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#3626 Re: Puzzles and Games » Real life architectual problem » 2006-01-12 18:10:04

Triangle on the left is triangle A, triangle on the right is triangle B.  Angle x is at the top of triangle A and angle y is at the bottom right of triangle A.  I have to assume both triangles are  similar, so angle x is also at the bottom of triangle B and angle y is at the top left of triangle B.

Draw a line from the point of angle y in triangle B to the hypotenuse of triangle A, such that it hits the hypotenuse at a 90 degree angle.  Then, draw a line which connects the very top of triangle A to that of B.  Call this new triangle, triangle C.

The bottom most angle of triangle C is 90 degrees.  Since the hypotenuses in triangles A and B are parallel (assumed), the top left angle in triangle C is equal to angle y in triangle B.  Now add the top right angle of triangle C, the 90 degree angle just to the right of it, and angle y in triangle B.  Since all these angles put together form a line, they add up to 180 degrees.  But we also know that x + y + 90 = 180.  Thus, the top right angle in C is equal to angle x.  Therefore, triangle C is similar to triangle B and A (assuming these two are congruent).

However, we know that the side of triangle C is 2'.  Since the triangles are similar: 6 / 12 = 2 / x, 6x = 24, x = 4.  So the other side of triangle C is 4'.  This means the hypotenuse of triangle C is about 4.47'.

The length of the entire bottom is 12 feet, 12 - 4.47 = 7.53'.

#3627 Re: Puzzles and Games » Puzzle » 2006-01-12 13:20:17

Heres one:  what comes next in this sequence:

9, 10, 11, 12, 13, 14...

Once you've looked that, try to guess the pattern...

#3628 Re: Jokes » Optimism vs pessimism » 2006-01-12 13:11:31

I've always said an optimist hopes for the best and gets disappointed.  A pessimist thinks the worst will come and gets pleasantly surprised.

#3630 Re: Help Me ! » vectors » 2006-01-12 13:04:28

f2=170°

You gave a direction, but not a magnitude.  You need both.

#3631 Re: Help Me ! » logarithmic differentiation » 2006-01-12 13:03:02

Since your taking the derivative with respect to x, dx = 1.  But you aren't taking the derivative with respect to y, so dy doesn't have to be 1.  For example:

y = x

Take the derivative of both sides:

y' = x'

But x' = 1 (dx = 1)

y' = 1

#3632 Re: Help Me ! » logarithmic differentiation » 2006-01-11 16:19:37

The only way I know how to solve this is implicit differentiation:

Let y = x^x.

ln(y) = ln(x^x)

ln(y) = x*ln(x)

The derivative of ln(y) is 1/y, but since it's implicit, we need to use the chain rule on y, which means we multiply it by the derivative of y.  It then becomes 1/y * y'

The derivative of x*ln(x) using the product rule is x*1/x + ln(x) * 1.  This is the same as 1 + ln(x)

So we have:

y' / y = 1 + ln(x)

But we know that y = x^x

y' / x^x = 1 + ln(x)

y' = (1 + ln(x)) * x^x

Now just go through the same steps using 2's.

Edit: To check your answer, the answer to 2(x)^(2x) is 4(ln(x)+1)x^(2x) and the answer to (2x)^(2x) is (2ln(2x)+2)*(2x)^(2x), I'm not sure which one you meant.

#3633 Re: Help Me ! » confusing trig problem... » 2006-01-11 16:02:58

I have put it on my list to either create or find a website with even the most obscure trig identities.  So far, wikipedia (as always) is the best:

http://en.wikipedia.org/wiki/Trig_identities

#3634 Re: Help Me ! » A really obvious but interesting question » 2006-01-11 15:54:56

Consider a / b.  This could be written as a * 1/b.  Now let a = 3/2 and b = 1/2.  (3/2) / (1/2)

Take the 1/2.  What we want is 1 / (1/2), which would be 1 / b.  Multiply this by 2/2 (which is multiplying it by 1, thus not changing the value).  You get (1 * 2) / ( (1/2)*2 ) = 2 / 1

So 1 / b = 2 / 1.  a * 1 / b = 3/2 * 2 / 1.

#3635 Re: Help Me ! » confusing trig problem... » 2006-01-11 14:00:04

That's pi.  Phi is normally the "vertical" (not sure exactly what to call it) angle when you are in spherical coordinates.

To:

First step, as always, is to simplify:

Now it becomes fairly simple, just use the trig identity:

Where a = x and b = nx

Which simplifies to:

#3636 Re: Help Me ! » inflection points » 2006-01-11 10:18:01

f'''(x) = 6, is not equal to 0 when x = 0

God, I have never heard of that.  On wikipedia, it states:

One also needs the lowest-order non-zero derivative to be of odd order (third, fifth, etc.). If the lowest-order non-zero derivative is of even order, the point is not a point of inflection. (An example of such a function is y = x^4 -x).

It sounds similar, but not quite the same.

Edit:

But the wikipedia article doesn't sound right either.  According to it:

f(x) = x^4 - 5x^2 + 4

Doesn't have any inflection points, yet it appears to change concavity.

#3637 Re: Help Me ! » Long term behaviour of sequences » 2006-01-11 09:59:39

Think of it this way:

Where c is not 0, then basically what you are doing is:

Any number that you add an infinite amount of times goes to infinity.  And since An approaches this c, you are adding it an infinite amount of times.

