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August 21, 2014

August 21, 2014

Posted by **Craig** on Tuesday, November 25, 2008 at 7:54pm.

r'(t): 5.7 4.0 2.0 1.2 0.6 0.5

The volume of a spherical hot air balloon expands as the air inside the balloon is heated. The radius of the balloon in feet is modeled by a twice-differentiable function of r of time t, where t is measured in minutes. For 0<t<12, the graph of r is concave down. The table above give selected values of the rate of change, r'(t), of the radius of the balloon over the time interval 0<t<12. The radius of the balloon is 30ft when t=5.

a)est. radius of balloon when t=5.4 using tan line approximated at t=5. Why?

b)Find rate of change of the volume of the balloon w/ respect to time when t=5. what is the units of measure?

- Calculus -
**Craig**, Tuesday, November 25, 2008 at 7:56pmsorry about that: when t=0, r't=5.7;

t=2, r't=4; t=5, r't=2; t=7, r't=1.2; t=11, r't=.6; t=12, r't=.5

- please help!!! -
**Craig**, Tuesday, November 25, 2008 at 9:55pmi really don't know how to start this problem

- Calculus -
**Reiny**, Tuesday, November 25, 2008 at 10:49pmI have not answered your question since I am somewhat puzzled by the wording of your question.

by <<The radius of the balloon in feet is modeled by a twice-differentiable function of r of time t>> do you mean that r is a quadratic function of t, a cubic or what?

As I see it, I could differentiate a function any number of times, just because I might reach a derivative of zero does not mean I couldn't do it once more.

e.g y = x^3 + ...

y' = 3x^2 ....

y'' = 6x ....

y''' = 6

y'''' = 0

y ''''' = 0

- Calculus -
**Craig**, Tuesday, November 25, 2008 at 11:00pmi'm guessing that twice differentiable means y''

- Calculus -
**Reiny**, Tuesday, November 25, 2008 at 11:33pmagree, but what level is the function?

e.g. If I knew that y = ... is quadratic, I could say y = at^2 + bt + c and go from there, but how do I know it's not a cubic?

- Calculus -
- Calculus -
**Damon**, Wednesday, November 26, 2008 at 6:23amIt does not matter I think.

We are not fitting the entire data set with a function but doing linear, then quadratic, interpolation.

For part a use radius at 5 and dr/dt at 5

r(5.4) = r(5) + .4 r'(5)

r(5.4) = 30 + .4 (2)

r(5.4) = 30 + .8

r(5.4) = 30.8

- Calculus -
**Damon**, Wednesday, November 26, 2008 at 6:29amV = Volume = (4/3) pi r^3 in ft^3

DV/dt = (4/3) pi (3) r^2 dr/dt in ft^3/sec

at t = 5

dV/dt = 4 pi r^2 (2)

= 8 pi (30)^2 = 240 pi ft^3/second

NOTE - the rate of change of volume is the surface area, 4 pi r^2 times the rate of change of radius - think about it.

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