# Multi-Calc Help

• Apr 14th 2007, 04:00 AM
Ideasman
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?
• Apr 14th 2007, 04:06 AM
topsquark
Quote:

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
• Apr 15th 2007, 01:20 PM
Ideasman
Quote:

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.
• Apr 16th 2007, 05:37 AM
topsquark
Quote:

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