Posted by Craig on Tuesday, November 25, 2008 at 7:54pm.
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?
Responses
* Calculus - Craig, Tuesday, November 25, 2008 at 7:56pm
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
* please help!!! - Craig, Tuesday, November 25, 2008 at 9:55pm
i really don't know how to start this problem
* Calculus - Reiny, Tuesday, November 25, 2008 at 10:49pm
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
* Calculus - Craig, Tuesday, November 25, 2008 at 11:00pm
i'm guessing that twice differentiable means y''
o Calculus - Reiny, Tuesday, November 25, 2008 at 11:33pm
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?
* Calculus - Damon, Wednesday, November 26, 2008 at 6:23am
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
* Calculus - Damon, Wednesday, November 26, 2008 at 6:29am
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.
Sorry about the spelling :(
@ Damon on part B
You made an error when squaring 30, it seems like you didn't square. the answer would be 7200pi ft^3/min
To estimate the radius of the balloon when t=5.4 using the tangent line approximation at t=5, you can use linear interpolation. This involves finding the equation of the line tangent to the graph at t=5 and using it to estimate the radius at t=5.4.
1. Find the slope of the tangent line at t=5:
The slope of the tangent line is equal to the rate of change of the radius, r'(t), at t=5. From the table, the value of r'(t) at t=5 is 2.
2. Find the radius at t=5:
The given information states that the radius of the balloon is 30 ft when t=5.
3. Write the equation of the tangent line:
Using the slope-intercept form, the equation of the tangent line is:
r - r(5) = r'(5)(t - 5)
Plugging in the values:
r - 30 = 2(t - 5)
r - 30 = 2t - 10
r = 2t + 20
4. Estimate the radius at t=5.4:
Plug in t=5.4 into the equation of the tangent line:
r = 2(5.4) + 20
r ≈ 31.8 ft
Therefore, the estimated radius of the balloon when t=5.4 using the tangent line approximation at t=5 is approximately 31.8 feet.
To find the rate of change of the volume of the balloon with respect to time when t=5, you can use the volume formula for a sphere and differentiate it with respect to time.
1. Volume formula for a sphere:
The volume V of a sphere is given by the formula:
V = (4/3)πr^3, where r is the radius of the sphere.
2. Find the rate of change of volume with respect to time:
Differentiate the volume formula with respect to time (t):
dV/dt = (4/3)π(3r^2)(dr/dt)
3. Plug in the given values:
From the table, the value of r'(t) at t=5 is 2, and the radius of the balloon when t=5 is 30.
r = 30, dr/dt = 2
dV/dt = (4/3)π(3(30)^2)(2)
= (4/3)π(3)(900)(2)
= 240π ft^3/second
Therefore, the rate of change of the volume of the balloon with respect to time when t=5 is 240π ft^3/second.