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Author Topic: Maths/Physics help  (Read 23627 times)

Offline radam

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Re: Maths/Physics help
« Reply #60 on: March 26, 2009, 06:29:29 pm »
This is an extract from the Physics specimen 09 HL paper 2:

"A digital camera is used to take a photograph of a plant. The CCD in the camera has 1.6*10^-7 square pixels. Each pixel has an area of 2.3*10^-10m^2. A particular leaf of the plant has an area of 3.5*10^-2 m^2. The image formed on the CCD is 1*10^-3 m^2. Two indentations on the lead are seperated by 0.50mm. Deduce that it is unlikely that the images of the two indentations will be resolved"

The marking scheme states that you should calculate the magnification as (1*10^-3/2.5*10^-2)=0.04, and so my question is, isnt this formula the area magnication, in which case the linear magnification would be sqrt(0.04)= 0.2? The only problem is that if you use the square-root then you get the distance of seperation as greater than the minimum distance to be just-resolved... is their a problem in the question or is my method incorrect?

Thanks a lot in advance :)


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Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #61 on: March 26, 2009, 06:55:59 pm »
Have been very busy all day. Will do this when I come back tonight.
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Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #62 on: March 27, 2009, 08:30:26 am »
Quote from: radam on March 26, 2009, 06:29:29 pm
This is an extract from the Physics specimen 09 HL paper 2:

"A digital camera is used to take a photograph of a plant. The CCD in the camera has 1.6*10^-7 square pixels. Each pixel has an area of 2.3*10^-10m^2. A particular leaf of the plant has an area of 3.5*10^-2 m^2. The image formed on the CCD is 1*10^-3 m^2. Two indentations on the lead are seperated by 0.50mm. Deduce that it is unlikely that the images of the two indentations will be resolved"

The marking scheme states that you should calculate the magnification as (1*10^-3/2.5*10^-2)=0.04, and so my question is, isnt this formula the area magnication, in which case the linear magnification would be sqrt(0.04)= 0.2? The only problem is that if you use the square-root then you get the distance of seperation as greater than the minimum distance to be just-resolved... is their a problem in the question or is my method incorrect?
Thanks a lot in advance :)
The distance between the indentations is 0.2*0.5mm=0.1mm=10^(-4), and the length of a side of a pixel is sqrt(2.3*(10^-10))=1.5*10^(-5). I see the same problem. I think it must be a mistake in the question.
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Offline radam

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Re: Maths/Physics help
« Reply #63 on: March 27, 2009, 01:08:36 pm »
Hmm i figured as much... thanks a lot for your help :)
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Offline omarsubei

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Re: Maths/Physics help
« Reply #64 on: April 10, 2009, 09:28:39 pm »
Hi it's been a while.
As usual I'm getting a different answer to the markscheme,
"A circle has equation x^2 + (y-3)^2 = 4 . Knowing that y= +/- (4 - x^2)^0.5 , find the volume of revolution of the shaded region about the x-axis.
The mark scheme says 144, and I got 237
PLEASE HELP!
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Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #65 on: April 10, 2009, 09:30:27 pm »
Quote from: omarsubei on April 10, 2009, 09:28:39 pm
Hi it's been a while.
As usual I'm getting a different answer to the markscheme,
"A circle has equation x^2 + (y-3)^2 = 4 . Knowing that y= +/- (4 - x^2)^0.5 , find the volume of revolution of the shaded region about the x-axis.
The mark scheme says 144, and I got 237
PLEASE HELP!


Surely y= 3+/- (4 - x^2)^0.5
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Offline omarsubei

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Re: Maths/Physics help
« Reply #66 on: April 10, 2009, 09:33:24 pm »
Ya I know I just wanted to make it less time consuming for you
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Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #67 on: April 10, 2009, 09:37:53 pm »
What region is shaded? If you cant say which paper the question is on can you describe it?
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Offline omarsubei

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Re: Maths/Physics help
« Reply #68 on: April 10, 2009, 09:56:05 pm »
The whole circle is shadad
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Offline omarsubei

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Re: Maths/Physics help
« Reply #69 on: April 10, 2009, 10:04:18 pm »
It's an integration question, to do with integral of the function squared multiplied by pie, if that helps
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Offline omarsubei

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Re: Maths/Physics help
« Reply #70 on: April 10, 2009, 10:08:56 pm »
Do you mind if we do this on msn, I think it'd be much easier
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Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #71 on: April 10, 2009, 10:22:17 pm »
V=pi\int(3+sqrt(4-x^2))^2dx-pi\int(3-sqrt(4-x^2))^2dx
     =pi\int12sqrt(4-x^2)dx
Use the substition x=2sint then dx=2costdt
     =12pi\int2cost.2costdt=12pi\int(2+2cos2t)dt=12pi(2t+sin2t)

The limits are changed from \pm 2 to \pm pi/2

We have V=12pi((2pi/2+sinpi)-(-2pi/2+sin-pi))==24pi^2

« Last Edit: April 10, 2009, 10:51:11 pm by astarmathsandphysics »
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Offline omarsubei

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Re: Maths/Physics help
« Reply #72 on: April 10, 2009, 10:27:29 pm »
WHAT???!! 6pi?? I just got it anyways, my bounds were wrong, they need to be from -1 to 1, and I was doing -2 to 2. Thanks anyways. Appreciate it.
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Offline astarmathsandphysics

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Re: Maths/Physics help
« Reply #73 on: April 10, 2009, 10:32:51 pm »
I dont understand. If the limits are +-1 how can the whole circle be shaded?

I looked up the volume of a tours here. It is consistent with my answer.
http://whistleralley.com/torus/torus.htm
« Last Edit: April 10, 2009, 10:38:06 pm by astarmathsandphysics »
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Offline omarsubei

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Re: Maths/Physics help
« Reply #74 on: April 10, 2009, 10:38:24 pm »
I guess you have to split it into the two equations + and -, that way you can have the whole circle shaded
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