The edge of a cube is increasing at a rate of 2 inches per minute. At the instant when the volume is 27 cubic inches, how fast is the volume changing? indicate the units of measure. V=x^3
dx/dt = 2 inches/min
V = x^3
so,
dV/dt = 2^3 = 8 cubic inches/min
V = x^3
dV/dt = 3x^2 dx/dt
when V= 27, x = 3, dx/dt = 2
dV/dt = 3(3^2)(2) cubic inches/min
= 54 cubic inches/min
To find how fast the volume is changing, we can use the formula for the volume of a cube: V = x^3, where x is the length of an edge.
Given that the edge of the cube is increasing at a rate of 2 inches per minute, we can find the rate of change of the volume by differentiating the volume function with respect to time (t).
dV/dt = d/dt (x^3)
To find dV/dt, we need to know the value of x when the volume is 27 cubic inches. Since the volume is given as V = 27, we can solve for x:
27 = x^3
Take the cube root of both sides:
∛27 = ∛x^3
3 = x
Now we have the value of x when the volume is 27 cubic inches: x = 3.
Next, differentiate V with respect to t:
dV/dt = d/dt (x^3)
dV/dt = 3x^2(dx/dt)
Substituting the known values:
dV/dt = 3(3^2)(dx/dt)
dV/dt = 27(dx/dt)
Since dx/dt represents the rate at which the edge of the cube is increasing (which is given as 2 inches per minute), we have:
dV/dt = 27(2)
dV/dt = 54
Therefore, at the instant when the volume is 27 cubic inches, the volume is changing at a rate of 54 cubic inches per minute.
To find how fast the volume is changing, we need to differentiate the volume function with respect to time (t) using the chain rule.
Given: V = x^3, where V is the volume and x is the length of the edge of the cube.
Since the volume is expressed in cubic inches and the rate of change is in inches per minute, we need to express both the volume and the rate of change in those units.
Differentiating the volume equation with respect to time, we have:
dV/dt = d(x^3)/dt
To evaluate this derivative, we can use the chain rule. The derivative of x^3 with respect to x is 3x^2, and the derivative of x with respect to t is dx/dt. Therefore, we can rewrite the equation as:
dV/dt = 3x^2 * dx/dt
Now we have an expression involving both the rate of change of the volume (dV/dt) and the rate of change of the edge length (dx/dt). We need to find the values of x and dx/dt at the instant when the volume is 27 cubic inches.
Given that V = 27, we can substitute this value into the volume equation:
27 = x^3
Taking the cube root of both sides, we find:
x = 3 inches
Now we need to find the rate of change of the edge length (dx/dt). We are given that the edge of the cube is increasing at a rate of 2 inches per minute, so:
dx/dt = 2 inches/minute
Substituting these known values into the equation for dV/dt, we have:
dV/dt = 3(3^2) * 2
Simplifying, we find:
dV/dt = 54 cubic inches per minute
Therefore, the volume is changing at a rate of 54 cubic inches per minute when the volume is 27 cubic inches.