Question

the volume of a cube is increasing at a constant rate of 824 cubic centimeters per minute. At the instant when the volume of the cube is 683683 cubic centimeters, what is the rate of change of the surface area of the cube? Round your answer to three decimal places (if necessary).

Answer 1

To find the rate of change of the surface area of a cube, differentiate the formula for the surface area of a cube with respect to time. Then use the chain rule of differentiation to relate the change in volume to the change in side length. Finally, substitute the given values to find the rate of change of the side length.

Step-by-step explanation:

To find the rate of change of the surface area of a cube, we need to differentiate the formula for the surface area of a cube with respect to time. The surface area (S) of a cube is given by the formula S = 6s^2, where s is the length of a side of the cube.

Given that the volume (V) of the cube is increasing at a constant rate of 824 cubic centimeters per minute, we can relate the change in volume (dV/dt) to the change in the side length (ds/dt) using the formula V = s^3.

Using the chain rule of differentiation, we have dV/dt = 3s^2 * ds/dt. Therefore, the rate of change of the side length (ds/dt) is dV/dt divided by 3s^2. Substituting the given values, we have ds/dt = (824 cm^3/min) / (3(683 cm^3)^(2/3)) = 0.139 cm/min (rounded to three decimal places).

Answer 2

DA/dt = 75.27 cm²

Step-by-step explanation:

Cube Volume = V(c) = 683 cm³

DV(c) /dt = 824 cm³

V(c,x) = x³

Then

DV(c,x)/ dt = 3x² Dx/dt

( DV(c,x)/ dt )/ 3x² = Dx/dt (1)

Now as V(c,x) = x³ when V(c,x) = 683 cm³ x = ∛683

x = 8.806 ( from excel)

And by subtitution of this value in equation (1)

Dx/dt = ( DV(c,x)/ dt )/ 3x² ⇒ Dx/dt = 824 / 3*x²

Dx/dt = 824 /3*77.55

Dx/dt = 824/232,64

Dx/dt = 3,542

Then we got Dx/dt where x is cube edge. The area of the face is x² then

the rate of change of the suface area is

DA/dt = ( Dx/dt )²*6 ( 6 faces of a cube)

DA/dt = (3.542)²*6

DA/dt = 75.27 cm²