# Maths test question

• Jan 24th 2010, 07:01 AM
HNCMATHS
Maths test question - Engineering Mathematics

The radius of a Cylinder is reduced by 4% and its height is increased by 2%

(a)Determine the approximate change in the cylinders volume

(b)Determine the curved surface area (volume = pi r(squared)h and curved surface area = 2pi rh)
• Jan 24th 2010, 07:21 AM
e^(i*pi)
Quote:

Originally Posted by HNCMATHS

The radius of a Cylinder is reduced by 4% and its height is increased by 2%

(a)Determine the approximate change in the cylinders volume

(b)Determine the curved surface area (volume = pi r(squared)h and curved surface area = 2pi rh)

$V_0 = \pi r_0^2 h$

$V_1 = \pi (r_0 -0.04)^2(h+0.02)$

$\Delta r = \frac{V_1}{V_0} = \frac{\pi (r_0 -0.04)^2(h+0.02)}{\pi r_0^2 h} = \frac{(r_0 - 0.04)^2(h+0.02)}{r_0^2 h}$
• Jan 24th 2010, 07:25 AM
skeeter
Quote:

Originally Posted by HNCMATHS
The radius of a Cylinder is reduced by 4% and its height is increased by 2%

(a)Determine the approximate change in the cylinders volume

change in V = final V - initial V

$\textcolor{red}{\Delta V = \pi (.96r)^2 (1.02h) - \pi r^2 h =}$ ?

(b)Determine the curved surface area

do the same procedure to find the change in the lateral surface area

(volume = pi r(squared)h and curved surface area = 2pi rh)

...
• Jan 24th 2010, 07:36 AM
HNCMATHS
Arghhhhh now i'm confused haha
• Jan 24th 2010, 08:10 AM
skeeter
Quote:

Originally Posted by HNCMATHS
Arghhhhh now i'm confused haha

$\Delta V = \pi (.96r)^2 (1.02h) - \pi r^2 h
$

$\Delta V = \pi r^2 h[(.96)^2(1.02) - 1]$

$\Delta V = \pi r^2 h[(.96)^2(1.02) - 1]$

$\Delta V = \pi r^2 h (-0.06) = -0.06V$

the volume has decreased by about 6%

unconfused now?
• Jan 25th 2010, 12:38 AM
HNCMATHS
Quote:

Originally Posted by skeeter
$\Delta V = \pi (.96r)^2 (1.02h) - \pi r^2 h
$

$\Delta V = \pi r^2 h[(.96)^2(1.02) - 1]$

$\Delta V = \pi r^2 h[(.96)^2(1.02) - 1]$

$\Delta V = \pi r^2 h (-0.041) = -0.06V$

the volume has decreased by about 6%

unconfused now?

Hello Skeeter,

Not all together sure where the values .96 & 1.02 come from ?

Regards,
Chris
• Jan 25th 2010, 02:51 AM
hi Chris,

the radius being reduced by 4% causes the radius to decrease by 0.04 of it's original length,

as 4% is $\frac{4}{100}=0.04$

Therefore, the new radius length can still be expressed using the old radius by writing

$new\ radius=r-0.04r=r(1-0.04)=0.96r$

the height increases by 2%, so the new height is h+2% of h

$h+\frac{2}{100}h=h+0.02h=h(1+0.02)=1.02h$
• Jan 25th 2010, 03:08 AM
HallsofIvy
Since the original problem said "approximate", I wonder if this isn't for a Calculus class!

$V= \pi r^2h$ so $dV= \pi (2r h dr+ r^2 dh)$

Then $\frac{dV}{V}= \pi\frac{2r h dr+ r^2 dh}{\pi r^2 h}$
$= \frac{2rh dr}{r^2 h}+ \frac{r^2 dh}{r^2 h}= 2 \frac{dr}{r}+ \frac{dh}{h}$

"dr/r" is approximately -4%= -.04, "dh/h" is approximately +2%= +.02, so, approximately, dV/V= 2(-.04)+ .02= -.06.

This differs from what you would get using the others' method because it does not include the products of the percents multiplied together.

This is the same as the old mechanics rule of thumb: "When quantities are added, their errors add; when quantities are multiplied, their relative errors add." Here we have "r*r*h". Their "relative errors" are -.04 and .02 so the relative error in V is -.04- .04+ .02.
• Jan 25th 2010, 10:29 AM
HNCMATHS
Quote:

Originally Posted by HNCMATHS
Hello Skeeter,

Not all together sure where the values .96 & 1.02 come from ?

Regards,
Chris

I think I follow this now, does this apply to part b also?

could someone please run through part b?

Thanks to you all for your help!

Chris
• Jan 25th 2010, 11:54 AM
Quote:

Originally Posted by HNCMATHS

The radius of a Cylinder is reduced by 4% and its height is increased by 2%

(b)Determine the curved surface area (curved surface area = 2pi rh)

Original curved surface area = $2{\pi}rh$

New surface area = $2{\pi}[r-0.04r][h+0.02h]=2{\pi}(0.96r)(1.02h)$

$=[0.96(1.02)]2{\pi}rh=0.96(1.02)[original\ curved\ surface\ area]$
• Jan 25th 2010, 02:09 PM
HNCMATHS
Quote:

Original curved surface area = $2{\pi}rh$

New surface area = $2{\pi}[r-0.04r][h+0.02h]=2{\pi}(0.96r)(1.02h)$

$=[0.96(1.02)]2{\pi}rh=0.96(1.02)[original\ curved\ surface\ area]$

Is the answer therfore 2% reduction in curved surface area?
• Jan 25th 2010, 02:16 PM
HNCMATHS
Quote:

Originally Posted by skeeter
$\Delta V = \pi (.96r)^2 (1.02h) - \pi r^2 h
$

$\Delta V = \pi r^2 h[(.96)^2(1.02) - 1]$

$\Delta V = \pi r^2 h[(.96)^2(1.02) - 1]$

$\Delta V = \pi r^2 h (-0.06) = -0.06V$

the volume has decreased by about 6%

unconfused now?

Archie, could you please explain part (b) like above? Thanks in advance Chris
• Jan 25th 2010, 02:43 PM
HNCMATHS
Ah Ha...

(0.96)(1.02)-1 = -0.02

Thus the reduction in curved surface area is 2%

Correct?

Thanks Guys... Chris
• Jan 25th 2010, 03:49 PM
$0.96(0.96)(1.02)=0.94$ or 94% of the original volume,
as volume has the $r^2$ factor.