r^2 * h
Since we are given that the altitude is always equal to the radius of the base, we can substitute h = r into the volume formula:
V = (1/3) * π * r^2 * r => V = (1/3) * π * r^3
Now, let's find the rate of change of the volume with respect to time, dV/dt:
dV/dt = d/dt[(1/3) * π * r^3]
dV/dt = (1/3) * π * d/dt(r^3)
dV/dt = (1/3) * π * 3r^2 * dr/dt
dV/dt = π * r^2 * dr/dt
Since dr/dt represents the rate of change of the radius with respect to time, and we are given that the height and radius are always equal, dr/dt can be replaced by dh/dt.
dV/dt = π * r^2 * dh/dt
But we know that dV/dt is constant and equal to 10 m^3/min, so we can replace dV/dt with 10:
10 = π * r^2 * dh/dt
Now, we can find the rate of change of the height (dh/dt) at the point where the height equals the radius by substituting h = r:
10 = π * r^2 * dh/dt
10 = π * r^2 * (dh/dt)|h=r
Simplifying this equation will give us the rate of change of the height with respect to time at the relevant point.