the volume of a cube is increasing at the rate of 1200 cm^3/min at the instant it edges are 20cm long. at what rate are the edges changing at the instant?
Is the cone angle staying the same? Is the edge length the slant height? What is the cone angle?
To find the rate at which the edges of the cube are changing, we can use the concept of related rates. The relationship between the volume and the edge length of a cube is given by V = s^3, where V is the volume and s is the edge length.
Given that dV/dt (the rate of change of volume) is 1200 cm^3/min, we need to find ds/dt (the rate of change of the edge length) at the instant when the edges are 20 cm long.
To solve this problem, we will differentiate both sides of the equation V = s^3 with respect to time t using the chain rule:
dV/dt = 3s^2 * ds/dt
Since dV/dt is given as 1200 cm^3/min and s is 20 cm at the instant in question, we can substitute these values into the equation:
1200 = 3(20^2) * ds/dt
Simplifying further:
1200 = 3(400) * ds/dt
Dividing both sides by 3(400):
1200 / (3(400)) = ds/dt
Now, we can solve for ds/dt:
ds/dt = 1200 / (3(400)) = 1 cm/min
Therefore, the edges of the cube are changing at a rate of 1 cm/min at the instant when the edges are 20 cm long.