Asked by Craig
t: 0 2 5 7 11 12
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?
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?
Answers
Answered by
Craig
sorry 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
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
Answered by
Craig
i really don't know how to start this problem
Answered by
Reiny
I 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
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
Answered by
Craig
i'm guessing that twice differentiable means y''
Answered by
Reiny
agree, 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?
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?
Answered by
Damon
It 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
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
Answered by
Damon
V = 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.
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.