sphere's surface area changes as volume changes

[7777ft]

This question comes from a textbook.
A spherical balloon is increasing in volume. If, when its radius is [MATH]r[/MATH] feet, its volume is increasing at the rate of 4 cubic feet per second, at what rate is its surface then increasing? Textbook's answer is "At the rate of [MATH]8\sqrt{\pi}[/MATH] square feet per second."

Part 1: differentiate the volume to find how fast the radius is changing.
[MATH] V=\frac{4}{3}\pi r^3\\ \begin{align*} \frac{dV}{dt}&=\frac{dV}{dr}*\frac{dr}{dt}\\ &=4\pi r^2*\frac{dr}{dt} \end{align*} [/MATH]The volume is increasing at 4 cubic feet per second, therefore [MATH]\frac{dV}{dt}=4\frac{ft^3}{sec}[/MATH].
[MATH] 4=4\pi r^2*\frac{dr}{dt}\\ \frac{dr}{dt}=\frac{1}{\pi r^2}\\ [/MATH]Part 2: how fast is the surface area changing relative to radius' change.
[MATH] A=4\pi r^2\\ \begin{align*} \frac{dA}{dt}&=\frac{dA}{dr}*\frac{dr}{dt}\\ &=8\pi r*\frac{dr}{dt}\\ &=8\pi r*\frac{1}{\pi r^2}\\ &=\frac{8}{r}\frac{ft^2}{sec} \end{align*} [/MATH]My answer differs from the textbook. Why did I go wrong?

 

[MarkFL]

I think I would begin with:

[MATH]V=A\left(\frac{r}{3}\right)[/MATH]
Hence:

[MATH]\frac{dV}{dt}=\frac{dA}{dt}\cdot\frac{r}{3}+A\cdot\frac{1}{3}[/MATH]
[MATH]12=\frac{dA}{dt}r+A[/MATH]
[MATH]\frac{dA}{dt}=\frac{12-A}{r}[/MATH]
Now, using your result, we find:

[MATH]A=4\implies r=\frac{1}{\sqrt{\pi}}[/MATH]
Hence:

[MATH]\frac{dA}{dt}=8\sqrt{\pi}[/MATH]

 

[7777ft]

[MATH]A=4\implies r=\frac{1}{\sqrt{\pi}}[/MATH]
Hence:

[MATH]\frac{dA}{dt}=8\sqrt{\pi}[/MATH]

Is [MATH]A=4[/MATH] proper? [MATH]A[/MATH] is square feet, and [MATH]4[/MATH] is cubic feet per second. Equating different types is fudging. The resulting [MATH]8\sqrt{\pi}[/MATH] could be a "happy accident," as Bob Ross used to say.

 

[MarkFL]

The result:

[MATH]A=4[/MATH]
Comes from equating:

[MATH]12-A=8[/MATH]
From:

[MATH]\d{A}{t}=\frac{12-A}{r}=\frac{8}{r}[/MATH]
They must be dimensionally equivalent.

 

[topsquark]

Is [MATH]A=4[/MATH] proper? [MATH]A[/MATH] is square feet, and [MATH]4[/MATH] is cubic feet per second. Equating different types is fudging. The resulting [MATH]8\sqrt{\pi}[/MATH] could be a "happy accident," as Bob Ross used to say.

Ahhhh! Happy trees! ?

-Dan

 

[7777ft]

I think I would begin with:

[MATH]V=A\left(\frac{r}{3}\right)[/MATH]
Hence:

[MATH]\frac{dV}{dt}=\frac{dA}{dt}\cdot\frac{r}{3}+A\cdot\frac{1}{3}[/MATH]

Differentiate [MATH]V=A\left(\frac{r}{3}\right)[/MATH] with respect to time using product rule
\begin{align*}
\frac{dV}{dt} &=A\cdot\frac{d}{dt}\left(\frac{r}{3}\right)+\frac{r}{3}\cdot\frac{d}{dt}A\\
&=A\cdot\frac{1}{3}\cdot\frac{dr}{dt}+\frac{r}{3}\cdot\frac{dA}{dt}
\end{align*}

How come my result is has a [MATH]\frac{dr}{dt}[/MATH] and yours doesn't? When [MATH]\frac{d}{dr}\left(\frac{r}{3}\right) = \frac13[/MATH], and so there is no [MATH]\frac{dr}{dt}[/MATH]. However implicit differentiation of [MATH]\frac{d}{dt}\left(\frac{r}{3}\right) = \frac13\cdot\frac{dr}{dt}[/MATH]. And wouldn't that change stuff that follows?

 

[MarkFL]

Yes, I should have written:

[MATH]\frac{dV}{dt}=\frac{dA}{dt}\cdot\frac{r}{3}+A\cdot\frac{1}{3}\frac{dr}{dt}[/MATH]
And this leads to:

[MATH]\frac{dA}{dt}=\frac{8}{r}[/MATH]
This is what you had,

Now, consider that:

[MATH]\frac{dV}{dr}=A[/MATH]
[MATH]\frac{dV}{dt}=4\pi r^2[/MATH]
And this implies:

[MATH]r^2=\frac{1}{\pi}[/MATH]
Or:

[MATH]r=\frac{1}{\sqrt{\pi}}[/MATH]
Hence:

[MATH]\frac{dA}{dt}=8\sqrt{\pi}[/MATH]