# Wikipedia:Reference desk/Archives/Mathematics/2009 February 13

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# February 13

## Finding x in logs

log_{2} x + log_{2} (x+5) = log_{2} 9

I use the law of logs and multiply x into x+5, then I raise 2 to both sides and end up with x^2+5x=9 but I can't factor that and get nice numbers, and I know I'm not supposed to use the quadratic formula. What am I doing wrong here? 98.221.85.188 (talk) 03:40, 13 February 2009 (UTC)

Complete the square from first principles like an honest man? Algebraist 03:46, 13 February 2009 (UTC)
5 doesn't divide evenly... 98.221.85.188 (talk) 04:19, 13 February 2009 (UTC)
Your problem does not have a nice round answer. So either you have to accept that or you wrote down the wrong equation above. Dragons flight (talk) 04:24, 13 February 2009 (UTC)
Well it's a webwork problem meaning we have to type in the answer on the web, and then it tells us if we got it right or wrong. So it's supposed to have an exact answer, but I keep getting an approximation of 1.4. The problem looks like it's written correctly. I don't know where the problem is. 98.221.85.188 (talk) 04:27, 13 February 2009 (UTC)
Nvm, I put sqrt15.25-2.5 and it says I was correct 98.221.85.188 (talk) 04:29, 13 February 2009 (UTC)
You missed something: You need to reject the extraneous root. You can't take the logarithm of a negative number. Michael Hardy (talk) 22:52, 13 February 2009 (UTC)
${\displaystyle x^{2}+5x=9\,}$

To complete the square, you divide 5 by 2 and then square, and add that amount to both sides:

${\displaystyle x^{2}+5x+{\frac {25}{4}}=9+{\frac {25}{4}}={\frac {61}{4}}.}$

Then:

${\displaystyle \left(x+{\frac {5}{2}}\right)^{2}={\frac {61}{4}},}$

so that

${\displaystyle x+{\frac {5}{2}}=\pm {\frac {{}\ {\sqrt {61}}\ {}}{2}},}$

etc. Now the complication: One of the solutions is negative. You need to reject that one since there is no base-2 logarithm of a negative number. Michael Hardy (talk) 22:51, 13 February 2009 (UTC)

Reals-chauvinist! —Tamfang (talk) 06:42, 18 February 2009 (UTC)

## Finding distance given initial speed and friction

I'm making a simple game that involves throwing balls around a complex level. The ground has some friction on the balls that slows them down. I'm not going for a perfect simulation, so as it is I'm just multiplying their velocity to a constant f slightly less than 1. So, given an initial position and speed, and a certain friction, I'm trying to predict where it will stop so I can plug in some AI code. I'm guessing it'll involve some calculus, but it's been a while... What I have looks like:

${\displaystyle P=P_{0}+v\times t}$
${\displaystyle v=v_{0}\times f^{t}}$

But I'm kinda lost there. If anyone could point out the way on how to solve this thing, I'd be very grateful. Thanks! -- JSF —Preceding unsigned comment added by 189.112.59.185 (talk) 13:34, 13 February 2009 (UTC)

I don't think you want ${\displaystyle P=P_{0}+v\times t}$, you probably want ${\displaystyle P=P_{0}+\int v\,\mathrm {d} t}$. That gives you:
${\displaystyle P=P_{0}+\int _{0}^{\infty }v\,\mathrm {d} t}$
${\displaystyle =P_{0}+\int _{0}^{\infty }v_{0}f^{t}\,\mathrm {d} t}$
${\displaystyle =P_{0}+{\frac {v_{0}}{\log {f}}}{\big [}f^{t}]_{t=0}^{\infty }}$
${\displaystyle =P_{0}-{\frac {v_{0}}{\log {f}}}}$
(Note, since f<1, log(f) is negative, so that minus sign does make sense.) That's if it's done continuously, if you are actually simulating it using discrete time steps, you'll get a slightly different answer (but not too far off if the time steps are small enough). --Tango (talk) 14:31, 13 February 2009 (UTC)
Watch out for the calculus because it tells you that with constant friction the ball never stops completely. Possibly you want a procedure like this pseudocode:
Enter P0 = start position (distance units e.g. inches)
V0 = start velocity (distance per time step)
F  = friction (velocity change per time step)
p=P0
v=V0
NEX:
p = p + v*(1+F)/2
v = v * F
if v > 0.01 goto NEX
REM The ball stops at position p.


