+0  
 
+1
87
7
avatar+193 

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

 

My question:

 

 

Answer : 0.5mm/sec

 

Kindly advise thanks.

 Dec 7, 2020

Best Answer 

 #1
avatar+112000 
+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.... frown

 

 

---------------------------------------------

 

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}\\~\\
When\qquad  \frac{dr}{dt}=0.5\\

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

 Dec 7, 2020
 #1
avatar+112000 
+2
Best Answer

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.... frown

 

 

---------------------------------------------

 

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}\\~\\
When\qquad  \frac{dr}{dt}=0.5\\

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

Melody Dec 7, 2020
 #2
avatar+114096 
+1

Very nice, Melody   !!!!!

 

 

cool cool cool

CPhill  Dec 7, 2020
 #3
avatar+112000 
+1

Thanks Chris. :)

Melody  Dec 7, 2020
 #4
avatar+193 
+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
avatar+112000 
+1

Thanks Oldtimer,

 

I am fond of you too and I admire your perseverance.

(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
avatar+193 
+1

 

 

Thanks again Melody!! All the best to the Team!

 Dec 11, 2020
 #7
avatar+112000 
0

Thanks very much OldTimer.

And a merry Christmas to you too!

 

Melody  Dec 16, 2020

26 Online Users

avatar
avatar
avatar