Respuesta :
I've GOT to assume that this is a calculus problem, so it's OK for me to use it to solve the problem.
Volume of a sphere = 4/3 π R³
Rate of change of the volume = d(Vol)/dt = (4/3 π) (3R² · dR/dt)
We know the rate of change of the volume ... 4 ft³ / sec
4 ft³/sec = (4/3 π) (3R²) dR/dt
After 3 minutes (180 seconds), the volume of the balloon is
(4 ft³/sec) · (180 sec) = 720 ft³
What is the balloon's radius at that time ?
Volume = 4/3 π R³
720 ft³ = 4/3 π R³
R³ = 720 ft³ / (4/3 π)
R = ∛ (720 ft³ / 4/3π)
R = 5.56 ft
Plug this back into the equation with the rates of change:
4 ft³/sec = (4/3 π) (3R²) dR/dt
4 ft³/sec = (4/3 π) (3) · (5.56ft)² · dR/dt
dR/dt = 4 / (4/3 π · 3 · 5.56²)
I get dR/dt = 0.000103 ft/sec. Seems weird. You'd better check this.
"At what time is the radius increasing at a rate of 140 feet per second?" Well, think about it. That's a huge huge rate of expansion. So it has to be right at the beginning, when the first tiny sniff of gas enters the balloon, and the volume jumps from zero to a-little-something.
Start back at our equation for dR/dt :
4 ft³/sec = (4/3 π) (3R²) dR/dt
We want to find out what ' R ' is when dR/dt is 140 ft/sec. Then, when we know 'R', we can calculate the volume, and we'll know how long the gas has been blowing in, because we know that it comes in at 4ft³/sec.
4 ft³/sec = (4/3 π) (3R²) dR/dt
4 ft³/sec = (4/3 π) (3R²) (140 ft/sec)
3 R² = 4 / (4/3 π · 140)
R² = 4 / (4 π · 140)
R = √ (1 / 140π)
R = 0.0477 ft.
Volume = 4/3 π R³ ft³
Time = (Volume/4 ft³/sec) seconds
Time = (4/3 π R³ / 4) seconds
Time = π R³ / 3 seconds
Time = π (0.0476)³/3 sec
I get Time = 0.000114 second . You'd REALLY better check it.
The rate of change function of a function describes how the value of the function changes with the input at a specified point or interval
The required values are;
- After 3 minutes, the radius is increasing at 1.03 × 10⁻² ft./s
- The time, after start, at which the radius is increasing at 140 feet per second approximately 3.614 × 10⁻⁵ seconds
The reasons the above values are correct are given as follows;
Known parameter:
The constant rate at which helium is pumped into the spherical balloon, r = 4 ft.³/s
Required:
The rate at which the radius of the balloon is increasing after 3 minutes
We have;
Solution:
[tex]\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}[/tex]
Which gives;
[tex]\dfrac{dr}{dt} = \dfrac{ \dfrac{dV}{dt} }{\dfrac{dV}{dr} }[/tex]
The volume of the balloon after 3 minutes (180 seconds), V₁₈₀, is given as follows;
V₁₈₀ = 180 s × 4 ft.³/s = 720 ft.³
The radius of the balloon at 720 ft.³, r, is given as follows;
[tex]V = \dfrac{4}{3} \cdot \pi \cdot r^3[/tex]
[tex]{\dfrac{dV}{dr} }=4 \cdot \pi \cdot r^2[/tex]
[tex]r = \sqrt[3]{ \dfrac{3}{4 \cdot \pi} \cdot V }[/tex]
Which gives;
[tex]r_{180} = \sqrt[3]{ \dfrac{3}{4 \cdot \pi} \cdot 720 } \approx 5.56[/tex]
[tex]\dfrac{dr}{dt} = \dfrac{ \dfrac{dV}{dt} }{\dfrac{dV}{dr} }=\dfrac{4}{4 \cdot \pi \cdot r^2}[/tex]
At t = 3 minutes, (180 seconds), r ≈ 5.56 ft.
[tex]\dfrac{dr}{dt} \approx \dfrac{4}{4 \cdot \pi \cdot 5.56^2} \approx 1.03 \times 10^{-2}[/tex]
After 3 minutes, the radius is increasing at 1.03 × 10⁻² ft./s
Second part:
Required: The time at which the volume is increasing at 140 feet per second
[tex]\dfrac{dr}{dt} =140[/tex]
Therefore;
[tex]140=\dfrac{4}{4 \cdot \pi \cdot r^2}[/tex]
[tex]r = \sqrt{\dfrac{1}{\pi \times 140} }[/tex]
The volume when the radius is increasing at 140 feet per second is therefore;
[tex]V = \dfrac{4}{3} \times \left( \sqrt{\dfrac{1}{\pi \times 140} }\right)^3[/tex]
r × t = V
[tex]t = \dfrac{V}{r}[/tex]
∴ The time at which the radius is increasing at 140 feet per second, t, is given as follows;
[tex]t = \dfrac{ \dfrac{4}{3} \times \left( \sqrt{\dfrac{1}{\pi \times 140} }\right)^3}{4} \approx 3.614 \times 10^{-5}[/tex]
The time at which the radius is increasing at 140 feet per second, t ≈ 3.614 × 10⁻⁵ seconds
Learn more about rate of change here:
https://brainly.com/question/16875614
