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Offline Queen Victoria

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Mathematical problem
« on: March 12, 2013, 07:20:53 PM »
PA is trying to solve this.  He almost has it, but can't quite solve it.  Can you provide the steps?  He asks me to remind you that he hasn't had algebra since the 1840's.

Problem: A = pi * r * sq rt of r2 + h2.  Solve for r.

Answer
r = sq rt of [sq rt of pi2 * h4 + 4A2] - pi * h2 over 2 pi.

The 2 pi is divided into the under lined equation.  There is a nested square root in the answer and the brackets are mine to indicate the nesting. 

Thanks.
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Offline Al Swearegen

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Re: Mathematical problem
« Reply #1 on: March 12, 2013, 10:04:42 PM »
Will have to look tonight as am at wok.
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I2 today is not i2 of yesteryear. It is a knitting circle. Those that participate be they nice or asshats know their place and the price to be there. Odeon is the overlord

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Offline earthboundmisfit

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Re: Mathematical problem
« Reply #2 on: March 12, 2013, 10:16:17 PM »


Quote from: Al Swearengen on March 12, 2013, 10:04:42 PM
Will have to look tonight as am at wok.





 :zoinks:

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P7PSP

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Re: Mathematical problem
« Reply #3 on: March 13, 2013, 12:19:19 AM »
 :lol1:
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Offline Al Swearegen

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Re: Mathematical problem
« Reply #4 on: March 13, 2013, 04:44:38 AM »
Sorry your Highness. Too many years for me.

Though I would imagine that

a = pi * r * (sq rt of r2) + h2

a =  pi * r * r + h2

a = pi * r2 + h2

Then? I dunno

sq rt a = sq rt pi * sq rt r2 + sq rt h2

sq rt a = sq rt pi * r + h

Now isolating r?

Need to - h and - Sq rt pi on both sides of the ledger and to make

sq rt a - h = sq rt pi * r

r = sq rt a - h - sq rt pi

Bah...Dunno. Sorry your Highness. Humble effort
« Last Edit: March 14, 2013, 07:35:04 AM by Al Swearengen »
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I2 today is not i2 of yesteryear. It is a knitting circle. Those that participate be they nice or asshats know their place and the price to be there. Odeon is the overlord

.Benevolent if you toe the line.

Think it is I2 of old? Even Odeon is not so delusional as to think otherwise. He may on occasionally pretend otherwise but his base is that knitting circle.

Censoring/banning/restricting/moderating myself, Calanadale & Scrapheap were all not his finest moments.

How to apologise to Scrap

Offline Icequeen

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Re: Mathematical problem
« Reply #5 on: March 13, 2013, 09:23:51 AM »
 :zombiefuck: :runaway:
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Offline Lestat

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Re: Mathematical problem
« Reply #6 on: March 14, 2013, 01:11:20 AM »
Just looking at that made my eyes ache (is if they didn't enough already)
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Offline Parts

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Re: Mathematical problem
« Reply #7 on: March 14, 2013, 05:07:21 AM »
The answer is 5 :M
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Offline Calavera

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Re: Mathematical problem
« Reply #8 on: March 14, 2013, 05:49:40 AM »
Fuck this. I give up. :(

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Offline Lestat

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Re: Mathematical problem
« Reply #9 on: March 14, 2013, 07:05:36 AM »
I figured it would be a lost cause even before clicking the thread title.

I struggle like hell with basic math, I'm about as dyscalculic as they come.
Actually, even for the ONE bit of math, counting change on my phone aside, that I ever, ever use on a regular basis, figuring out the actual weight of (insert compound XYZ here) from a molar percentage...I have to relearn it every single fucking time I need to do it. And it gives me a right splitting headache on a bad day, on a really bad day, it just won't compute.
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Offline Al Swearegen

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Re: Mathematical problem
« Reply #10 on: March 14, 2013, 08:10:40 AM »
Oh shit I feel like an idiot.

