The math help thread

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

A very big difference:


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

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

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

A very big difference:


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

Ugh...
I should revive my first equation then.

Ugh...
I should revive my first equation then.

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

Ugh...
I should revive my first equation then.

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



Noah

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

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

answer: Do the following:

Set


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

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


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

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

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

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


we want to find


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...



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


And the big triangle is of size


We know that


Which means that


Therefore, let's just differentiate the height of the big triangle!


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

Edit: Ninja'd, it seems.

Noah

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?

inches/ hour
This was the answer I got

and my friend got
in/ second

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

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

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.

Well, here's what I did.





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





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.

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

 Spoiler

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

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

Ahh, I see now, thanks a lot guys!

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

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