Q:

Rate of change of a cubes area in respect to the space diagonal

The space diagonal of a cube shrinks with $0.02\rm m/s$. How fast is the area shrinking when the space diagonal is $0.8\rm m$ long?

I try:
Space Diagonal = $s_d = \sqrt{a^2+b^2+c^2}=\sqrt{3a^2}$ Where $a$ is the length of one side.
  Area = $a^2$
Rate of change for $s_d$ with respect to $a$
$${\mathrm d\over \mathrm da}s_d={\mathrm d\over \mathrm da}\sqrt{3a^2}={\sqrt{3}a \over \sqrt{a^2}}$$
Rate of change for $\rm area$ with respect to $a$
$${\mathrm d\over \mathrm da}\mathrm{area}={\mathrm d\over \mathrm da}a^2={2a}$$
Im stuck when it comes to calculating one thing from another thing! However I have no problem when it comes to position, velocity and acceleration! Can anybody solve this? 

A:

You want $dA/dt$. By the chain rule, $$\frac{dA}{dt}=\frac{dA}{da}\cdot\frac{da}{ds}\cdot\frac{ds}{dt}$$ which is an equation that applies at all points in time, for all sizes of the cube.
At the moment in question, you know $s$ directly. You can use algebra (no calculus) to find $a$ and $A$ at that moment too. Then you have already worked out a formula for the first factor. You have a formula for the reciprocal of the second factor. And the value of the third is a constant negative number given in the problem.

