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# Cylinder - Rate of Change of Diameter

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Hi Dear Experts - Seasons Greetings & thanks for your support!!!

My question: Dec 7, 2020

#1
+2

Hi Old Timer it is good to see you again.

Let R be the outer radius at any given point in time

Ler r be the inner radius at any given point in time

Let k be the length which is a constant.

Volume is also a constant so we can work that out first.

$$Volume=(\pi R^2-\pi r^2)k\\ Volume=\pi k (R^2-r^2)\\ When\;\; R=0.05\;\; r=0.03\\ so\\ Volume=\pi k (0.0025-0.0009)\\ Volume=0.0016k\pi\\$$

Now I will need to know what

NOW:

WE need to find

$$\frac{dR}{dt}\quad when \quad \frac{dr}{dt}=0.5mm/sec$$

$$\frac{dR}{dt}=\frac{dR}{dr}\times \frac{dr}{dt}$$

So I need to find dR/dr

$$V=\pi k (R^2-r^2)\\ 0.0016k \pi =\pi k (R^2-r^2)\\ 0.0016= (R^2-r^2)\\ R^2=0.0016+r^2\\ R=(0.0016+r^2)^{0.5}\\ \frac{dR}{dr}=0.5(0.0016+r^2)^{-0.5}*2r\\ \frac{dR}{dr}=(0.0016+r^2)^{-0.5}*r\\ \frac{dR}{dr}=\frac{r}{\sqrt{0.0016+r^2}}\\$$

$$\frac{dR}{dt}=\frac{dR}{dr}\times \frac{dr}{dt}\\ \frac{dR}{dt}=\frac{r}{\sqrt{0.0016+r^2}}\times \frac{dr}{dt}\\~\\ When\qquad \frac{dr}{dt}=0.5\\ \frac{dR}{dt}=\frac{r}{\sqrt{0.0016+r^2}}\times 0.5\\~\\$$

r = 0.03   we were told that in the question.

$$\frac{dR}{dt}=\frac{0.03}{\sqrt{0.0016+0.0009}}\times 0.5\\~\\ \frac{dR}{dt}=\frac{0.03}{\sqrt{0.0025}}\times 0.5\\~\\ \frac{dR}{dt}=\frac{0.03}{0.05}\times 0.05\\~\\ \frac{dR}{dt}=0.03 mm/sec\\~\\$$

If I have not done anything careless then I think my answer is ok but maybe I turned a molehill into a mountain.... ---------------------------------------------

LaTex:

Volume=(\pi R^2-\pi r^2)k\\
Volume=\pi k (R^2-r^2)\\
When\;\; R=0.05\;\; r=0.03\\
so\\
Volume=\pi k (0.0025-0.0009)\\
Volume=0.0016k\pi\\

V=\pi k (R^2-r^2)\\
0.0016k \pi =\pi k (R^2-r^2)\\
0.0016= (R^2-r^2)\\
R^2=0.0016+r^2\\
R=(0.0016+r^2)^{0.5}\\
\frac{dR}{dr}=0.5(0.0016+r^2)^{-0.5}*2r\\
\frac{dR}{dr}=(0.0016+r^2)^{-0.5}*r\\
\frac{dR}{dr}=\frac{r}{\sqrt{0.0016+r^2}}\\

\frac{dR}{dt}=\frac{dR}{dr}\times \frac{dr}{dt}\\
\frac{dR}{dt}=\frac{r}{\sqrt{0.0016+r^2}}\times \frac{dr}{dt}\\~\\

\frac{dR}{dt}=\frac{r}{\sqrt{0.0016+r^2}}\times 0.5\\~\\

Dec 7, 2020

#1
+2

Hi Old Timer it is good to see you again.

Let R be the outer radius at any given point in time

Ler r be the inner radius at any given point in time

Let k be the length which is a constant.

