The math help thread - Printable Version

The math help thread - Printable Version

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The math help thread - Nikkey - 2009-11-16

Hazzy Wrote:*facepalm*
There's a difference between 2^(n^2) and (2^n)^2?

A very big difference:
[Image: yf6hgrq.png]

That means, (2^n)^2 = 2^(2n)

The math help thread - Matt - 2009-11-17

Never mind my previous problem, I made an error differentiating.

The math help thread - Horusmaster - 2009-11-17

sry read wrong:

2^(n^2) = 2^(n*n) = 2^n^n
so the equation should have been 2^n^n-2^n=5 if that helps
i still can't solve it... if I let a=2^n then i get 2 variables. a^n-a=5

The math help thread - Hazzy - 2009-11-17

Devil's Sunrise Wrote:A very big difference:

That means, (2^n)^2 = 2^(2n)

Ugh...
I should revive my first equation then.
$[Image: eq.latex?%282%5En%29%5E2-2%5En=5]$

The math help thread - Horusmaster - 2009-11-17

Hazzy Wrote:Ugh...
I should revive my first equation then.
$[Image: eq.latex?%282%5En%29%5E2-2%5En=5]$

OMG then we already solved it..... read my first post about this.

The math help thread - Noah - 2009-11-17

Hazzy Wrote:Ugh...
I should revive my first equation then.
$[Image: eq.latex?%282%5En%29%5E2-2%5En=5]$

Then, you should go Horusmaster's way of solving it:

[Image: yzjl458.png]

Noah

The math help thread - Russt - 2009-11-17

Oh but you also can't separate log(1+-sqrt(21)) into log(1) +- log(sqrt(21)). heh.

And since log of a negative number is complex, the only real solution would be n = log2(1 + sqrt(21)) = ln(1 + sqrt(21))/ln(2) =~ 1.481

The math help thread - Noah - 2009-11-17

Russt Wrote:Oh but you also can't separate log(1+-sqrt(21)) into log(1) +- log(sqrt(21)). heh.

And since log of a negative number is complex, the only real solution would be n = log2(1 + sqrt(21)) = ln(1 + sqrt(21))/ln(2) =~ 1.481

That certainly is correct. I was thinking about it at university when I had my math-class. I don't understand how I manage to do the same math-error twice, but oh well.

[IMGspoiler]http://i36.tinypic.com/33ljl0k.jpg[/IMGspoiler]

Anywho, for those who are interested in the
[Image: y8ad8nd.png]

answer: Do the following:

Set
[Image: yz66355.png]

First off, prove that there exist a real answer (A prof. and I discussed the topic, and we got off to imaginary numbers. Yehaw.):
As
[Image: yea2tvl.png]

is true for all real a, the function is continuous. (Or use Cauchy, if you're that unsure that this is a continuous function)
Set n=1, then f(n) = 0. If we set n = 2, then
[Image: yhrwqne.png]

Therefore, there is a root 1<n<2 such that f(n) = 5.

To find the exact answer, you can turn it to a series expansion or repeat Newton's method infinitely many times (Well...).

Noah

The math help thread - Matt - 2009-11-17

Another question -
A light moving at 3 ft/s approaches a 6 ft. man standing 12 ft from a wall. The light is 3 ft above the ground. How fast is the tip P of the man's shadow moving when the light is 24 ft from the wall?

The shadow is on the wall, if you didn't understand my wording =x

The math help thread - KajitiSouls - 2009-11-17

I'm not entirely sure the way I did it was correct, so that's your warning.

We need the equation that describes the location of tip P on the wall. So let's figure it out!

We'll call x the distance of the light from the wall. At any moment, the rays of light running lower than the angle of arctan(height/run) are blocked by the man, and makes the shadow. Therefore, P on the wall is described by:

Code:
f(t) = x - 3t

dx/dt = v(t) = -3

P(t) = (x - 3t)*(3/(x - 3t - 12)) + 3

     = (3*(x - 3t))/(x - 3t - 12) + 3

P'(t) = 3 * [-3*(x - 3t - 12) - -3*(x - 3t)] / (x - 3t - 12)^2

      = 108 / (x - 3t - 12)^2

x = 24 ft

P'(0) = 108/144 = .75 ft/s

The math help thread - Noah - 2009-11-17

Matt Wrote:Another question -
A light moving at 3 ft/s approaches a 6 ft. man standing 12 ft from a wall. The light is 3 ft above the ground. How fast is the tip P of the man's shadow moving when the light is 24 ft from the wall?

