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## #1 2006-01-16 13:07:39

mikau
Member
Registered: 2005-08-22
Posts: 1,504

### Seriously need help! x_x

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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## #2 2006-01-16 13:31:39

mikau
Member
Registered: 2005-08-22
Posts: 1,504

### Re: Seriously need help! x_x

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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## #3 2006-01-16 14:18:40

irspow
Member
Registered: 2005-11-24
Posts: 457

### Re: Seriously need help! x_x

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)

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## #4 2006-01-16 15:28:22

mikau
Member
Registered: 2005-08-22
Posts: 1,504

### Re: Seriously need help! x_x

.....

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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## #5 2006-01-17 09:14:13

irspow
Member
Registered: 2005-11-24
Posts: 457

### Re: Seriously need help! x_x

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)

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## #6 2006-01-17 16:11:06

mikau
Member
Registered: 2005-08-22
Posts: 1,504

### Re: Seriously need help! x_x

Yeal well never underestimate the value of substitution. :-)

A logarithm is just a misspelled algorithm.

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