INTENSITY²
Start here => Games => Topic started by: Queen Victoria on March 12, 2013, 07:20:53 PM
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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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Will have to look tonight as am at wok.
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Will have to look tonight as am at wok.
(http://zuramode.files.wordpress.com/2012/09/classical-wok.jpg)
:zoinks:
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:lol1:
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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
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:zombiefuck: :runaway:
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Just looking at that made my eyes ache (is if they didn't enough already)
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The answer is 5 :M
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Fuck this. I give up. :(
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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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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
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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.
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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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.
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Have you tried Wolfram?
http://www.wolframalpha.com/input/?i=A+%3D+pi+ (http://www.wolframalpha.com/input/?i=A+%3D+pi+)*+r+*+sq+rt+of+r2+%2B+h2
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Have you tried Wolfram?
http://www.wolframalpha.com/input/?i=A+%3D+pi+ (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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Wolfram? theres an equation solver online?
There was me thinking it was only good for making armor-piercing ammunition, lightbulb filaments, and really bloody heavy stuff. (wolfram being a foreign name for tungsten)
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I think I may have solved it. Hold on. There's a neat trick.
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Step 1 (rewriting original equation for clarification):
A = Pi * r * sqrt(r2 + h2)
Step 2 (squaring both sides is safe because both sides are definitely positive):
A2 = Pi2 * r2 * (r2 + h2)
Step 3 (switch both sides around and then subtract A2 from both sides to get a zero on the right side):
Pi2 * r2 * (r2 + h2) - A2 = 0
Step 4 (time for some expanding):
Pi2 * r4 + Pi2 * r2 * h2 - A2 = 0
Step 5 (neat trick: adjust the equation so that it resembles a quadratic equation with r as the main variable):
Pi2 * r2 * r2 + Pi2 * h2 * r * r - A2 = 0
Step 6 (now focusing on the discriminant and fixing it a bit to make life easier):
discriminant = Pi4 * h4 * r2 + 4 * Pi2 * r2 * A2
Step 7 (factoring the discriminant):
discriminant = Pi2 * r2 * (Pi2 * h4 + 4 * A2)
Step 8 (pull one of the factors of the discriminant out of the square root sign):
discriminant = Pi * r * sqrt(Pi2 * h4 + 4 * A2)
Step 9 (back to the bigger picture and solving for r using the infamous quadratic formula, rejecting the negative square root option because r must be positive):
r = [-(Pi2 * h2 * r) + discriminant] / (2 * Pi2 * r2)
Step 10 (focusing now on the numerator of the fraction on the right side and switching the addition operands around thanks to the valuable property of commutativity):
numerator = discriminant - Pi2 * h2 * r
Step 11 (replace discriminant with its expression value and factor out the common factors):
numerator = Pi * r * (sqrt(Pi2 * h4 + 4 * A2) - Pi * h2)
Step 12 (back to the overall equation and simplifying both the numerator and the denominator of the fraction on the right side of the equation):
r = (sqrt(Pi2 * h4 + 4 * A2) - Pi * h2) / (2 * Pi * r)
Step 13 (multiplying both sides by r to get rid of the r on the right side):
r2 = (sqrt(Pi2 * h4 + 4 * A2) - Pi * h2) / (2 * Pi)
Step 14 (square root both sides to get the solution):
r = sqrt[(sqrt(Pi2 * h4 + 4 * A2) - Pi * h2) / (2 * Pi)]
And voila! Am I the king of the universe or what?