Tim Cinel » Maths

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Parallelograms and Vectors

Tim Cinel — Sat, 15 Mar 2008 14:40:53 +0000

“Again with the maths…” Thursday on the train home, I was trying to solve a maths exercise – I am given A, B and C. I was asked to solve D so that ABCD formed a parallelogram. Where did I start? Here’s what I originally thought:

kBC = AD (The top and bottom edges should be parallel)
|AB| = |CD| (The two sides would be the same length)

I couldn’t think of any more rules to help me, so I gave made a note of it, then moved on. It wasn’t long until I’d be at it again, and this time I’d solve it.

Paul (again) mentioned that you could use the midpoint between the opposing vertices (A and C ) and (B and X). And that’s how I solved it

Here is the proof…

If A, B and C are known, solve D so ABCD forms a parallelogram.
Mathamatical Methods for Engineers and Scientists, p7, 1.2.3 Q3

Blast my inferior logic!

Vectors – Finding a Midpoint Between Two Points

Tim Cinel — Sat, 15 Mar 2008 12:12:02 +0000

Being the complete twit I am, over the last three days I’ve spent about … one hour, maybe two … trying to work out the coordinates of the midpoint between two points on a 3 dimensional Cartesian plane, using vector addition. It’s all better now, though.

In an example, I was given A(1,-1,3) and B(-2,7,-2) and I needed to find the midpoint, P.

I was trying to work it out as if P was the same as 1/2(AB) but it’s quite obviously not because that’s only the vector position from point A to point P (halfway Between A and B). Rather, we were looking for the vector position from the origin to the point P. If it wasn’t for Paul, I might still be trying to calculate it like this!

Therefore, the absolute vector position of point P (midpoint between a and B) can be deducted like:
P = OA + 1/2 * (AB)

Here are the workings for the CORRECT solution

I Are Okay at Implicit Differentiation (Now)

Tim Cinel — Fri, 14 Mar 2008 14:09:45 +0000

Last week I got stuck working on basic differentiation – it was the implicit differentiation that threw me. This sentence in particular:

“In these cases [when an equation can't be transposed to make y the subject and you are finding the derivative of y with respect to x], y needs to be solved as a function of x”

Now I get it. You have to treat y as if it were x, but a little bit differently. So if:
y = 2y + 5

Instead of saying:
dy/dx = 2 (wrong)

You need to do this:
dy/dx = 2y’

So that problem I couldn’t solve last week – I totally solved it. I smashed the shit out of it. I can’t be bothered getting it and scanning it though, so if you don’t believe me go to hell.

So I said to the guy, “Hey, that’s my daughter!”
Group laughs

I Are Suck at Implicit Differentiation!

Tim Cinel — Sat, 08 Mar 2008 14:34:50 +0000

As far as I understand, Implicit Differentiation is useful when y’ can’t directly be solved with respect to x, like in this equation:

x² – y³ = 8

The first way to try and get around this is to to transpose the equation, making y the subject, ie:

y = cuberoot(x² – 8 )

At this point, you can solve y’ with respect to x as per usual.

Sometimes, though, you can’t make solve for y.

“In these cases, y needs to be solved as a function of x” – I don’t completely get this – I’m going to ask about it though… Here’s where I get stuck:

The answer’s supposed to be 4y!!!