Respuesta :
The rate of change of area is how the surface area changer per time.
The rate of change of the surface area is 36 square meters per hour.
The given parameters are:
[tex]\mathbf{V = 8m^3}[/tex] --- the volume of the cube
[tex]\mathbf{V' = 18m^3h^{-1}}[/tex] --- rate of change of the cube per hour
Let the side length be l.
So, the volume of the cube is:
[tex]\mathbf{V = l^3}\\[/tex]
Make l, the subject in [tex]\mathbf{V = l^3}\\[/tex]
[tex]\mathbf{l =V^\frac{1}{3}}[/tex]
And the surface area is:
[tex]\mathbf{A = 6l^2}[/tex]
Substitute [tex]\mathbf{l =V^\frac{1}{3}}[/tex] in [tex]\mathbf{A = 6l^2}[/tex]
[tex]\mathbf{A =6V^\frac{2}{3}}[/tex]
Differentiate
[tex]\mathbf{A' =\frac{2}{3} \times 6V^\frac{-1}{3} \times V'}[/tex]
Substitute values for V and V'
[tex]\mathbf{A' =\frac{2}{3} \times 6 \times 8^\frac{-1}{3} \times 18}[/tex]
[tex]\mathbf{A' =\frac{2}{3} \times 6 \times \frac{1}{2} \times 18}[/tex]
[tex]\mathbf{A' =36}[/tex]
Hence, the rate of change of the surface area is 36 square meters per hour.
Read more about rates of change at:
https://brainly.com/question/18904995
Using implicit differentiation, it is found that the rate of change of the surface area of the cube at that instant is of 36 square meters per hour.
The volume of a cube of side length a is given by:
[tex]V = a^3[/tex]
In this problem, volume of 8 m³, thus:
[tex]a^3 = 8[/tex]
[tex]a = \sqrt[3]{8}[/tex]
[tex]a = 2[/tex]
The rate of change of the volume is:
[tex]\frac{dV}{dt} = 3a^2\frac{da}{dt}[/tex]
Since it is of 18 m³/h, we have that:
[tex]18 = 3(2)^2\frac{da}{dt}[/tex]
[tex]12\frac{da}{dt} = 18[/tex]
[tex]\frac{da}{dt} = \frac{3}{2}[/tex]
The surface area is:
[tex]S = 6a^2[/tex]
The rate of change is:
[tex]\frac{dS}{dt} = 12a\frac{da}{dt}[/tex]
We already have the values, thus:
[tex]\frac{dS}{dt} = 12(2) \times \frac{3}{2} = 36[/tex]
The rate of change of the surface area of the cube is of 36 square meters per hour.
A similar problem is given at https://brainly.com/question/18402765
