At a certain instant, each edge of a cube is 5 inches long and the volume is increasing at the rate of 2 cubic inches per minute. How fast is the surface area of the cube increasing?

V = s^3

dV/dt = 3v^2 ds/dt
when s = 5, dV/dt = 2
2 = 3(25) ds/dt
ds/dt = 2/75

A = 6s^2
dA/dt = 12s ds/st
= 12(5) (2/75) = 8/5 inches^2 / min

To find how fast the surface area of the cube is increasing, we can use the relationship between the volume and the surface area of a cube. Let's call the length of each edge of the cube "s" and the volume "V".

The volume of a cube with edge length "s" is given by:
V = s^3

Differentiating both sides of this equation with respect to time t, we get:
dV/dt = 3s^2 * ds/dt

Given that dV/dt (the rate of change of volume) is 2 cubic inches per minute, we can substitute this value:
2 = 3s^2 * ds/dt

Now let's find the surface area "A" of the cube. The surface area of a cube with edge length "s" is given by:
A = 6s^2

Differentiating both sides of this equation with respect to time t, we get:
dA/dt = 12s * ds/dt

To find how fast the surface area is increasing, we need to determine ds/dt, the rate at which the edge length is changing.

Given that each edge of the cube is 5 inches long, we have s = 5.

Now we can substitute the known values into the equation for dA/dt:
dA/dt = 12 * 5 * ds/dt
dA/dt = 60 * ds/dt

So, the rate at which the surface area of the cube is increasing is 60 times the rate at which the edge length is changing.

To find ds/dt, we can rearrange the equation we obtained earlier:
2 = 3 * 5^2 * ds/dt

Simplifying this equation, we have:
ds/dt = 2 / (3 * 5^2)
ds/dt = 2 / 75

Substituting this value back into the equation for dA/dt, we get:
dA/dt = 60 * (2 / 75)
dA/dt = 8 / 5

Therefore, the surface area of the cube is increasing at a rate of 8/5 (or 1.6) square inches per minute.

Well, let's dive into this cube conundrum, shall we? To find the rate at which the surface area of the cube is increasing, we can use a little calculus manipulation.

First, let's call the length of the cube's side "s". Since we are given that each edge is 5 inches long, we can say that s = 5 inches.

Now, let's differentiate the equation for the volume of a cube with respect to time (t), so we can get an expression for how it changes over time. In this case, the volume, V, is increasing at a rate of 2 cubic inches per minute, so dV/dt = 2 in^3/min.

The volume of a cube with side length s is given by V = s^3. Taking the derivative of both sides with respect to time, we get:

dV/dt = 3s^2 * ds/dt.

Plugging in the given values, we have:

2 = 3(5^2) * ds/dt.

Simplifying, we have:

2 = 3(25) * ds/dt,
2 = 75 * ds/dt,
ds/dt = 2/75 inches/minute.

So, the surface area of the cube is increasing at a rate of 2/75 square inches per minute.

To find how fast the surface area of the cube is increasing, we need to relate the surface area and volume of a cube.

The surface area (A) of a cube is given by the formula:

A = 6s^2

Where s is the length of each side of the cube.

The volume (V) of a cube is given by the formula:

V = s^3

Given that each edge of the cube is 5 inches long and the volume is increasing at a rate of 2 cubic inches per minute, we can find the rate at which the surface area is changing.

We can differentiate the volume equation with respect to time (t) to find how the volume is changing:

dV/dt = 3s^2 * ds/dt

Given that dV/dt = 2 cubic inches/minute, and s = 5 inches, we can solve for ds/dt:

2 = 3(5^2) * ds/dt

2 = 3(25) * ds/dt

ds/dt = 2 / (3 * 25)

ds/dt = 2 / 75 inches/minute

So, the rate at which the side length of the cube is changing is 2/75 inches/minute.

To find how fast the surface area (A) is changing, we can differentiate the surface area equation with respect to time (t):

dA/dt = 12s * ds/dt

Given that ds/dt = 2/75 inches/minute, and s = 5 inches, we can solve for dA/dt:

dA/dt = 12(5) * (2/75)

dA/dt = 10/5 inches/minute

dA/dt = 2/3 inches/minute

So, the surface area of the cube is increasing at a rate of 2/3 inches/minute.

a = 6s^2

da/dt = 12s ds/dt
da/dt = 12*5*2 = 120 in^2/min