1. ## Related Rates?

1) Air is leaking out of a spherical balloon at the rate of 3 cubic cm/sec. When the radius is 5 cm, how fast is the surface area decreasing?

2) A cone shaped paper cup is being filled with water at the rate of 3 cubic cm/sec. The height of the cup is 10 cm and the radius of its base is 5 cm. How fast is the water level rising when the level is 4 cm?

3) A ship B is moving westward toward a fixed point A at a speed of 12km/h. At the moment when the ship is 72 km away from A, a ship C passes through A, heading due south at 10km/h. What is the rate of change in the distance between the ships two hours after ship C passed through A?

2. Originally Posted by duskmantle
1) Air is leaking out of a spherical balloon at the rate of 3 cubic cm/sec. When the radius is 5 cm, how fast is the surface area decreasing?
The volume when the radius is $r$ is:

$V(r)=\frac{4}{3}\pi r^3$

So:

$\frac{dV}{dt}=\frac{4}{3} \pi (3r^2)\frac{dr}{dt}$

You are told $dV/dt$ so now find $dr/dt$ when $r=5$

CB

3. Originally Posted by duskmantle
2) A cone shaped paper cup is being filled with water at the rate of 3 cubic cm/sec. The height of the cup is 10 cm and the radius of its base is 5 cm. How fast is the water level rising when the level is 4 cm?
Write the formula for the volume of water in the cup if the level is d. Now differentiate to get the rate of change of volume with time as a function of water level and the rate of change of water level. You are told the rate of change of volume, and the level, so solve for the rate of change of level.

CB

4. ok thanks
can someone help me with 3? thanks =)
and im not sure but is bumping an old post against the rules?

5. Well to start we have to calculate the distance Ship C has traveled from A, the new distance Ship B is away from A, and the distance between the two.

New distance for Ship B = $72 - 12*2 = 48 km$
New distance for Ship C = $10*2 = 20 km$
Distance between the two = $\sqrt{48^2 + 20^2} = 52$

Now we can use the Pythagorean theorem to calculate the rate of change between the two ships:

$d(a^2 + b^2 = c^2) \Rightarrow 2a\,da + 2b\,db = 2c\,dc$

Well we know the rate of change for the two ships:
$da = -12$ (this is negative because Ship B is going towards A, making the distance smaller)
$db = 10$

And we plug in what we know:
$(2 * 48 * -12) + (2 * 20 * 10) = (2 * 52 * dc)$

Then you solve for dc which is the rate of change between the two ships .

$dc = \frac{-94}{13}$