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: . **[1]**

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 **[1]**: .

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?

Code:

* R
* :
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: . **[2]**

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 **[2]**: . m/sec.