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#1 Re: Help Me ! » Splitting ugly polynomial divisions into fractions » 2010-11-11 13:56:06

I guess it would.  usually when solving for a bunch of unknowns I would have just as many equations but this time I only got the one and my brain suddenly didn't know what to do.

#2 Help Me ! » Splitting ugly polynomial divisions into fractions » 2010-11-11 12:44:16

leadfoot
Replies: 3

I'm doing this chapter on splitting big ugly polynomial divisions into simpler fractions.  At one point in the chapter we come to an example that got reduced to:

Anyone have a guess as to how the author got 2C = 16?  Did the author simply lift

right out of the equation, throw away
, ignored the other stuff, and solved for C?

#3 Re: Help Me ! » Maximum cylinder in a cone » 2010-11-08 08:11:30

of course you're right.  i forgot about the quadratic formula.

#4 Re: Help Me ! » Maximum cylinder in a cone » 2010-11-08 04:31:20

TheDude wrote:

... if you solved

you probably got 1/3 R, right?

I didn't get 1/3 R.

Did I make a math error to arrive at a quadratic?  Is factoring possible without knowing the true value of R?

#5 Re: Help Me ! » Maximum cylinder in a cone » 2010-11-07 17:31:54

Let me revise my last post.

Working with my previous equation

yielded 
.  A quadratic ewww...

So I decided to solve for the cylinder radius instead with


The V'(h) = 0 did give the correct maxima of
.

Shouldn't the answer workout correctly regardless of whether I solve for the radius or the height?  Why did it work out solving for the radius? How can I know in advance whether to solve for the radius or the height?

#6 Help Me ! » Maximum cylinder in a cone » 2010-11-07 16:37:28

leadfoot
Replies: 5

Problem:
Inscribe in a given cone, the height of which is equal to the radius of the base, a cylinder whose volume is a maximum.

This is a calculus maxima problem.  Having drawn a diagram I noticed that


R = b + h or
b = R - h

The generic cylinder volume formula is


Which gives us

After that I did

.  Set
.  Solve for h to get the max volume which does not match the answer of
.

Was I wrong in the initial observation R = b + h ??

#7 Re: Help Me ! » Diff y^5 with respect to y^2 » 2010-11-05 10:57:52

I'm back.  I've tried to solve the same problem using an alternate method without all the substituting.

which happens to be the answer.  The question: is this a valid method of solving the problem or did it work out by accident?

#8 Re: Help Me ! » Diff y^5 with respect to y^2 » 2010-11-05 08:10:15

thanks for the speedy solution.  I never would have thought to substitute like that although I should have known swapping directly 1 for 1 is incorrect.

#9 Help Me ! » Diff y^5 with respect to y^2 » 2010-11-05 06:17:00

leadfoot
Replies: 8

Yes both symbols are y.

In any normal exercise things would be y with respect to x with an equation like
y = f(x) = something involving x

Following the same pattern I tried

I guess that would be too easy because the official text book answer is

Anyone know how they came up with that?

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