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Math Help - Maths test question

  1. #1
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    Maths test question - Engineering Mathematics

    Hello Can someone please help me out with the following qustion...

    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)
    Last edited by HNCMATHS; January 24th 2010 at 07:17 AM.
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  2. #2
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    Quote Originally Posted by HNCMATHS View Post
    Hello Can someone please help me out with the following qustion...

    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}
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  3. #3
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    Quote Originally Posted by HNCMATHS View Post
    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)
    ...
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  4. #4
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    Arghhhhh now i'm confused haha
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  5. #5
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    Quote Originally Posted by HNCMATHS View Post
    Arghhhhh now i'm confused haha
    \Delta V = \pi (.96r)^2 (1.02h) - \pi r^2 h<br />

    \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?
    Last edited by skeeter; January 25th 2010 at 10:06 AM. Reason: fix an unneeded square
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  6. #6
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    Quote Originally Posted by skeeter View Post
    \Delta V = \pi (.96r)^2 (1.02h) - \pi r^2 h<br />

    \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
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  7. #7
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    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
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  8. #8
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    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.
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  9. #9
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    Quote Originally Posted by HNCMATHS View Post
    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
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  10. #10
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    Quote Originally Posted by HNCMATHS View Post

    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]
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  11. #11
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    Quote Originally Posted by Archie Meade View Post
    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?
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  12. #12
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    Quote Originally Posted by skeeter View Post
    \Delta V = \pi (.96r)^2 (1.02h) - \pi r^2 h<br />

    \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
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  13. #13
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    Ah Ha...

    (0.96)(1.02)-1 = -0.02

    Thus the reduction in curved surface area is 2%

    Correct?


    Thanks Guys... Chris
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  14. #14
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    That's it, Chris,

    you have it.

    In skeeter's calculation for the decrease in volume,
    the new volume is

    0.96(0.96)(1.02)=0.94 or 94% of the original volume,

    as volume has the r^2 factor.
    This is a 6% reduction in volume.
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