205.23 related rates sphere volume

[karush]

$\tiny {205.23} $
$\text{ The volume }\displaystyle V=\frac{4}{3}\pi{r}^{3}
\text{ of a spherical balloon changes with the radius.} $
$\text{a) at what rate } \displaystyle \frac{ft^3}{ft} \\$
$\text{does the volume change with respect to the radius when $r=2 ft.$} $

$$\d{V}{f}=4\pi{r}^{2}\d{r}{f}\implies\d{V}{r}=4\pi{r}^{2} $$
$\text{at $r=2ft $ then} $
$$\d{V}{r}=4\pi{2}^{2} =16\pi \frac{ft^3}{ft} $$

$\text{b)how much does the volume increase when the}$
$\text{radius increases from $2$ to $2.3 ft$}$

$\displaystyle 4\pi(2.3^2)-16\pi\approx16.21 \frac{ft^3}{ft} $
$\text{so far ??}$

 

[MarkFL]

The units you are using for the rate of change of the volume should be the cube of a linear measure divided by a time measure, such as ft³ / s. If we choose to use seconds as the time measure, then:

\(\displaystyle \d{V}{t}=4\pi r^2\,\frac{\text{ft}^3}{\text{s}}\)

Hence:

\(\displaystyle \left.\d{V}{t}\right|_{r=2}=4\pi (2)^2\,\frac{\text{ft}^3}{\text{s}}=16\pi\,\frac{\text{ft}^3}{\text{s}}\)

To find the exact change in volume when the radius increases from $r=2$ to $r=2.3$, we could simply write:

\(\displaystyle \Delta V=V(2.3)-V(2)=\frac{4}{3}\pi(2.3^3-2^3)\text{ ft}^3=\frac{4}{3}\pi\cdot\frac{4167}{1000}\text{ ft}^3=\frac{1389}{250}\pi\text{ ft}^3\)

If we wish to use a linear approximation, we could write:

\(\displaystyle \d{V}{r}=4\pi r^2\)

\(\displaystyle \Delta V\approx\d{V}{r}\Delta r=4\pi r^2\Delta r\)

With $r=2$ and $\Delta r=\frac{3}{10}$, we obtain:

\(\displaystyle \Delta V\approx\frac{24}{5}\pi\text{ ft}^3\)

 

[karush]

View attachment 6071

I thought so too but the problem actually says
\(\displaystyle \frac{ft^3}{ft}\) don't see any time unit given

 

[karush]

$\tiny {205.23} $
$\text{ The volume }\displaystyle V=\frac{4}{3}\pi{r}^{3}
\text{ of a spherical balloon changes with the radius.} $
$\text{a) at what rate } \displaystyle \frac{ft^3}{ft} \\$
$\text{does the volume change with respect to the radius when $r=2 ft.$} $
$\d{V}{f}=4\pi{r}^{2}\d{r}{f}\implies\d{V}{r}=4\pi{r}^{2} $
$\text{at $r=2ft $ then} $
$$\d{V}{r}=4\pi{(2)}^{2} =16\pi \frac{ft^3}{ft} $$
$\text{b)how much does the volume increase when the}$
$\text{radius increases from $2$ to $2.3 ft$}$
$\displaystyle V(2.3)-V(2) \approx 17.46 {ft}^{3} $

 

[MarkFL]

At first I thought the problem as given had a typo, but now I believe I understand what is intended. We are just to consider a change in the radius, not a change in the radius with respect to time. I apologize for the confusion.

 

[karush]

quite understandable, normally I'm sure its related to time

but your post still was very helpful, as always giving clearer insight to what seems so daunting at times. which is missing to often in the class room.,

 

[HallsofIvy]

$\tiny {205.23} $
$\text{ The volume }\displaystyle V=\frac{4}{3}\pi{r}^{3}
\text{ of a spherical balloon changes with the radius.} $
$\text{a) at what rate } \displaystyle \frac{ft^3}{ft} \\$
$\text{does the volume change with respect to the radius when $r=2 ft.$} $

$$\d{V}{f}=4\pi{r}^{2}\d{r}{f}\implies\d{V}{r}=4\pi{r}^{2} $$
$\text{at $r=2ft $ then} $
$$\d{V}{r}=4\pi{2}^{2} =16\pi \frac{ft^3}{ft} $$

Yes, that is the correct rate of change of volume with respect to the radius.

$\text{b)how much does the volume increase when the}$
$\text{radius increases from $2$ to $2.3 ft$}$

$\displaystyle 4\pi(2.3^2)-16\pi\approx16.21 \frac{ft^3}{ft} $
$\text{so far ??}$

No, that is how much the rate of volume increase increases!

When r= 2, the volume is $$(4/3)\pi r^3= (4/3)\pi (2^3)= (4/3)\pi (8)= (32/3)\pi \approx 25.13$$ cubic feet. When r= 2.3 the volume is $$(4/3)\pi r^3= (4/3)\pi (2.3^3)= (4/3)\pi (12.162)\approx 50.96$$ cubic feet. The volume increases by approximately 50.96- 25.13= 25.82 cubic feet.