A right circular cone has a radius that enlarges at a rate of 5 cm/sec.. The
equation h = 2r relates the radius and height of a cone. How fast is the volume
of the cone changing at the instant when the radius is 8 centimeters.
Given Information:
dr/dt = 5 cm/s
h = 2r
r = 8
V = (π*r^2*h)/3 --> V = (π*r^2*2r)/3 --> V = (2πr^3)/3
dV/dt = ?
Calculation:
V = (2πr^3)/3
dV/dt = 2πr^2(dr/dt)
dV/dt = 2π(8)^2(5)
dV/dt = 10π(64)
dV/dt = 640π cm^3/s
So, the volume of the cone changes at the instant when the radius is 8 centimeters at a rate of 640π cm^3/s.
v = 1/3 πr^2 h = 2π/3 r^3
dv/dt = 2πr^2 dr/dt = 2π*8^2 * 5 = 640π cm^3/s
Well, well, well, it seems we have a growing concern here! Now, let's break it down step by step.
First, we need to find the equation for the volume of a cone. Remember, the volume of a cone is given by V = (1/3)πr²h.
Now, we know that the equation h = 2r relates the radius and height of the cone. So, we can substitute this equation into the volume equation: V = (1/3)πr²(2r).
Next, let's differentiate both sides of the equation with respect to time (t), because we're looking for how the volume changes over time.
dV/dt = (1/3)π(2r)(2r(dr/dt) + r²(dh/dt)).
But don't worry, we know some values! The radius (r) is 8 centimeters, and we're given that dr/dt (the rate of change of the radius) is 5 cm/sec.
So, substituting in those values, we get:
dV/dt = (1/3)π(2(8))(2(8)(5) + (8²)(dh/dt).
Now, because we're looking for how fast the volume is changing when the radius is 8 centimeters, we need to find dh/dt (the rate of change of the height) at that specific moment. Unfortunately, we don't have that information here. So, we'll just have to leave it as dh/dt for now.
Therefore, the expression for how fast the volume is changing becomes:
dV/dt = (1/3)π(2(8))(2(8)(5) + (8²)(dh/dt).
And that, my friend, is as far as we can go without knowing the rate of change of the height. The volume is changing with a speed that depends on how quickly the height is changing.
I hope I didn't leave you feeling circular and confused! If there's anything else I can help with, feel free to ask!
To find the rate at which the volume of the cone is changing, we need to use the formula for the volume of a cone:
V = (1/3) * π * r^2 * h
Given that the equation h = 2r relates the radius and height of the cone, we can substitute this into the volume formula:
V = (1/3) * π * r^2 * (2r)
Simplifying this equation, we get:
V = (2/3) * π * r^3
To find how fast the volume is changing, we can differentiate both sides of the equation with respect to time (t):
dV/dt = (2/3) * π * d(r^3)/dt
Now, we need to find the rate at which the radius is changing (d(r)/dt). We are given that the radius is enlarging at a rate of 5 cm/sec, so d(r)/dt = 5 cm/sec.
Substituting this value into the equation, we get:
dV/dt = (2/3) * π * d(r^3)/dt
= (2/3) * π * d(r^3)/dr * d(r)/dt
= (2/3) * π * 3r^2 * d(r)/dt
Now, we can substitute the given radius value (r = 8 cm) into the equation:
dV/dt = (2/3) * π * 3(8)^2 * 5
= (2/3) * π * 3 * 64 * 5
= 640π cm^3/sec
Therefore, the volume of the cone is changing at a rate of 640π cm^3/sec when the radius is 8 centimeters.
To find how fast the volume of the cone is changing, we need to differentiate the volume equation with respect to time and then substitute the given values to find the rate of change.
The volume of a right circular cone can be given by the formula:
V = (1/3) * π * r^2 * h
We are given the equation h = 2r, so we can substitute this into the volume equation to express the volume purely in terms of r:
V = (1/3) * π * r^2 * (2r)
V = (2/3) * π * r^3
Now, let's differentiate both sides of the equation with respect to time (t):
dV/dt = (2/3) * π * 3r^2 * (dr/dt)
dV/dt = 2π * r^2 * (dr/dt)
We are given that the radius is enlarging at a rate of 5 cm/sec or (dr/dt = 5 cm/sec), and we want to find how fast the volume is changing at the instant when the radius is 8 centimeters or (r = 8 cm).
Substituting these values into the equation, we get:
dV/dt = 2π * (8^2) * 5
dV/dt = 2π * 64 * 5
dV/dt = 640π cm^3/sec
Therefore, the volume of the cone is changing at a rate of 640π cubic centimeters per second when the radius is 8 centimeters.