Respuesta :
Answers:
A) 1 ft/min
B) 40pi ft^2/min
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Work Shown:
We'll use these two equations
Surface area of sphere = S = 4*pi*r^2
Volume of sphere = V = (4/3)pi*r^3
along with this given info
dV/dt = rate of change of volume
dV/dt = 100pi cubic feet per minute
r = 5 ft is the radius
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Part A
V = (4/3)pi*r^3
dV/dt = 3*(4/3)pi*r^2*dr/dt ... apply derivative
dV/dt = 4pi*r^2*dr/dt
100pi = 4pi*5^2*dr/dt ... substitution
100pi = 4pi*25*dr/dt
100pi = 100pi*dr/dt
1 = 1*dr/dt ... divide both sides by 100pi
1 = dr/dt
dr/dt = 1
When the radius is 5 feet, the radius is increasing at 1 foot per minute.
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Part B
S = 4*pi*r^2
dS/dt = 2*4*pi*r*dr/dt
dS/dt = 8*pi*r*dr/dt
dS/dt = 8*pi*5*1
dS/dt = 40pi
When the radius is 5 feet, the surface area is increasing at a rate of 40pi square feet per minute.
Both rates for parts A and B are instantenous rates of change.
The rate at which the balloon's radius increases at the instant the radius is 5 ft is [tex]1ft/min[/tex]
The surface area is increasing at a rate of 40π ft²/min at the instant the radius is 5 ft.
The formula for calculating the volume of the spherical balloon is expressed as:
[tex]V =\frac{4}{3} \pi r^3[/tex]
r is the radius of the spherical balloon = 5 ft
A) The rate at which the balloon is increasing is expressed as:
[tex]\frac{dV}{dt} = \frac{dV}{dr}\cdot \frac{dr}{dt}[/tex]
[tex]\frac{dV}{dr} = 4 \pi r^2\\ \frac{dV}{dt}= 4 \pi r^2\frac{dr}{dt}[/tex]
Substitute the given parameters:
[tex]100 = 4 \pi (5)^2 \cdot \frac{dr}{dt}\\100 = 100 \pi \cdot \frac{dr}{dt}\\1 \pi= \pi \cdot \frac{dr}{dt}\\ \frac{dr}{dt} = 1ft/min[/tex]
Hence the rate at which the balloon's radius is increasing at the instant the radius is 5 ft is [tex]1ft/min[/tex]
B) The formula for finding the surface area of the spherical balloon is expressed as;
[tex]S = 4 \pi r^2[/tex]
[tex]\frac{dS}{dt} = \frac{dS}{dr} \cdot \frac{dr}{dt} \\\frac{dS}{dt} = 8 \pi r \cdot \frac{dr}{dt} \\\frac{dS}{dt} = 8 \pi (5) \cdot 1 \\\frac{dS}{dt} =40 \pi ft^2/min[/tex]
This shows that the surface area is increasing at a rate of 40π ft²/min at the instant the radius is 5 ft.
Learn more here: https://brainly.com/question/10211249
