Question

You are blowing air into a spherical balloon at a rate of (4*pi)/3 cubic inches per second. Assume the radius of your balloon is zero at time zero. Let r(t), A(t), and V(t) denote the radius, the surface area, and the volume of your balloon at time t, respectively. (Assume the thickness of the skin is zero.) All of your answers below are expressions in t:
r'(t) = ? inches per second,
A'(t) = ? square inches per second, and
V'(t) = ? cubic inches per second.
Hint: The surface area A and the volume V of a sphere of radius r are given by
A = 4*pi*r^2 V = (4*pi*r^3)/3

Answer 1

V(t) is the volume of the balloon at time t

V'(t) is the rate of change of the volume of the balloon at time t. Basically how fast/slow the balloon is increasing/decreasing in volume

Because we are told that "You are blowing air into a spherical balloon at a rate of (4*pi)/3 cubic inches per second" this means that the balloon is increasing in volume at a rate of (4*pi)/3 cubic inches per second. So V'(t) = 4pi/3 for all defined t values.

Answer 2

makes sense.

Answer 3

@jim_thompson5910 could come to my queastion please

Answer 4

we know the volume of any sphere is

\[\Large V = \frac{4}{3}\pi r^3 \]

what happens when we differentiate both sides with respect to t ?

Answer 5

is it simply dt/dr = 4/3*pi*r^3 , since there is no t value in the actual equation? I'm not quite sure actually.

Answer 6

it should be

\[\Large V = \frac{4}{3}\pi r^3\]

\[\Large \frac{dV}{dt} = \frac{4}{3}*3*\pi r^2*\frac{dr}{dt}\]

\[\Large \frac{dV}{dt} = 4\pi r^2*\frac{dr}{dt}\]

Answer 7

dV/dt is the same as V'(t)

dr/dt is the same as r'(t)

Answer 8

recall that above, we found that dV/dt was 4pi/3, so let's plug that in

\[\Large \frac{dV}{dt} = 4\pi r^2*\frac{dr}{dt}\]

\[\Large \frac{4\pi}{3} = 4\pi r^2*\frac{dr}{dt}\]

what do you get when you isolate dr/dt ?

Answer 9

1/(3r^2) = dr/dt

Answer 10

good,

\[\Large \frac{dr}{dt} = \frac{1}{3r^2}\]

\[\Large r \ '(t) = \frac{1}{3r^2}\]

Answer 11

so we can see that the speed of the radius depends on what the radius is

Answer 12

the larger the radius, the slower it grows (r gets larger, 1/(3r^2) gets smaller)

Answer 13

Finally, the surface area A(t) can be found using this formula

\[\Large A = 4\pi*r^2\]

derive both sides with respect to t to get

\[\Large \frac{dA}{dt} = 4\pi*2*r^1*\frac{dr}{dt}\]

\[\Large \frac{dA}{dt} = 8\pi r*\frac{dr}{dt}\]

\[\Large \frac{dA}{dt} = 8\pi r* \left(\frac{1}{3r^2}\right)\]

\[\Large \frac{dA}{dt} = \frac{8\pi}{3r}\]

the surface area rate of change depends on the radius. As the radius gets larger, the rate of change of the surface area gets smaller

Answer 14

and again dA/dt is the same as A'(t), right?

Answer 15

correct, two ways of saying the same thing

Answer 16

Alright, but my answers can't contain the value r, only t, so how would I express that?

Answer 17

that's the issue I initially ran into: how to express r(t) as a function of t

Answer 18

let me think for a sec

Answer 19

have you learned about integration?

Answer 20

i'm not sure, maybe.

Answer 21

I guess if you haven't learned about integrals, then it's not possible to find r(t) in terms of t

Answer 22

so I probably did, since they would not give me a question to do that I can't answer.

Answer 23

however, this is how you do it

\[\Large \frac{dr}{dt} = \frac{1}{3r^2}\]

\[\Large 3r^2 dr = 1*dt\]

\[\Large \int 3r^2 dr = \int 1*dt\]

\[\Large r^3 = t+C\]

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At time t = 0, the radius is 0. So r = 0 and t = 0. Solve for C


\[\Large r^3 = t+C\]

\[\Large 0^3 = 0+C\]

\[\Large 0 = C\]

\[\Large C = 0\]


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Plug in C = 0 and solve for r


\[\Large r^3 = t+C\]

\[\Large r^3 = t+0\]

\[\Large r^3 = t\]

\[\Large r = \sqrt[3]{t}\]


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So the function r(t) is \[\Large r(t) = \sqrt[3]{t}\]

Answer 24

Now that you know r(t), you can plug that into r of r'(t) and A'(t) to get functions in terms of t (instead of r)

Answer 25

so for r'(t) it would be dr/dt = 1/(3*(t^(1/3))^2. Is this correct?

Answer 26

which can be written as

\[\Large \frac{dr}{dt} = \frac{1}{3t^{2/3}}\]

Answer 27

and A'(t) would be (8 pi)/(3 t^(1/3)) , right?

Answer 28

correct

Answer 29

Awesome! Thank you very much.

Answer 30

you're welcome