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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: tarquin008 on August 26, 2008, 05:35:30 pm

Title: maths help required
Post by: tarquin008 on August 26, 2008, 05:35:30 pm: need help on this question:

the position of an objectat any time , t , is given as follows r(t): 3t²i + t³j - 6tk
find when the object is moving in a direction parallel to the unit vector k

any help much appreciated
Title: Re: maths help required
Post by: shinny on August 26, 2008, 05:47:02 pm: Uhh, perhaps I've misunderstood the question/theory, but for some reason I'm getting for no values of t will the particle be moving parallel to k. Anyone got any ideas? Or is that the actual answer...
Title: Re: maths help required
Post by: bigtick on August 26, 2008, 05:58:20 pm: At t =0.
Title: Re: maths help required
Post by: Collin Li on August 26, 2008, 06:09:32 pm: Differentiate the position vector with respect to t to get your velocity vector. Find t such that the i and j components are zero (and k non-zero). The answer is t = 0.
Title: Re: maths help required
Post by: tarquin008 on September 29, 2008, 09:53:24 pm: a complex number question i need help on .... i know, i suck at maths

if z= x + yi, find the values of x and y such that ( z - 1) divided by ( z + 1 ) = i
Title: Re: maths help required
Post by: shinny on September 29, 2008, 10:13:29 pm: $\frac{z-1}{z+1}=i$

$\frac{z-1+1-1}{z+1}=i$

$\frac{z+1}{z+1}-\frac{2}{z+1}=i$

$1-\frac{2}{z+1}=i$

$z+1-2=i(z+1)$

$z-iz=i+1$

$z(1-i)=i+1$

$z=\frac{1+i}{1-i}$

$z=\frac{(1+i)^{2}}{(1-i)(1+i)}$

$z=\frac{2i}{2}$

$z=i$

$x+iy=i$

$\mbox{Therefore x=0, y=1}$

Given the simplicity of the answer, there might be a much faster way of approaching this but meh =T
Title: Re: maths help required
Post by: Collin Li on September 29, 2008, 10:16:27 pm: $\frac{z-1}{z+1} = i$

$\implies z-1 = i(z+1)$

$\implies x-1 + yi = i(x+1 + yi)$

$\implies x-1 + yi = -y + i(x+1)$

Equating real and imaginary parts:

$x-1 = -y \implies y = 1-x$ --(1)

$y = x+1$ --(2)

Equating them:

$x+1 = 1-x$

$\implies 2x = 0$

$\implies x = 0$

$\implies y = 1$
Title: Re: maths help required
Post by: tarquin008 on September 29, 2008, 10:22:15 pm: yeah thanks
Title: Re: maths help required
Post by: Mao on September 29, 2008, 10:40:37 pm: Quote from: shinjitsuzx on September 29, 2008, 10:13:29 pm
$\frac{z-1}{z+1}=i$

$\frac{z-1+1-1}{z-1}=i$

$\frac{z+1}{z-1}-\frac{2}{z-1}=i$

$1-\frac{2}{z-1}=i$

$z+1-2=i(z+1)$

there's some major confusion with + and - there

corrected version:

$\frac{z-1}{z+1}=i$

$\frac{z-1+1-1}{z+1}=i$

$\frac{z+1}{z+1}-\frac{2}{z+1}=i$

$1-\frac{2}{z+1}=i$

$z+1-2=i(z+1)$
Title: Re: maths help required
Post by: shinny on September 29, 2008, 10:50:54 pm: Whoops, confused it near the start of transcribing and I thought I fixed it back up. Guess not =T