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

Show answers

Answers

Answer from: 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.

Learn more here: link

Answer from: 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.

Answer from: Quest

He will need 2,898 pieces of paper

Answer from: Quest

step-by-step explanation:

because 1/3 is a fraction. we can tell because there is a divisor symbol between the 1 and the 3. if we were to add 2 fractions together, we would add the numerators, which is the number on top, but not the denominators, which is the number on the bottom

1/3+1/3=2/3

hope this !

Another question on Mathematics

Mathematics, 21.06.2019 14:00

Aright triangle has an area of 18 square inches. b: if the triangle is not an isosceles triangle, what are all the possible lengths of the legs, if the lengths are whole numbers. answer !

Answers: 1

Answer

Mathematics, 21.06.2019 16:20

7.(03.01 lc)which set represents the range of the function shown? {(-1, 5), (2,8), (5, 3), 13, -4)} (5 points){-1, 2, 5, 13){(5, -1), (8, 2), (3,5), (-4, 13)){-4, 3, 5, 8}{-4, -1, 2, 3, 5, 5, 8, 13}

Answers: 3

Answer

Mathematics, 21.06.2019 19:30

Sundar used linear combination to solve the system of equations shown. he did so by multiplying the first equation by 5 and the second equation by another number to eliminate the y-terms. what number did sundar multiply the second equation by? 2x+9y=41 3x+5y=36

Answers: 1

Answer

Mathematics, 21.06.2019 19:30

Select the margin of error that corresponds to the sample mean that corresponds to each population: a population mean of 25, a standard deviation of 2.5, and margin of error of 5%

Answers: 2

Answer

You know the right answer?

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

Questions

Mathematics, 27.11.2019 21:31

Aball is thrown into the air, and its position is given by h(t) = −4.9t2 + 62t + 4 m, where t is the time in seconds since the ball was thrown. find t...