The shadow is on the wall, if you didn't understand my wording =x

y = the height in feet of the shadow (as the triangle ABC is proportional with the triangle ADE (double in size), we can deduce that the height of the triangle is 6 feet. Add 9 in the end.)
x = the length in feet from the wall

we have that
[Image: yduc5uh.png]

we want to find
[Image: y8sv9uy.png]

For y, how can we make a function that's describing its height? At least we know that for y(0) = 9. We also know that the triangles ABC and ADE are proportional to eachother. However, the proportionality changes. If we find a general formula for the proportionality to those triangles...

[Image: yedhcyz.png]

So in reality, all we have to do is to find a way of describing Dy, then we've solved this issue.

The small triangle is of size
[Image: yds36xp.png]

And the big triangle is of size
[Image: ykjo543.png]

We know that
[Image: ykvm2xx.png]

Which means that
[Image: yk6kvff.png]

Therefore, let's just differentiate the height of the big triangle!
[Image: ygxxbwg.png]

Insert for t = 0 and you'll receive 3/4, or 0.75 ft/s.

Edit: Ninja'd, it seems.

Noah

The math help thread - Matt - 2009-11-17

Thanks guys.

One more question -
The minute hand of a clock is 8 in. long and the hour hand is 4 in. long. How fast is the distance between the tips of the hands changing at one o'clock?

[Image: gif&s=16]

inches/ hour
This was the answer I got

and my friend got
[Image: gif&s=15]

in/ second

Are these answers equivalent? Or did one of us screw up somewhere?

The math help thread - Russt - 2009-11-17

No, the answers are not equivalent - yours is equal to your friend's, but with a 1pi on the top instead of 11pi.

The math help thread - KajitiSouls - 2009-11-17

Russt Wrote:No, the answers are not equivalent - yours is equal to your friend's, but with a 1pi on the top instead of 11pi.

I think what you meant to say was 39600/5400 =/= 2/3 >_>

So... I take it you want to know who's correct o.O Brb...

Code:
Parametric equation time!

t is in seconds.

π = Pi.  Because this font is really unkind to Pi.

Minute Hand Tip's position

xm(t) = -8*cos(t/3600*2π + π/2)

ym(t) = 8*sin(t/3600*2π + π/2)

Hour Hand Tip's position

xh(t) = -4*cos(t/43200*2π + 2π/3)

yh(t) = 4*sin(t/43200*2π + 2π/3)

f(t) = distance between the tips of the two hands

     = sqrt([xm(t) - xh(t)]^2 + [ym(t) - yh(t)]^2)

     = sqrt([-8*cos(t/3600*2π + π/2) + 4*cos(t/43200*2π + 2π/3)]^2 + [8*sin(t/3600*2π + π/2) - 4*sin(t/43200*2π + 2π/3)]^2)

f'(t) = ...

*pulls out graphing calculator*

f'(0) = -11π/(5400 * sqrt(5 - 2*sqrt(3)))

Looks like I got your friend's answer.

The math help thread - Matt - 2009-11-17

Well, here's what I did.
[Image: ya4konq.png]

at one o'clock, the angle between the hands is pi/6 rads
cos(A) = sqrt(3)/2
sin(A) = 1/2

dA/dt = 2pi radians in 12 hours = pi/6 rads/h

[Image: ylexpzt.png]

He had his on a test and had it marked right, so if our answers don't match, mine must be wrong, and if so, I'm not sure where I went wrong in my steps.

The math help thread - Russt - 2009-11-17

I just did it and got your friend's answer:

Spoiler

The math help thread - KajitiSouls - 2009-11-17

Matt Wrote:Well, here's what I did.

<---- ERROR!!!!!!

at one o'clock, the angle between the hands is pi/6 rads
cos(A) = sqrt(3)/2
sin(A) = 1/2

dA/dt = 2pi radians in 12 hours = pi/6 rads/h<---- ERROR!!!!!!

He had his on a test and had it marked right, so if our answers don't match, mine must be wrong, and if so, I'm not sure where I went wrong in my steps.

sqrt(4) = 2, not 4.

If angle A is the angle between the minute hand and the hour hand, dA/dt is NOT 2*π radians per 12 hours. Rather, you should have found the difference dA/dt minute hand - dA/dt hour hand = net dA/dt. That's 2*π radians per hour - 2*π radians per 12 hours, or 2π/3600 rad/second - 2π/43200 = 11π/21600. Your real dA/dt was really -11π/21600 in/sec

And why didn't I think of Law of Cosines >.< lmao

The math help thread - Russt - 2009-11-17

^ ah, also. The minute hand moves too, so dC/dt (dA/dt in his post) is -11pi/6, not just pi/6.

The math help thread - Matt - 2009-11-17

Ahh, I see now, thanks a lot guys!

The math help thread - DarkPwnage - 2009-11-19

http://people.math.gatech.edu/~geronimo/ma1502/pt5.pdf

Would anyone happen to know how to do 3 and 4?