Edit: God, I'm starting to think you're right.  I've gotten into a habit of hearing converge and diverge, and automatically thinking infinite series.  The An is also common when doing infinite series.

Wonder why (s)he didn't object to my restating of the question using summation though...

Edit #2:

It sorta sounds silly, but only because you misunderstood the question.

Starting to find that quite ironic right now.

#3638 Re: Help Me ! » inflection points » 2006-01-11 09:50:17

Another way of putting it is that the function changes concavity.

Yes.

What defines an inflection point is that both the first and second derivatives equal zero.

It is not possible to change concavity when the slope is 0.

From http://en.wikipedia.org/wiki/Inflection_point:

a point on a curve at which the second derivative changes sign. This is very similar to the previous definition, since the sign of the curvature is always the same as the sign of the second derivative, but note that the curvature is not the same as the second derivative.

Or:

a point (x,y) on a function, f(x), at which the first derivative, f'(x), is at an extremum, i.e. a minimum or maximum. (This is not the same as saying that y is at an extremum, and in fact implies that y is not at an extremum).

Or:

if f'(x) is not zero, the point is a non-stationary point of inflection

#3640 Re: Help Me ! » Long term behaviour of sequences » 2006-01-11 08:38:49

Not quite.  An heads to 1/4 as n heads to infinity, that part was right.  But this means the series diverges.

#3641 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2006-01-11 08:34:24

Yea, I guess they are.  They looked completely different at first glance.

#3642 Re: Jai Ganesh's Puzzles » Problems and Solutions » 2006-01-11 04:16:41

I'm not sure, this is what I get:

#3643 Re: Help Me ! » inflection points » 2006-01-11 04:02:20

You're right, points of inflection have f'(x) and f''(x) both equal to 0.

Huh?  Inflection points don't have to have a slope of 0.  On the contrary, they can't have a slope of 0.

Check this over, make sure I didn't make any mistakes:

f(x) = ax^3 + bx^2 + cx + d
f'(x) = 3ax^2 + 2bx + c
f''(x) = 6ax + 2b

We want an inflection point at (1, -2).

f''(1) = 6a + 2b = 0

3a + b = 0, or b = -3a

Now we have a max at (0,0):

f'(0) = c = 0

Finally, we want the graph to pass through the point (0, 0):

f(0) = d = 0

Since c is 0 in the 2nd derivative, c is 0 in the original function making it:

f(x) = ax^3 + bx^2

The only restricting factor on this equation is:

b = -3a

Trying it for a = 1:

b = -3, a = 1:

f(x) = x^3 - 3x^2

This passes through the points (0, 0) and (1, -2), has a local max at (0, 0), and an inflection at (1, -2).

#3644 Re: Help Me ! » Tangent to Normal. » 2006-01-10 15:32:54

Don't worry about it, on a math test, I wrote:

2x = 2
x = 2

#3645 Re: Help Me ! » 2nd Fundamental Theorem of Calculus » 2006-01-10 15:30:54

I dare you to integrate:

Without using the 2nd Fundamental Thoerm of Calculus.

To do so with it:



#3646 Re: Help Me ! » Tangent to Normal. » 2006-01-10 15:25:04

dy/dx = 6x - 5

dy/dx = 4x - 5, slope at (1, 2) is 3

Try it now.

#3647 Re: Help Me ! » Long term behaviour of sequences » 2006-01-10 15:16:43

It sorta sounds silly, but only because you misunderstood the question.

His question is:

First, you have to find whether it converges or diverges.  If you have a sequence An, and:

Then the series diverges.  Try to take this limit.

#3648 Re: Help Me ! » Very interesting problems.. » 2006-01-10 08:11:52

Quite odd, mine works completely fine, even overnight.

#3649 Re: Help Me ! » Very interesting problems.. » 2006-01-10 05:48:12

For every circle of size n, you must be able to contruct a Hamilton circuit.  So arrange the numbers around in a cirlce, and then draw a line for every two numbers that add up to a perfect square.

To be a ciruit, every number has to have at least two lines connected to it.  It can be observed that 2 won't have two lines going in until 14:

2 + x = perfect square, where x is a natural number not equal to two.  The lowest two perfect squares are 9 and 16 (2 and 1 don't count since x is natural and not 2).  So to get two lines connected to 2:

2 + x = 16, x = 14.

Size 14 is a minimum, when only considering 2.

For n ≥ 14, consider 8.

8 + x = perfect square

The two lowest perfect squares are 9 and 25.

8 + x = 25, x = 17

So size 17 is a minimum.

For n ≥ 17, consider 16.

16 + x = perfect square

The two lowest perfect squares are 25 and 36.

16 + x = 36, x = 20.

This pattern will continue till 32, where every natural number up to and including 32 has a least two lines connected to it.

#3650 Re: Help Me ! » Very interesting problems.. » 2006-01-10 04:31:05

Would you mind posting your code?  I'm kind of curious as to why it bombed out.

As for a list of 62:

1 3 6 10 15 21 4 5 11 14 50 31 18 7 2 34 30 19 45 36 13 51 49 32 17 47 53 28 8 41 40 60 61 20 29 52 48 33 16 9 55 26 23 58 42 39 25 56 44 37 12 24 57 43 38 62 59 22 27 54 46 35

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