If the volume of an expanding cube is increasing at the rate of 7m^3/sec,how fast is it surface area increasing when the surface area is 24cm^2.

Answer 1

To find the rate at which the surface area is increasing, we need to find the derivative of the surface area with respect to time.

Let the side length of the cube be denoted by x. The volume of the cube is given by x^3, and the surface area is given by 6x^2.

We are given that the volume is increasing at a rate of 7 m^3/sec, so we have dV/dt = 7. Taking the derivative of both sides of the volume equation, we get:

dV/dt = d/dt (x^3) = 3x^2 (dx/dt)

Since we are given that dV/dt = 7, we can solve for dx/dt:

7 = 3x^2 (dx/dt)

dx/dt = 7 / (3x^2)

Now, we are given that the surface area is 24 cm^2, which is equal to 0.0024 m^2. So we have:

6x^2 = 0.0024

x^2 = 0.0004

x = 0.02 (since side length cannot be negative)

Now we can use the value of x to find the rate at which the surface area is increasing:

dx/dt = 7 / (3(0.02)^2)

dx/dt = 7 / (3(0.0004))

dx/dt ≈ 58333.33 m^2/sec

Therefore, the surface area is increasing at a rate of approximately 58333.33 m^2/sec when the surface area is 24 cm^2.