v = 1/3 πr^2 h = 2π/3 r^3
dv/dt = 2πr^2 dr/dt = 2π*8^2 * 5 = 640π cm^3/s
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.
2 answers
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.
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.
2 answers