All edges of a cube are expanding at the rate of 3 centimeters per second. How fast is the volume chang ing when each edge is (a) 1 centimeter and (b) 10 centimeters?
let the edge be x cm long
V = x^3
dV/dt = 3x^2 dx/dt
= 3x^2 (3) = 9x^2 cm^3/s
so when x = 1
dV/dt = 9(1) cm^3/s
do the 2nd part in the same way
To find the rate of change of the volume of a cube as the edges expand, we'll use the formula for the volume of a cube, which is V = s^3, where s is the length of each edge.
Let's start with part (a), where each edge is 1 centimeter. We want to find how fast the volume is changing with respect to time, so we need to find dV/dt. To do this, we'll take the derivative of the volume formula with respect to time. Let's differentiate both sides of the equation:
dV/dt = d/dt (s^3)
Using the chain rule, we have:
dV/dt = 3s^2 * ds/dt
Given that ds/dt = 3 cm/s (since each edge is expanding at a rate of 3 cm/s), and s = 1 cm, we can substitute these values in to find dV/dt:
dV/dt = 3(1^2) * 3 = 9 cm^3/s
So, when each edge is 1 centimeter, the volume is changing at a rate of 9 cubic centimeters per second.
Now, let's move on to part (b), where each edge is 10 centimeters. Following the same steps as before, we have:
dV/dt = 3s^2 * ds/dt
Given that ds/dt = 3 cm/s and s = 10 cm, we can substitute these values in to find dV/dt:
dV/dt = 3(10^2) * 3 = 900 cm^3/s
So, when each edge is 10 centimeters, the volume is changing at a rate of 900 cubic centimeters per second.
To find the rate at which the volume of a cube is changing, we need to use the formula for the volume of a cube: V = s^3, where V is the volume and s is the length of each edge.
Let's begin with part (a), where each edge is 1 centimeter.
Given:
ds/dt = 3 cm/s (rate of change of the edge length)
We are required to find dV/dt (rate of change of the volume) when s = 1 cm.
Using the volume formula, V = s^3, we can differentiate both sides with respect to time (t) using implicit differentiation:
dV/dt = d/dt (s^3)
Using the chain rule, d/dt (s^3) = 3s^2(ds/dt)
Substituting the given values:
dV/dt = 3(1 cm)^2(3 cm/s)
= 3(1)(3) cm^3/s
= 9 cm^3/s
Therefore, when each edge is 1 centimeter, the volume is changing at a rate of 9 cm^3/s.
Moving on to part (b), where each edge is 10 centimeters.
Given:
ds/dt = 3 cm/s (rate of change of the edge length)
We are required to find dV/dt (rate of change of the volume) when s = 10 cm.
Using the same volume formula, V = s^3, and differentiating both sides with respect to time (t):
dV/dt = d/dt (s^3)
Using the same chain rule, d/dt (s^3) = 3s^2(ds/dt)
Substituting the given values:
dV/dt = 3(10 cm)^2(3 cm/s)
= 3(100)(3) cm^3/s
= 900 cm^3/s
Therefore, when each edge is 10 centimeters, the volume is changing at a rate of 900 cm^3/s.