Last Minute Maths

Burt

Burt

Burt

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Joined
30 Mar 2004
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686
Location
Edinburgh
Been doing some maths work and figured as a last resort I'd try here before confessing to my lecturer I'm a lazy fool.

I have two questions, both of which I understand the basics in, but both of which have elements that throw me.

1) The gravitational attraction at the point (x, y) due to point masses in the plane is G(x,y)= 1/x + 4/y + 9/(4-x-y).

Use partial differentiation to calculate the minimum value of this attraction.

((This one seems pretty clear - find Gx=0 and Gy=0 to determine the minimum root, right? But I'm a little hazy on how to deal with the terms. Is it the quotient rule or something to that effect? Can't seem to get an equation I can solve.))

2) Determine the image coordinates of the point (1, -1, 2) when it is rotated through 45 degrees about the line: r= -i + 2j + k + t(i-2k).

((Think I get the idea behind this one, but the 't(i-2k)' throws me off. What do I do with that?))
 
I can do both of those questions, and i'd love to help you, but its far too late for my head to get round either of them.

So to sum this reply up, im just saying i can do them but im not helping you. I'm in work tomorrow at 11am, if im up early enough to collect my thoughts i'll check here and see if i can help more then.

I really need time to check over my folder work too, its about 10 months since i touched maths and im horrbily rusty at it :(.
 
You need to use the quotient rule on the first one. The quotient rule is something like

d/dx (u/v) = (v du/dx - u dv/dx) / v^2

The second one I could do if I sat down and thought about it, but I really don't want to.
 
in the latter question, you may want to think of r= -i + 2j + k + t(i-2k) like y=mx + c, where

-i+2j + k == c

i-2k == m.

I'm still thinking on the rest of it though :)

first I would think of some transforms to get the line of rotation to be an axis, eg

1)r-> r+i-2j-k

2) then rotate in the plane to line up i-2k with an axis

3) rotate 45degrees about the axis

4)put the initial point into your new formula :)


edit: better would be:

1)translate
2)rotate for alignment
3)rotate degree from question)
4)inverse alignment rotation
5)inverse translation

so

r0 def= -i+2j+k

r def= (1, -1, 2)

[+r0][Rz_inverse(atan(2))][Rx(-45)] [Rz(atan(2))] [-r0] operating on [r]

where Rx, Rz are rotation matrices.

the atan(2) comes from tan(theta) = dy/dx

and since r = t(i-2k), after translation r->r-r0, the angle between the y axis and the new 'r' is atan(-2) so you want to rotate -atan(-2) = atan(2) degrees to line up the line and the x axis (increasing in the same direction).

then rotate *the point* 45deg, so rotate the axis -45deg.

then use the inverse rotation/translation to get back to the starting space.


hth, and hope i'm right! :)
 
Last edited:
Amleto said:
in the latter question, you may want to think of r= -i + 2j + k + t(i-2k) like y=mx + c, where

-i+2j + k == c

i-2k == m.

I'm still thinking on the rest of it though :)

first I would think of some transforms to get the line of rotation to be an axis, eg

1)r-> r+i-2j-k

2) then rotate in the plane to line up i-2k with an axis

3) rotate 45degrees about the axis

4)put the initial point into your new formula :)


edit: better would be:

1)translate
2)rotate for alignment
3)rotate degree from question)
4)inverse alignment rotation
5)inverse translation

so

r0 def= -i+2j+k

r def= (1, -1, 2)

[+r0][Rz_inverse(atan(2))][Rx(-45)] [Rz(atan(2))] [-r0] operating on [r]

where Rx, Rz are rotation matrices.

the atan(2) comes from tan(theta) = dy/dx

and since r = t(i-2k), after translation r->r-r0, the angle between the y axis and the new 'r' is atan(-2) so you want to rotate -atan(-2) = atan(2) degrees to line up the line and the x axis (increasing in the same direction).

then rotate *the point* 45deg, so rotate the axis -45deg.

then use the inverse rotation/translation to get back to the starting space.


hth, and hope i'm right! :)
Click to expand...

I was just about to post that!
 
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