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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: ggxoxo on October 29, 2012, 09:45:24 pm

Title: centre of motion
Post by: ggxoxo on October 29, 2012, 09:45:24 pm

What is the centre of motion???

For r= cos(t)i - (2sin(t))j +3k

State the position of the centre of motion.

Apparently the answer is 3k??? How???

Title: Re: centre of motion
Post by: BubbleWrapMan on October 29, 2012, 09:52:01 pm

The path is an ellipse with its centre at (0,0,3)

Title: Re: centre of motion
Post by: BigAl on October 30, 2012, 07:16:45 pm

notice that k component is not a variable but something constant..so you work out as if there is no k..u get an ellipse..and then put k back there

Title: Re: centre of motion
Post by: rife168 on October 30, 2012, 07:50:51 pm

Quote from: ggxoxo on October 29, 2012, 09:45:24 pm

What is the centre of motion???

For r= cos(t)i - (2sin(t))j +3k

State the position of the centre of motion.

Apparently the answer is 3k??? How???

Imagine if that k "coefficient" was instead zero. Then we would have an ellipse centred about the origin, or (0,0,0). Adding 3k simply shifts the whole curve 3 units in the positive direction of the z axis. Thus we get (x,y,z) -> (x,y,z+3) for all the points on the curve. So it follows that the 'centre', that is, the point (0,0,0), gets shifted up to (0,0,3)

This process holds for any curve where the "coefficient" (or "scale factor" or whatever you want to call it) of any unit vector i, j or k are constant.