Mathematics Corner :)

Lol, it's like impossible to set out vector questions...but for the first part, you first have to see that since the point C lies on l, its position vector can be defined as :
(1+3t)i + (7-t)j + (-5+2t)k ....read t as lamda, lol

So OC is just ^that^ minus (0 0 0), which is still (1+3t)i + (7-t)j + (-5+2t)k.

If OC is perpendicular to l, the dot/scalar product of OC with the direction vector of l, (3i - j + 2k), should give you zero. From there, you find t, put that into OC, and you have the position vector of C.

The ratios should be okay...AC = |C-A|, CB = |B-C| (the modulus is where you do the pythagoras-type stuff, obviously)...

The difference between position and direction vectors, lol...I dunno what the best way to explain it is, but if you have a vector, OA = (1 1 1) maybe, then that gives you the position vector of the point A = (1 1 1), and the direction vector for every line parallel to the line between O and A. To get the equation of any of these other lines, you add a multiple of the direction vector to the position vector of some particular point on that line...

format_quote Originally Posted by ~Raynn~

Lol, it's like impossible to set out vector questions...

I think I did rather well. O_O

but for the first part, you first have to see that since the point C lies on l, its position vector can be defined as
(1+3t)i + (7-t)j + (-5+2t)k ....read t as lamda, lol

So OC is just ^that^ minus (0 0 0), which is still (1+3t)i + (7-t)j + (-5+2t)k.

If OC is perpendicular to l, the dot/scalar product of OC with the direction vector of l, (3i - j + 2k), should give you zero. From there, you find t, put that into OC, and you have the position vector of C.

The ratios should be okay...AC = |C-A|, CB = |B-C| (the modulus is where you do the pythagoras-type stuff, obviously)...

I see... Somewhat making sense. But not fully. Why "OC is just ^that^ minus (0 0 0)"?

format_quote Originally Posted by The Ruler

I think I did rather well. O_O

Totallyyy, I meant answers...

format_quote Originally Posted by The Ruler

I see... Somewhat making sense. But not fully. Why "OC is just ^that^ minus (0 0 0)"?

Lol, ^that^ refers to the position vector of C. And the direction OC is simply C - O.

man, i was solving that, i go write the answer and there it is, umm, good job Ryann , heh..

anyway, any other questions?

Okay... I still have difficulty fully understanding that, but I think in a while, I should be able to understand.

Could you integrate the following:

integral of: (e^3x)/(1 + e^x)

use the substitution u = 1 + e^x

The answer should be like this:

0.5e^2x - e^x + ln(1 + e^x) + k

where k is a constant.

I get all of that... except I also get an additional -1.5

format_quote Originally Posted by The Ruler

Okay... I still have difficulty fully understanding that, but I think in a while, I should be able to understand.

Try drawing a little diagram too...a line l, a point C on it, and a point O off it...may help see which two directions you're taking the dot product of...

The integration...

u = 1 + e^x >> du/dx = e^x = u - 1, dx = 1/ (u - 1) du
and e^3x = (u - 1)^3, so...

[Integral] e^3x / (1+ e^x) dx becomes
[Integral] (u - 1)^3 /u * 1/ (u - 1) du

Cancel out a (u-1) from the top and bottom, then expand the (u - 1)^2, so you'll get:

[Integral] (u^2 -2u + 1)/u du
= [Integral] u - 2 + 1/u du
= 0.5u^2 - 2u + lnu + k....and you can substitute back in, lol...

... Aaaah. This: dx = 1/ (u - 1) du is not something I do often.

And another one:

In an experiment, a scientise considered the loss of mass of a collection of picked leaves. The mass M grams of a single leaf was measures at times t days after leaf was picked.

The scientist attempted to find a relationship between M and t. In a preliminary model she assumed that the rate of loss of mass was proportional to the mass M grams of the leaf.

a) write down a differential equation for the rate of change of mass of the leaf, using this model.

- dM/dt = -kM where k>0

b) Show, by differentiation, that M = 10(0.98)^t satisfies this differential equation.

What exactly is the question asking for?

This is an interesting question, lol...basically, if you differentiate:

M = 10(0.98)^t, which can be written as:

M = 10 e^ln(0.98^t) = 10 e^(tln0.98)

dM/dt = 10 ln0.98e^(tln0.98)

= 10 ln0.98 (0.98^t)

= ln0.98 * (10 (0.98^t))

= ln0.98 * M

...which is in the form dM/dt = -kM, where k = -ln0.98.

That may take a while to decipher, lol...

No, that was easier for me to understand than any of the vector stuff. It's just that I've not come across a question like that until now. And in a mock paper too. Not an actual past paper. Tch.

do high math enough, you forget basic 2nd grade level math?

put these numbers in order from highest to lowest, in less than one minute:


0.1010
0.0525
0.676
0.0505
0.578
0.1111
0. 1727
0.0100
1.0101

lol.. that is brilliant you deserve reps, insha'Allah..
actually you came in very early, the timing is in competition with your own self not with the forum...

what is a quick way to differentiate? which is bigger and which is smaller? when you are dealing with decimals? believe it or not get me at the end of the day, and I get confused.. this was an exercise I started to get myself into 'normal lab values' of some pathological disorders.. when you stare at numbers long enough, you become confused...

no it is my pleasure, thank you, I always did it the hard way, which is you'd subtract one from the other and the +ve number is bigger but your method is better..
tis possible indeed to have your brain cells fried on the simplest of things lol like figuring out a tip at a restaurant..

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