# The volume of a cube is increasing at the rate of 20 cubic centimeters per second. How fast, in square centimeters per second, is the surface area of the cube increasing at the instant when each edge of the cube is 10 centimeters long?

Consider that the edge of the cube varies with time so that is a function of time

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To find the rate at which the surface area of the cube is increasing when each edge is 10 centimeters long, we'll use the formula for the surface area of a cube, which is ( 6s^2 ), where ( s ) is the length of the side of the cube.

Given that the volume of the cube is increasing at a rate of 20 cubic centimeters per second, we can use the formula for the volume of a cube, which is ( s^3 ), where ( s ) is the length of the side of the cube.

Differentiating the volume formula with respect to time ( t ), we get: [ \frac{dV}{dt} = 3s^2 \frac{ds}{dt} ]

Given that ( \frac{dV}{dt} = 20 ) cubic centimeters per second, and when ( s = 10 ), we can find ( \frac{ds}{dt} ): [ 20 = 3(10)^2 \frac{ds}{dt} ] [ 20 = 300 \frac{ds}{dt} ] [ \frac{ds}{dt} = \frac{20}{300} = \frac{1}{15} ] centimeters per second.

Now, using the formula for the surface area of the cube, when ( s = 10 ): [ A = 6(10)^2 = 600 ] square centimeters.

Differentiating the surface area formula with respect to time ( t ), we get: [ \frac{dA}{dt} = 12s \frac{ds}{dt} ]

Substituting the values ( s = 10 ) and ( \frac{ds}{dt} = \frac{1}{15} ) into the formula: [ \frac{dA}{dt} = 12(10) \left( \frac{1}{15} \right) ] [ \frac{dA}{dt} = \frac{8}{5} ] square centimeters per second.

Therefore, the surface area of the cube is increasing at a rate of ( \frac{8}{5} ) square centimeters per second when each edge of the cube is 10 centimeters long.

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