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yusufgovani
 Post subject: Derivation
PostPosted: Sun, 10 Apr 2011 22:34:21 UTC 
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A conical tank has base radius of 25mm and the height 50mm. It is filled with water till 40mm.

How much does the volume of the water increase if the level is raised by 2mm.

Please help me to solve it.[/u]
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Shadow
 Post subject: Re: Derivation
PostPosted: Sun, 10 Apr 2011 22:38:19 UTC 
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yusufgovani wrote:
A conical tank has base radius of 25mm and the height 50mm. It is filled with water till 40mm.

How much does the volume of the water increase if the level is raised by 2mm.

Please help me to solve it.[/u]


the change is the final minus the initial. Use that the volume of a cylinder is \pi ^2*h where h is the height.

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outermeasure
 Post subject: Re: Derivation
PostPosted: Sun, 10 Apr 2011 22:48:17 UTC 
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Shadow wrote:
yusufgovani wrote:
A conical tank has base radius of 25mm and the height 50mm. It is filled with water till 40mm.

How much does the volume of the water increase if the level is raised by 2mm.

Please help me to solve it.[/u]


the change is the final minus the initial. Use that the volume of a cylinder is \pi ^2*h where h is the height.


The volume of a cone is given by \frac{1}{3}\pi r^2 h...

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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Shadow
 Post subject: Re: Derivation
PostPosted: Sun, 10 Apr 2011 22:50:13 UTC 
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outermeasure wrote:
Shadow wrote:
yusufgovani wrote:
A conical tank has base radius of 25mm and the height 50mm. It is filled with water till 40mm.

How much does the volume of the water increase if the level is raised by 2mm.

Please help me to solve it.[/u]


the change is the final minus the initial. Use that the volume of a cylinder is \pi ^2*h where h is the height.


The volume of a cone is given by \frac{1}{3}\pi r^2 h...


:oops: cone, cylinder...they both start with c. X_X

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yusufgovani
 Post subject: The conical tank
PostPosted: Mon, 11 Apr 2011 12:38:44 UTC 
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Thank you for help. I worked out the final volume minus the intial volume but I do not get the answer which is 2.5ml.

I used the formula for volume of cone 1/3 r^2 pi h. I do not unerstand where i have gone wrong.

Any tips.

Thank you
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outermeasure
 Post subject: Re: The conical tank
PostPosted: Mon, 11 Apr 2011 13:16:58 UTC 
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yusufgovani wrote:
Thank you for help. I worked out the final volume minus the intial volume but I do not get the answer which is 2.5ml.

I used the formula for volume of cone 1/3 r^2 pi h. I do not unerstand where i have gone wrong.

Any tips.

Thank you


How can we pinpoint where you have gone wrong if we haven't seen what you actually did nor what answer you got?

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\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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Shadow
 Post subject: Re: The conical tank
PostPosted: Mon, 11 Apr 2011 14:56:50 UTC 
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yusufgovani wrote:
Thank you for help. I worked out the final volume minus the intial volume but I do not get the answer which is 2.5ml.

I used the formula for volume of cone 1/3 r^2 pi h. I do not unerstand where i have gone wrong.

Any tips.

Thank you


I agree with outermeasure. Also, your eventual answer will have a \pi in it.

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yusufgovani
 Post subject: conical tank
PostPosted: Thu, 14 Apr 2011 12:42:11 UTC 
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I used : (pi/3*2.1*2.1*4.2)- (pi/3*2.0*2.0*4.0)=2.6

Radius calculation: 40/50*2.5= 2 INTIAL
42/50*2.5= 2.1 FINAL

Is this correct method?

Thanks

Yusuf
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outermeasure
 Post subject: Re: conical tank
PostPosted: Thu, 14 Apr 2011 13:20:25 UTC 
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yusufgovani wrote:
I used : (pi/3*2.1*2.1*4.2)- (pi/3*2.0*2.0*4.0)=2.6

Radius calculation: 40/50*2.5= 2 INTIAL
42/50*2.5= 2.1 FINAL

Is this correct method?

Thanks

Yusuf


Method is correct, answer is not. Read Shadow's post again.

_________________
\begin{aligned} Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\ Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\ Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\ Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\ Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\ Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4 \end{aligned}
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yusufgovani
 Post subject:
PostPosted: Thu, 14 Apr 2011 14:34:17 UTC 
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Joined: Wed, 6 Apr 2011 11:26:16 UTC
Posts: 8
Thank you

Yusuf
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