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Ok, I'm supposed to find the value of R for which dA/Dr is zero. (a turning point on the graph)
I've differentiated this several times but it always ends up producing nightmarish 5, 6, or 7 degree polynomials which I have no clue how to solve.
Some problems in my book are supposed to be solved with a graphing calculator but usually it tells you when to use it. Anyway this can be solved without a graphing calculator?
A logarithm is just a misspelled algorithm.
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WHEW! Nevermind I just got it. I think what helped is I created a constant C to equal (300/pi r^2)^2 and just kept the fact that it was a constant in mind. Then the problem becomes:
pi r sqrt(r^2 + c r^-4)
Which is much easier to work with. Give it a shot, its a fun problem!
A logarithm is just a misspelled algorithm.
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mikau,
I just differentiated your problem the old fashioned way and got;
2π²r^6 + 90000
r²√(π²r^6 + 90000)
Which would only equal zero if;
r^6 = -90000 / 2π²
Since nothing raised to the 6th power can be negative, I would think that there is no real solution.
Please post with what you found the solution to be.
Last edited by irspow (2006-01-16 14:20:52)
I am at an age where I have forgotten more than I remember, but I still pretend to know it all.
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.....
Okay..lol. I come here asking you guys to help me do this problem, and I end up helping you do it.
In this problem it helps to substitute the constant (300/pi)^2 with C and the expression inside the radical with u.
Here we are:
A logarithm is just a misspelled algorithm.
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I am sorry mikau, you were right. I found my earlier mistake that made me put my foot in my mouth.
Here is exactly how I did the problem:
A = πr √[r² + (300/πr²)²]
A = πr √[r² + (90000/π²r^4)] ; squared the inside term as indicated
A = πr √[(π²r^6 + 90000) / (π²r^4)] ; made the denominators the same
A = (1/r) √(π²r^6 + 90000) ; moved the denominator outside the radical sign
We know that the derivative of this function is; f(x)f'(g) + f'(x)f(g)
f(x) = 1/r; f(g) = √(π²r^6 + 90000);
(1/r) (1/[2√(π²r^6 + 90000)]) (6π²r^5) = f(x)f'(g)
f(x)f'(g) = 3π²r^4/(√(π²r^6 + 90000)
f'(x)f(g) = (-1/r²)(√(π²r^6 + 90000)) = -√(π²r^6 + 90000)/r²
So the derivative is;
3π²r^4 _ √(π²r^6 + 90000)
√(π²r^6 + 90000) r²
Making the denominators the same and simplifying gives;
2π²r^6 - 90000
r²√(π²r^6 + 90000)
So for this to equal zero r^6 = 90000/ (2π²)
You are correct, my mistake. (they are becoming more frequent with age.)
For some reason or other I had a positive sign before the constant when I integrated before. Nice move, your calculations seemed much neater.
Last edited by irspow (2006-01-17 09:16:04)
I am at an age where I have forgotten more than I remember, but I still pretend to know it all.
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Yeal well never underestimate the value of substitution. :-)
A logarithm is just a misspelled algorithm.
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