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 $\displaystyle x$ = side of the cube.

We are asked for the rate of change of the *volume*: .$\displaystyle \frac{dV}{dt}$

The volume of a cube is: .$\displaystyle V\;=\;x^3$

. . Differentiate with respect to time: .$\displaystyle \frac{dV}{dt}\;=\;3x^2\cdot\frac{dx}{dt}$ **[1]**

The left side is what we want. .If we know everything on the right side, we're golden.

We are told that: $\displaystyle x = 10$ . . . but they didn't tell us $\displaystyle \frac{dx}{dt}$

. . Instead they said the *surface area* is changing at 20 cm²/hr: .$\displaystyle \frac{dA}{dt} = 20$

Okay, we know the area of a cube . . . six faces, each with an area of $\displaystyle x^2$

. . So we have: .$\displaystyle A\:=\:6x^2\quad\Rightarrow\quad\frac{dA}{dt}\,=\,1 2x\cdot\frac{dx}{dt}$

. . Then: .$\displaystyle 12x\cdot\frac{dx}{dt}\:=\:20\quad\Rightarrow\quad \frac{dx}{dt}\,= \, \frac{5}{3x}$

Substitute into **[1]**: .$\displaystyle \frac{dV}{dt}\;=\;3x^2\cdot\frac{5}{3x}\:=\:15x$

Therefore, when $\displaystyle x = 10:\;\;\frac{dV}{dt}\;=\;15(10)\:=\:150$ 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 $\displaystyle SL = 3000$.

Let $\displaystyle y = RL$, the height of the rocket.

And let $\displaystyle x = SR$, the distance of the rocket from the spectator.

. . And we want $\displaystyle \frac{dx}{dt}$.

We need an equation that ties together all those factors.

We have a right triangle . . . How about Pythagorus?

Pythagorus says: .$\displaystyle x^2 \:=\:y^2 + 3000^2$

Differentiate with respect to time: .$\displaystyle 2x\cdot\frac{dx}{dt} \:= \:2y\cdot\frac{dy}{dt}$

So we have: .$\displaystyle \frac{dx}{dt} \:= \:\frac{y}{x}\cdot\frac{dy}{dt}$ **[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: $\displaystyle y = 4000$

. . and the rocket is rising at 300 m/sec: $\displaystyle \frac{dy}{dt} = 300$ . . . um, we don't know $\displaystyle x$.

But we can find $\displaystyle x$ . . . We have a right triangle with legs of 3000 and 4000.

. . Hence: .$\displaystyle x^2\:= \:3000^2 + 4000^2\:= \: 5000^2\quad\Rightarrow\quad x\,=\,5000$

Substitute into **[2]**: .$\displaystyle \frac{dx}{dt} \:= \:\frac{4000}{5000}(300) \:= \: 240$ m/sec.