Math Help - Multi-Calc Help

1. Multi-Calc Help

Suppose there are all these bubbles, where each bubble is shaped as a different type of surface. One of the bubbles is shaped as a circular cylinder, and it will keep that general shape over time, although the height increases at a rate of 3 cm/sec while the radius will decrease at a rate of 1 cm/sec (and when the bubble gets thin enough it will eventually pop).

1.) What rate is the volume, V, of this bubble changing when it is 10 cm high and 6 cm across at the bottom?

2.) What rate is the surface area of the bubble changing at the same instant?

2. Originally Posted by Ideasman
Suppose there are all these bubbles, where each bubble is shaped as a different type of surface. One of the bubbles is shaped as a circular cylinder, and it will keep that general shape over time, although the height increases at a rate of 3 cm/sec while the radius will decrease at a rate of 1 cm/sec (and when the bubble gets thin enough it will eventually pop).

1.) What rate is the volume, V, of this bubble changing when it is 10 cm high and 6 cm across at the bottom?

2.) What rate is the surface area of the bubble changing at the same instant?
1. V = (pi)r^2*h

dV/dt = (pi)*2rh*dr/dt + (pi)r^2*dh/dt

Just plug in the numbers.

2. SA = 2(pi)r^2 + 2(pi)r*h

d(SA)/dt = 2(pi)*2r*dr/dt + 2(pi)r*dh/dt

-Dan

3. Originally Posted by topsquark
1. V = (pi)r^2*h

dV/dt = (pi)*2rh*dr/dt + (pi)r^2*dh/dt

Just plug in the numbers.

2. SA = 2(pi)r^2 + 2(pi)r*h

d(SA)/dt = 2(pi)*2r*dr/dt + 2(pi)r*dh/dt

-Dan
Thanks, topsquark.

So using the info. given, it would be:

Pi*2*3*10*1 + Pi*3^2*3 = 60*Pi + 27*Pi = 87*Pi and so it is changing at a rate of 87*Pi cm/sec ?

I have the 3 in bold since when the prob. says "6 cm across" I assume it means the diameter, and therefore the radius is 3.

4. Originally Posted by Ideasman
Thanks, topsquark.

So using the info. given, it would be:

Pi*2*3*10*1 + Pi*3^2*3 = 60*Pi + 27*Pi = 87*Pi and so it is changing at a rate of 87*Pi cm/sec ?

I have the 3 in bold since when the prob. says "6 cm across" I assume it means the diameter, and therefore the radius is 3.
The radius is DECREASING at a rate of 1 cm/s, so dr/dt = -1 cm/s. The correct answer is:

Pi*2*3*10*(-1) + Pi*3^2*3 = -60*Pi + 27*Pi = -23*Pi (cm^3/s)
so the volume is decreasing at a rate of 23(pi) cm^3/s. (Beware the units!)

-Dan