Hello again, sooty!
Advice: make a sketch . . .
1. The surface area of a cube is increasing at a rate of 20 cm²/hr.
Find the rate (in cm³/hr) of change in the volume of the cube
and use it to compute the rate of change in the volume when the side of the cube is of length 10.
I won't make a sketch for the cube; we know what one looks like, right?
Let = side of the cube.
We are asked for the rate of change of the volume: .
The volume of a cube is: .
. . Differentiate with respect to time: . 
The left side is what we want. .If we know everything on the right side, we're golden.
We are told that: . . . but they didn't tell us
. . Instead they said the surface area is changing at 20 cm²/hr: .
Okay, we know the area of a cube . . . six faces, each with an area of
. . So we have: .
. . Then: .
Substitute into : .
Therefore, when cm³/hr.
2. A spectator standing at a distance of 3000 m from a lauch site, is observing a rocket launch.
The rocket is launched vertically and is rising at a rate of 300 m/sec.
When its altitude is 4,000 m, how fast is the distance between the rocket
and the spectator changing at that instant?
x * :
* : y
- * - - - - - - - - - - - * -
S 3000 L
The spectator is at S, the launch site is L, the rocket is at R.
We are told that .
Let , the height of the rocket.
And let , the distance of the rocket from the spectator.
. . And we want .
We need an equation that ties together all those factors.
We have a right triangle . . . How about Pythagorus?
Pythagorus says: .
Differentiate with respect to time: .
So we have: . 
The left side is what we want . . . Do we know everything on the right side?
We know that, at that particular instant . . . the height is 4000 m:
. . and the rocket is rising at 300 m/sec: . . . um, we don't know .
But we can find . . . We have a right triangle with legs of 3000 and 4000.
. . Hence: .
Substitute into : . m/sec.