Question on rates of change problem

[krackers]

Homework Statement

A spherical balloon is being inflated. Suppose the radius of the balloon is increasing at a rate of 2 centimeters per second.

a. Express the radius r of the balloon as a function of time t.
b. Express the volume V of the balloon as a function of time t

Homework Equations

$\frac{dr}{dt}\; =\; 2$

The Attempt at a Solution

Part A.

$\frac{dr}{dt}\; =\; 2$

Therefore, $dr=2\cdot dt$

$\int_{}^{}{}dr=\int_{}^{}{}2\cdot dt$

$r\left( t \right)=2t$Part B.

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

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

Substituting dr/dt = 2 gives:

$\frac{dV}{dt}\; =\; 8\pi r^{2}$

Since $r=2t$

$\frac{dV}{dt}=32\pi \cdot t^{2}$

$dV=32\pi \cdot t^{2}\cdot dt$

$V=\frac{32\pi \cdot t^{3}}{3}$

 

[verty]

Hint: what is the radius at time t = 0?

 

[krackers]

That would also be zero, right?

 

[STEMucator]

Your solution looks fine to me.

 

[1MileCrash]

I don't see anything wrong either, and you wrote your solution very nicely. What's your question, or were you just not confident that it was correct?

 

[krackers]

I wasn't sure if that was correct. This was actually a precalc problem, so I was not sure if integration/differentiation was needed and whether my answer was correct.

 

[haruspex]

I was not sure if integration/differentiation was needed

You're right, it isn't. It's sufficiently obvious that the radius is simply 2t cm after t seconds, so V = 4πr³/3 = 4π(2t)³/3.

 

[verty]

$\int_{}^{}{}dr=\int_{}^{}{}2\cdot dt$

$r\left( t \right)=2t$

This second line does not follow from the first without the additional assumption, nowhere written and not given in the question, that f(0) = 0. The question says, express the radius as "a" function of time t. The answer is correct but what is written here is not. I'm picking out the literal fallacy here.

 

[1MileCrash]

This second line does not follow from the first without the additional assumption, nowhere written and not given in the question, that f(0) = 0. The question says, express the radius as "a" function of time t. The answer is correct but what is written here is not. I'm picking out the literal fallacy here.

Good point.

An '$r_{0}$' should appear and be defined.

 

[krackers]

You're right, it isn't. It's sufficiently obvious that the radius is simply 2t cm after t seconds, so V = 4πr³/3 = 4π(2t)³/3.

That's a lot simpler.

Good point.

An '$r_{0}$' should appear and be defined.

You're right - I always forget to add the constant after anti-differentiation.