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=L^3
dv/dt=3L^2 * dL/dt
L=11cm and dL/dt=4cm/s
To find how fast the volume of the cube is increasing, we need to use the concept of related rates. Related rates deal with finding the rate of change of one quantity with respect to another, when the two quantities are related by an equation.
In this case, we know that the length of each edge of the cube is increasing at a constant rate of 4 cm/s. Let's call this rate of change dx/dt (the change in the length of the edge with respect to time).
We need to find how fast the volume of the cube (V) is changing with respect to time (dt). The volume of a cube is given by V = x^3, where x is the length of the edge.
Differentiating both sides of the equation with respect to time, we get dV/dt = 3x^2 * dx/dt.
Now we can substitute the given values:
dx/dt = 4 cm/s (since the edge is increasing at a constant rate of 4 cm/s)
x = 11 cm (given length of the edge)
Plugging these values into the equation, we get:
dV/dt = 3(11^2) * 4
= 3(121) * 4
= 363 * 4
= 1452 cm^3/s
Therefore, when the length of the edge is 11 cm, the volume of the cube is increasing at a rate of 1452 cm^3/s.
To find how fast the volume of the cube is increasing, we can use the formula for the volume of a cube: V = x^3, where x represents the length of the edge.
Taking the derivative of both sides of the equation with respect to time (t), we get:
dV/dt = 3x^2 * dx/dt
We are given that dx/dt, the rate at which the edge is increasing, is a constant 4 cm/s.
Substituting x = 11 cm and dx/dt = 4 cm/s into the equation, we have:
dV/dt = 3(11^2) * 4
dV/dt = 3(121) * 4
dV/dt = 363 * 4
dV/dt = 1452 cm^3/s
Therefore, when the length of an edge is 11 cm long, the volume of the cube is increasing at a rate of 1452 cm^3/s.