The volume of a cube increases by 2.0 m^3 / s, With what speed does the cube's side changes when the side is 6.0 m?
Thx for any help I can get here!
I hope you know basic calculus. I am answering your question with it.
Let the volume and a side of a cube be V m^{3} & a m. respectively.
So, V=a^{3}
Now differentiate both sides with respect to time t.
i.e. dV/dt = 3a^{2}da/dt
We have to find da/dt while a=6 & we know dV/dt=2.
i.e. 2 = 3 x 6^{2} x da/dt.
or, da/dt = 2/(3 x 36) = 1/54 m/s.
We know that $\displaystyle V = L^3$. Differentiating both sides with respect to time t,
$\displaystyle \frac{dV}{dt} = 3L^2 \frac{dL}{dt}$
Particularly, when $\displaystyle \frac{dV}{dt} = 2 \frac{m^3}{s}$ and $\displaystyle L = 6 m$,
$\displaystyle \frac{dL}{dt} = \frac{1}{54} \frac{m}{s}$