Respuesta :
EXPLANATION
Given that the increasing rate is 18cm^3/s, we can apply the following realtionship:
[tex]dV/dt=18[/tex]As the volume is equal to the multiplication of the sides and assuming that the edge is called "x" we have:
[tex]dV/dt=d(x^3)/dt\cdot d/dx[/tex]Applying derivatives:
[tex]dV/dt=3x^2\frac{dx}{dt}=18[/tex]Isolating dx/dt:
[tex]dx/dt=18/3x^2[/tex]Therefore:
[tex]dV_{\text{ }}/dt=\frac{18}{3x^2}=\frac{6}{x^2}[/tex]Now, if the measure of the edge is 2cm, we can plug this number into the velocity equation to obtain the speed:
[tex]\frac{dx}{dt}=\frac{6}{2^2}=\frac{6}{4}=\frac{3}{2}\frac{cm}{s}=\frac{1.5\operatorname{cm}}{s}[/tex]Hence, the speed is 1.5 cm/s
