The volume of a cube is increasing at a rate of 10cm^3/min. How fast is the surface area increasing when the length of an edge is 10cm?
I don't understand this question at all..
$\displaystyle V=a^3$
$\displaystyle S=6a^2$
$\displaystyle \frac{dV}{dt}=10cm^3/min$
$\displaystyle 3a^2\frac{da}{dt}=10$
$\displaystyle 3*10^2\frac{da}{dt}=10$
$\displaystyle \frac{da}{dt}=\frac{1}{30}$
$\displaystyle \frac{dS}{dt}=6*2a\frac{da}{dt}=12*10*\frac{1}{30} =4cm^2/min$
So, the surface area is increasing at a rate of $\displaystyle 4cm^2/min$.