a cube is expanding and its volume is increasing at the rate 96 cubic centimetres/second when the side is 4 cm.Find the rate at which the:
a.side is increasing when the side is 4 cm
b.surface area is increasing when the side is 4cm
V= x^3 A= 6x^2
dV/dt = 3x^2 dx/dt
96 = 3(4^2)dx/dt
dx/dt = 96/48 = 2 cm/second
A = 6x^2
dA/dt = 12x dx/dt
dA/dt = 12(4)(2) = 96 cm^2/sec
A spherical balloon is being inflated and its radius is changing at the rate of 2m/minute. When the radius is 4m, find the rate at which the:
(a).volume is increasing
V= 3/4 pie r^2
(b). surface area is increasing
A = 4 pie r^2
To find the rate at which the side of the cube is increasing when it is 4 cm, we can use the given information about the rate of change of the volume.
a. To find the rate at which the side is increasing, we need to differentiate the volume equation V = x^3 with respect to time. This will give us the equation dV/dt = 3x^2(dx/dt), where dV/dt is the rate of change of the volume and dx/dt is the rate of change of the side length.
Given that dV/dt = 96 cubic centimeters/second when x = 4 cm, we can substitute these values into the equation to solve for dx/dt.
96 = 3(4^2)(dx/dt)
96 = 48(dx/dt)
dx/dt = 96/48
dx/dt = 2 cm/second
Therefore, the side of the cube is increasing at a rate of 2 cm/second when the side length is 4 cm.
b. To find the rate at which the surface area is increasing, we can differentiate the surface area equation A = 6x^2 with respect to time. This will give us the equation dA/dt = 12x(dx/dt), where dA/dt is the rate of change of the surface area.
Given that dx/dt = 2 cm/second when x = 4 cm, we can substitute these values into the equation to solve for dA/dt.
dA/dt = 12(4)(2)
dA/dt = 96 cm^2/second
Therefore, the surface area of the cube is increasing at a rate of 96 cm^2/second when the side length is 4 cm.