The volume V of a cube with sides of length x in. is changing with respect to time. At a certain instant of time, the sides of the cube are 6 in. long and increasing at the rate of 0.3 in./sec. How fast is the volume of the cube changing at that instant of time?
V = x^3
dV/dt = 3x^2 dx/dt
given: when x = 6 , dx/dt = .3
dV/dt = 3(36)(.3) = 32.4 cubic inches/sec
Well, it seems like the cube is going through some growing pains! Let's figure out how fast its volume is changing.
We know that the volume V of a cube is given by V = x^3, where x is the length of its sides.
Now, we need to find out how fast the volume is changing with respect to time, which is represented as dV/dt. In other words, we want to find the derivative of the volume equation with respect to time.
Using the chain rule, we can write:
dV/dt = dV/dx * dx/dt
The first part, dV/dx, represents how much the volume changes with respect to the length of the sides. For a cube, it's easy, as we just differentiate V = x^3 with respect to x, giving us:
dV/dx = 3x^2
The second part, dx/dt, represents how fast the length of the sides is changing with respect to time. In this case, we know that dx/dt = 0.3 in./sec.
Now let's put it all together:
dV/dt = dV/dx * dx/dt
= 3x^2 * 0.3
= 0.9x^2 in.^3/sec
At the instant when the sides of the cube are 6 in. long, we can plug in x = 6 into the equation to find the rate of change:
dV/dt = 0.9 * 6^2
= 0.9 * 36
= 32.4 in.^3/sec
So, at that instant of time, the volume of the cube is changing at a rate of 32.4 in.^3/sec. It's certainly growing quite rapidly!
To find how fast the volume of the cube is changing at a certain instant of time, we need to find the derivative of the volume function with respect to time.
The volume V of a cube is given by the formula V = x^3, where x is the length of each side.
Given that the sides of the cube are increasing at a rate of 0.3 in./sec, we can say that dx/dt = 0.3 in./sec.
We need to find dV/dt, the rate at which the volume is changing with respect to time.
Using the chain rule, we can differentiate V = x^3 with respect to time:
dV/dt = (dV/dx) * (dx/dt)
To find dV/dx, the derivative of the volume with respect to x, we can differentiate V = x^3 with respect to x:
dV/dx = 3x^2
Now we substitute the given values:
dV/dt = (3x^2) * (dx/dt)
At the certain instant of time, the sides of the cube are 6 in. long, so x = 6.
Substituting x = 6 into the equation:
dV/dt = (3(6)^2) * (0.3)
Simplifying:
dV/dt = (3(36)) * (0.3)
dV/dt = 32.4 in.^3/sec
Therefore, the volume of the cube is changing at a rate of 32.4 in.^3/sec at that instant of time.
To find how fast the volume of the cube is changing at a certain instant of time, we need to use the chain rule of derivatives.
Given:
- The length of the sides of the cube, x = 6 in
- The rate of increase of the side length, dx/dt = 0.3 in/sec
We need to find the rate of change of volume, dV/dt, at that instant.
The volume of a cube is given by the formula V = x^3, where x is the length of the side.
Differentiating both sides of the equation with respect to time t, we get:
dV/dt = d/dt (x^3)
Using the chain rule, we can rewrite the derivative as:
dV/dt = d/dx (x^3) * dx/dt
Now, let's calculate each part separately:
1. d/dx (x^3):
To differentiate x^3 with respect to x, we bring down the exponent as the coefficient and reduce the exponent by 1:
d/dx (x^3) = 3x^2
2. dx/dt:
Given, dx/dt = 0.3 in/sec
Now, substitute the values we have in our equation:
dV/dt = (3x^2) * (dx/dt)
= (3 * 6^2) * (0.3) [Substituting x = 6]
= (3 * 36) * (0.3)
= 108 * 0.3
= 32.4 in^3/sec
Therefore, at that instant, the volume of the cube is changing at a rate of 32.4 in^3/sec.