If each edge of a cube is increasing at the constant rate of 4 cm/s. How fast is the volume of the cube increasing when the length x of an edge is
11 cm long?
v = x^3
dv/dx = 3 x^2
dv/dt = dx/dt * dv/dx
dv/dt = 4 * 3(121) cm^3/s
To find the rate at which the volume of the cube is changing, we can use calculus and the concept of related rates.
First, let's express the volume of the cube in terms of its edge length x. The volume of a cube is given by V = x^3.
Next, we need to find the relationship between the rate at which the edge length is changing (dx/dt) and the rate at which the volume is changing (dV/dt). We can use the chain rule to differentiate the volume with respect to time:
dV/dt = dV/dx * dx/dt
Now, we know that dx/dt is given as a constant rate of 4 cm/s. So dx/dt = 4 cm/s.
To find dV/dx, we differentiate the volume equation with respect to x:
dV/dx = 3x^2
Now we can substitute the values we know into the equation:
dV/dt = (3x^2) * (4 cm/s)
Given that the length x of an edge is 11 cm, we can calculate the rate at which the volume is increasing:
dV/dt = (3 * (11^2)) * (4 cm/s)
dV/dt = (3 * 121) * (4 cm/s)
dV/dt = 363 * (4 cm/s)
dV/dt = 1452 cm^3/s
Therefore, when the length of an edge is 11 cm, the volume of the cube is increasing at a rate of 1452 cm^3/s.