Volume is also a constant so we can work that out first.

$$Volume=(\pi R^2-\pi r^2)k\\ Volume=\pi k (R^2-r^2)\\ When\;\; R=0.05\;\; r=0.03\\ so\\ Volume=\pi k (0.0025-0.0009)\\ Volume=0.0016k\pi\\$$

Now I will need to know what

NOW:

WE need to find

$$\frac{dR}{dt}\quad when \quad \frac{dr}{dt}=0.5mm/sec$$

$$\frac{dR}{dt}=\frac{dR}{dr}\times \frac{dr}{dt}$$

So I need to find dR/dr

$$V=\pi k (R^2-r^2)\\ 0.0016k \pi =\pi k (R^2-r^2)\\ 0.0016= (R^2-r^2)\\ R^2=0.0016+r^2\\ R=(0.0016+r^2)^{0.5}\\ \frac{dR}{dr}=0.5(0.0016+r^2)^{-0.5}*2r\\ \frac{dR}{dr}=(0.0016+r^2)^{-0.5}*r\\ \frac{dR}{dr}=\frac{r}{\sqrt{0.0016+r^2}}\\$$

$$\frac{dR}{dt}=\frac{dR}{dr}\times \frac{dr}{dt}\\ \frac{dR}{dt}=\frac{r}{\sqrt{0.0016+r^2}}\times \frac{dr}{dt}\\~\\ When\qquad \frac{dr}{dt}=0.5\\ \frac{dR}{dt}=\frac{r}{\sqrt{0.0016+r^2}}\times 0.5\\~\\$$

r = 0.03   we were told that in the question.

$$\frac{dR}{dt}=\frac{0.03}{\sqrt{0.0016+0.0009}}\times 0.5\\~\\ \frac{dR}{dt}=\frac{0.03}{\sqrt{0.0025}}\times 0.5\\~\\ \frac{dR}{dt}=\frac{0.03}{0.05}\times 0.05\\~\\ \frac{dR}{dt}=0.03 mm/sec\\~\\$$

If I have not done anything careless then I think my answer is ok but maybe I turned a molehill into a mountain.... ---------------------------------------------

LaTex:

Volume=(\pi R^2-\pi r^2)k\\
Volume=\pi k (R^2-r^2)\\
When\;\; R=0.05\;\; r=0.03\\
so\\
Volume=\pi k (0.0025-0.0009)\\
Volume=0.0016k\pi\\

V=\pi k (R^2-r^2)\\
0.0016k \pi =\pi k (R^2-r^2)\\
0.0016= (R^2-r^2)\\
R^2=0.0016+r^2\\
R=(0.0016+r^2)^{0.5}\\
\frac{dR}{dr}=0.5(0.0016+r^2)^{-0.5}*2r\\
\frac{dR}{dr}=(0.0016+r^2)^{-0.5}*r\\
\frac{dR}{dr}=\frac{r}{\sqrt{0.0016+r^2}}\\

\frac{dR}{dt}=\frac{dR}{dr}\times \frac{dr}{dt}\\
\frac{dR}{dt}=\frac{r}{\sqrt{0.0016+r^2}}\times \frac{dr}{dt}\\~\\

\frac{dR}{dt}=\frac{r}{\sqrt{0.0016+r^2}}\times 0.5\\~\\

Melody Dec 7, 2020
#2
+1

Very nice, Melody   !!!!!   CPhill  Dec 7, 2020
#3
+1

Thanks Chris. :)

Melody  Dec 7, 2020
#4
+1

Brilliant Melody.....In all honesty I was getting the same answer (not in such fine style!!) as your solution!

This proves there is an error in the book!!

Therefore thanks a million and I wish you all the best for the festive season and assure you I am a great fan of yours!!

Dec 8, 2020
#5
+1

Thanks Oldtimer,

(I assume this is a retirement learning activity for you)

Maybe you had an easier method?

Anyway, I wish you a great Christmas and the best of new years.

Melody  Dec 8, 2020
#6
+1 Thanks again Melody!! All the best to the Team!

Dec 11, 2020
#7
0

Thanks very much OldTimer.

And a merry Christmas to you too! Melody  Dec 16, 2020