Each side of a cube is increasing at the rate of 9cm/s . When the lenght of each side is 5 cm , find the rate of inscrease of its volume and total surface area .
(9+5)^3/(5^3)=rate of increase in volume
V = x^3 , where x is the length of the side
dV/dt = 3x^2 dx/dt
for the given data,
dV/dt = 3(5^2)(9)
= 675 cm^3/s
S = 6x^2, where S is the surface area
dS/dt = 12x dx/dt
= 12(5)(9)
= 540 cm^2/s
To find the rate of increase of the volume and total surface area of a cube, we can use the formulas for calculating these values.
The volume of a cube is given by V = s^3, where s is the length of one side.
The total surface area of a cube is given by A = 6s^2.
Given that the length of each side is increasing at a rate of 9 cm/s, we can find the rate of increase of the volume and total surface area.
To find the rate of increase of the volume, we need to find dV/dt, which represents the derivative of the volume equation with respect to time (t).
Differentiating the volume equation, we get dV/dt = 3s^2(ds/dt).
Given that s = 5 cm and ds/dt = 9 cm/s, we can substitute these values into the derivative equation to find the rate of increase of the volume.
dV/dt = 3(5^2)(9) = 675 cm^3/s.
Therefore, the rate of increase of the volume of the cube is 675 cm^3/s.
To find the rate of increase of the total surface area, we need to find dA/dt, which represents the derivative of the surface area equation with respect to time (t).
Differentiating the surface area equation, we get dA/dt = 12s(ds/dt).
Given that s = 5 cm and ds/dt = 9 cm/s, we can substitute these values into the derivative equation to find the rate of increase of the total surface area.
dA/dt = 12(5)(9) = 540 cm^2/s.
Therefore, the rate of increase of the total surface area of the cube is 540 cm^2/s.