Mathematics, 25.12.2019 03:31, natalie2sheffield

Aspherical balloon is inflated with helium at the rate of 100pi ft^3/min.
a) how fast is the balloon's radius increasing at the instant the radius is 5 ft?
b) how fast is the surface area increasing at that instant?
the surface area is increasing at a rate of ft^2/min at the instant the radius is 5 ft.

Answers: 2

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answered: ashtonrieper2721

The rate at which the balloon's radius increases at the instant the radius is 5 ft is 1ft/min

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:

$V =\frac{4}{3} \pi r^3$

r is the radius of the spherical balloon = 5 ft

A) The rate at which the balloon is increasing is expressed as:

$\frac{dV}{dt} = \frac{dV}{dr}\cdot \frac{dr}{dt}$

$\frac{dV}{dr} = 4 \pi r^2\\ \frac{dV}{dt}= 4 \pi r^2\frac{dr}{dt}$

Substitute the given parameters:

$100 = 4 \pi (5)^2 \cdot \frac{dr}{dt}\\100 = 100 \pi \cdot \frac{dr}{dt}\\1 \pi= \pi \cdot \frac{dr}{dt}\\ \frac{dr}{dt} = 1ft/min$

Hence the rate at which the balloon's radius is increasing at the instant the radius is 5 ft is 1ft/min

B) The formula for finding the surface area of the spherical balloon is expressed as;

$S = 4 \pi r^2$

$\frac{dS}{dt} = \frac{dS}{dr} \cdot \frac{dr}{dt} \\\frac{dS}{dt} = 8 \pi r \cdot \frac{dr}{dt} \\\frac{dS}{dt} = 8 \pi (5) \cdot 1 \\\frac{dS}{dt} =40 \pi ft^2/min$

This shows that the surface area is increasing at a rate of 40π ft²/min at the instant the radius is 5 ft.

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answered: lilymessina94

Answers:

A) 1 ft/min

B) 40pi ft^2/min

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

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.

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.

answered: Guest

One shot. if 60% of 10 is 6, then what is 70%? we know it would be seven (since 60% is mathematically 60/100, or 6/10, then 70% is 70/100, or 7/10) but we can check anyway using ratios: 6 : 60% = x : 70%*inverse multiplicationx * 60% = 6 * 70%*divide both sides by 60%x = (6 * 70%) / 60%x = 7and since he has made 6 shots already, he would need just one more to get 7 shots or 70%. if the explanation is too confusing, comment and i will try to

answered: Guest

range =5

step-by-step explanation:

to find the range, we take the largest value and subtract the smallest value.

the largest value is 5

the smallest value is 0

the range is 5-0 =5

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Aspherical balloon is inflated with helium at the rate of 100pi ft^3/min.
a) how fast is the...