Sorry

A = pi * r * sq rt of r2 + h2

A= pi * r * (sq rt of r2 + h2)

A = pi * r * r + h

A = pi * r2 + h

(-h) A = pi * r2 + h (-h)

A - h = pi * r2

(/pi) A - h = pi x r2 (/pi)

(A - h )/ pi =  r2

sq rt ((A-h)/pi) = r

Finally.

Need to prove.

Assign values

pi = 3.14
r = 3
h = 2


A = 3.14 * 3 * sq rt of 9 + 4

A= 3.14 * 3 * (sq rt of 9 + 4)

A = 3.14 * 3 * 3 + 2

A = 3.14 * 9 + 2

30.26 = 3.14 * 9 + 2

(-2) 30.26 = 3.14 * 9 + 2 (-2)

30.26 - 2 = 3.14 * 9

(/3.14) (30.26-2) =3.14 x 9 (/3.14)

(30.26 - 2 )/ 3.14 =  9

(sq rt ((30.26 - 2)/3.14)) = sq rt of 9

(sq rt ((30.26 - 2)/3.14)) = 3

sq rt (28.26)/3.14) = 3

sq rt 9 = 3
« Last Edit: March 14, 2013, 08:44:26 AM by Al Swearengen »
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I2 today is not i2 of yesteryear. It is a knitting circle. Those that participate be they nice or asshats know their place and the price to be there. Odeon is the overlord

.Benevolent if you toe the line.

Think it is I2 of old? Even Odeon is not so delusional as to think otherwise. He may on occasionally pretend otherwise but his base is that knitting circle.

Censoring/banning/restricting/moderating myself, Calanadale & Scrapheap were all not his finest moments.

How to apologise to Scrap

Offline Queen Victoria

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Re: Mathematical problem
« Reply #11 on: March 14, 2013, 02:38:47 PM »
Thank you Sir Les for your answer.  Taking a look at my post below, I see that I've led you astray.  I owe you a beer for that the next time we meet each other.

Quote from: Queen Victoria on March 12, 2013, 07:20:53 PM
PA is trying to solve this.  He almost has it, but can't quite solve it.  Can you provide the steps?  He asks me to remind you that he hasn't had algebra since the 1840's.

Problem: A = pi * r * sq rt of r2 + h2.  Solve for r.

Answer
r = sq rt of [sq rt of pi2 * h4 + 4A2] - pi * h2 over 2 pi.

The 2 pi is divided into the under lined equation.  There is a nested square root in the answer and the brackets are mine to indicate the nesting. 

Thanks.

The entire question in the book is:  The formula for the lateral surface area A of a right circular cone with height h and base radius r is  A = pi * r * sq rt of (r2 + h2).  Solve for r in terms of A and h.

I neglected to indicate that r2 and h2 are both under the radical.

Since there is a 4A2 in the answer, PA was using the quadratic equation method to solve for r.
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Offline Dexter Morgan

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Re: Mathematical problem
« Reply #12 on: March 14, 2013, 07:43:40 PM »
Square the whole equation and you get the following coefficients:

a=pi^2
b=h^2*pi^2
c=-a^2


Using the quadratic equation, you get two solutions for r^2, which would give you 4 total solutions. The answer is wrong.
« Last Edit: March 14, 2013, 08:13:45 PM by Dexter Morgan »
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Offline Parts

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Re: Mathematical problem
« Reply #13 on: March 15, 2013, 04:41:49 AM »
Have you tried Wolfram?

http://www.wolframalpha.com/input/?i=A+%3D+pi+*+r+*+sq+rt+of+r2+%2B+h2
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Offline Queen Victoria

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Re: Mathematical problem
« Reply #14 on: March 15, 2013, 02:03:29 PM »
Quote from: Parts on March 15, 2013, 04:41:49 AM
Have you tried Wolfram?

http://www.wolframalpha.com/input/?i=A+%3D+pi+*+r+*+sq+rt+of+r2+%2B+h2

Wolfram and Hart?  The evil guys from Angel?
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A good monarch is a treasure. A good politician is an oxymoron.

My brain is both uninhibited and uninhabited.

:qv:
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