# Help !!! integration/differentiation

• Apr 29th 2008, 02:41 PM
Chez_
Help !!! integration/differentiation
(Doh)
Hey, thanks for looking.
I have this q/
A cylindrical tank, open at the top,has height hm and raduis rm.Its capacity is 1m^3.
i) show that h=1/pie r^2. Which i did..1m^3=pie r^2 multiply by h, 1m^3-pie r^2=h, 1mpier^2=h, then finally 1/pier^2=h. Is this the correct workin?

ii) Also, its total internal surface area is Sm^2. show that S=2/r +pier^2. Which iv done as,h multiplied by 2pier=2pie2h, pier^2+2pierh =pier^2+2pier.. but not sure where from here..

iii) finally...Determine the value of r which makes the surface area S as small as possible..For this i need to make dy/dx and the 2nd diff, or integration..im not sure, either way could you show me as not sure how to do it with fractions..
thank you
• Apr 29th 2008, 03:11 PM
icemanfan
Quote:

Originally Posted by Chez_
(Doh)
Hey, thanks for looking.
I have this q/
A cylindrical tank, open at the top,has height hm and raduis rm.Its capacity is 1m^3.
i) show that h=1/pie r^2. Which i did..1m^3=pie r^2 multiply by h, 1m^3-pie r^2=h, 1mpier^2=h, then finally 1/pier^2=h. Is this the correct workin?

ii) Also, its total internal surface area is Sm^2. show that S=2/r +pier^2. Which iv done as,h multiplied by 2pier=2pie2h, pier^2+2pierh =pier^2+2pier.. but not sure where from here..

iii) finally...Determine the value of r which makes the surface area S as small as possible..For this i need to make dy/dx and the 2nd diff, or integration..im not sure, either way could you show me as not sure how to do it with fractions..
thank you

For part 3, given $S = \frac{2}{r} + \pi{r^2}$, you want to minimize S. So we start by calculating dS/dr: $\frac{dS}{dr} = -\frac{2}{r^2} + 2\pi{r}$. So when is this equal to zero?
• Apr 29th 2008, 03:36 PM
Chez_
so if this =o then i need to find what r=so i need to rearrange...
How can this be re arranged to get r= ?
• Apr 29th 2008, 03:42 PM
icemanfan
Quote:

Originally Posted by Chez_
so if this =o then i need to find what r=so i need to rearrange...
How can this be re arranged to get r= ?

Fairly easily:

$2\pi{r} = \frac{2}{r^2}$

$\pi{r} = \frac{1}{r^2}$

$\pi{r^3} = 1$

$r^3 = \frac{1}{\pi}$
• Apr 29th 2008, 03:59 PM
Chez_
Following an eg, of 3-48/x^2=o ,48/x^2=3 then x^2=48/3 x=plus minus 4.....i did r^3=1/pie =0 , r=1/pie^3 but then it would be the third root of pie, which is 0.0322... but the answer sheet says r=0.683
• Apr 29th 2008, 04:03 PM
icemanfan
Quote:

Originally Posted by Chez_
Following an eg, of 3-48/x^2=o ,48/x^2=3 then x^2=48/3 x=plus minus 4.....i did r^3=1/pie =0 , r=1/pie^3 but then it would be the third root of pie, which is 0.0322... but the answer sheet says r=0.683

The third root of 1/pi is equal to $\frac{1}{\pi^{1/3}}$, not $\frac{1}{\pi^3}$. This might be where you are getting mixed up.
• Apr 30th 2008, 09:02 AM
Chez_
Got it now, and understand where i went wrong!!
THANK YOU!!