From these start values

P0, V0, F = 1, .1, .8


the ball rolled for 11 time steps (loops to NEX) and stopped at position p = 1.41. Cuddlyable3 (talk) 20:52, 13 February 2009 (UTC)

## College Math Problem Plea ....

A steel block of weight W rests on a horizontal surface in an inaccessible part of a machine. The coffecient of friction between the block and the surface is "u". To extract the block, a magnetic rod is inserted into the machine and this rod is used to pull the block at a constant speed along the surface with a force F. The magnetic force of attraction between the rod and the block is M. Explain why

(a) M > u X W (b) F = u X W

Math problem was under the chapter : "NEWTON'S THIRD LAW" in Mechanics M1 Book of Mathematics Course PLEASE DO HELP!!!!

—Preceding unsigned comment added by 202.72.235.204 (talk) 21:33, 13 February 2009 (UTC)

We are not going to do your homework for you. J.delanoygabsadds 21:37, 13 February 2009 (UTC)

Actually I was unable to do this one so I thought what better place for help than Wiki and ofcourse you guys!!!! —Preceding unsigned comment added by 202.72.235.204 (talk) 21:49, 13 February 2009 (UTC)

What bit are you stuck on? If you show us your working so far, we'll try and help you with the next bit, but we're not going to do the whole question for you. If you don't even know where to start, you should go and talk to your teacher. --Tango (talk) 21:50, 13 February 2009 (UTC)

Okay, WORKING:

The question's first part (a) suggests that magnetic attraction "M" should be greater than frictional force experienced by steel block "uW". When this happens, infact, the block will start to accelerate as the net force on block exceeds 0 (M > friction). But the block travels with steady speed as stated in the question. But as we see it is for granted that M is always constant thus object accelerates always.

Now, whats the deal with M and F. How does pulling the rod with a force of F change anything?

When the magnetic rod is in contact with the block there is a reactive force between them than partially cancels out M. If you weren't pulling the rod, it would completely cancel it out, by pulling the rod just the right amount you leave just enough resultant force on the block to counteract the friction allowing for constant velocity. Try drawing a diagram showing the block, the surface and the rod with all the forces (I count 7 forces in total). --Tango (talk) 22:43, 13 February 2009 (UTC)

CLARIFICATION...............................................................................................:

Okay, I didn't consider the steel block to be in contact with the magnet. There are few things for clarification though:

Lets take the steel block into consideration: Taking the steel block travels to the left

Forces to the right: The magnetic force M and the pull from the magnetic rod F / uW

Forces to the left (all horizontally): Reaction contact force R = M and friction uW

Resultant force horizontally = 0

My question is that it is logical that the contact reaction force on block from magnet would decrease INCREMENTALLY AS due to the pull increments but mathematically speaking, we can take reaction force constant and workout arithmetic summation to find net force taking that there is granted constant pull from magnet on block.

WHAT REALLY HAPPENS, DOES THE CONTACT FORCE DECREASES AS PULL EXIST OR INCREASE OR CONTACT FORCE REMAINS CONSTANT AS PULL ARISES??? —Preceding unsigned comment added by 202.72.235.208 (talk) 16:54, 14 February 2009 (UTC)

The pulling force from the rod isn't a separate force, it's just the difference between the magnetic attraction and the contact force. I suggest you include the rod in your diagram and consider the forces on it too (F is a force on the rod). --Tango (talk) 17:36, 14 February 2009 (UTC)
You're overthinking this Q. Yes, in reality any force which gets the block moving would also cause it to accelerate, but just ignore that. When they said it moves at a constant speed, what they really meant was "it will move at a slight acceleration, which is minimal enough that you need not consider it in your calculations". Similarly, there isn't a single coefficient of friction, but rather are two, a higher static one and a lower dynamic one. So, you would need to pull with a greater force to "break the block loose", then decrease the pulling force to prevent acceleration. StuRat (talk) 16:35, 15 February 2009 (UTC)