The sides of a cube are increasing in length at a constant rate of 5 cm/minute. At the moment that the surface area of the cube is 2400 cm^2, how fast is the VOLUME of the cube increasing?
when A = 2400, s = 20
v = s^3
dv/dt = 3s^2 ds/dt = 3*400*5 = 6000 cm^3/s
or
v = (A/6)^(3/2)
dv/dt = 1/4 (A/6)^(1/2) dA/dt = 1/4 (A/6)^(1/2) * 12s ds/dt
= 1/4 * 20 * 12 * 20 * 5 = 6000 cm^3/s
To find the rate at which the volume of the cube is increasing, we need to use the chain rule of differentiation.
Let's denote:
- x as the length of the sides of the cube (in cm)
- V as the volume of the cube (in cm^3)
- A as the surface area of the cube (in cm^2)
We are given that the sides of the cube are increasing at a constant rate of 5 cm/minute. This means that dx/dt = 5 cm/minute.
We also know that the surface area of the cube is 2400 cm^2. Hence, A = 2400 cm^2.
The surface area of a cube is given by the formula A = 6x^2. Therefore:
2400 = 6x^2
Simplifying, we find x^2 = 400
Taking the square root on both sides, we get x = 20 cm.
Now, we can find the relationship between the volume (V) and the side length (x) of the cube. The volume of a cube is given by V = x^3.
Differentiating V with respect to t (time) using the chain rule, we have:
dV/dt = dV/dx * dx/dt
To find dV/dt, we need to determine dV/dx and dx/dt.
The derivative of V with respect to x is:
dV/dx = 3x^2
We know dx/dt = 5 cm/minute.
Now we can substitute these values into the chain rule equation:
dV/dt = (3x^2) * (dx/dt)
dV/dt = (3(20)^2) * (5)
dV/dt = 3(400) * 5
dV/dt = 6000 cm^3/minute
Therefore, the volume of the cube is increasing at a rate of 6000 cm^3/minute when the surface area of the cube is 2400 cm^2.