Author Topic: VCE Specialist 3/4 Question Thread! (Read 2547242 times) Tweet Share

Homer · « **Reply #1080 on:** December 30, 2012, 10:45:01 pm »

Quote from: Calvin Climb on December 30, 2012, 10:36:42 pm

$\implies 8\sqrt{x^2+2x+1+y^2}=8-4x$

wouldn't it be -8-4x?

b^3 · « **Reply #1081 on:** December 30, 2012, 10:49:42 pm »

Are you sure thats the right equation? because the diagram shows something slightly different (x axis semi major length). Also I don't think the expressions you've got to is quite right, it nearly is, just one term is off, as long as I haven't made a mistake. Anyway, heres what I have

Spoiler

$\begin{alignedat}{1}|z+3|+|z+1| & =4 \\ \mathrm{Let}\; z=x+iy \\ |\left(x+3\right)+iy|+|\left(x+1\right)+iy| & =4 \\ \sqrt{\left(x+3\right)^{2}+y^{2}}+\sqrt{\left(x+1\right)^{2}+y^{2}} & =4 \\ \sqrt{\left(x+3\right)^{2}+y^{2}} & =4-\sqrt{\left(x+1\right)^{2}+y^{2}} \\ x^{2}+6x+9+y^{2} & =16-2\left(4\right)\sqrt{\left(x+1\right)^{2}+y^{2}}+x^{2}+2x+1+y^{2} \\ 4x-8 & =-8\sqrt{\left(x+1\right)^{2}+y^{2}} \\ 2-x & =2\sqrt{\left(x+1\right)^{2}+y^{2}} \\ 4-4x+x^{2} & =4\left(x^{2}+2x+1+y^{2}\right) \\ 0 & =3x^{2}+12x+4y^{2} \\ 3\left(x^{2}+4x+2^{2}-2^{2}\right)+4y^{2} & =0 \\ 3\left(\left(x+2\right)^{2}-4\right)+4y^{2} & =0 \\ 3\left(x+2\right)^{2}+4y^{2} & =12 \\ \frac{\left(x+2\right)^{2}}{4}+\frac{y^{2}}{3} & =1 \end{alignedat} $

EDIT: Beaten. but I typed all this out... soo..

Homer · « **Reply #1082 on:** December 30, 2012, 10:57:54 pm »

what I did was

$4x-8=-8\sqrt{(x+1)^2+y^2}$
$(4x-8)^2=64[{(x+1)^2+y^2}]$
$16x^2-64x+64=64x^2+128x+64+64y^2$
$48x^2-64x-64y^2=0$
$16(3x^2-4x-4y^2)=0$
$3x^2-4x-4y^2=0$
:/

BubbleWrapMan · « **Reply #1083 on:** December 30, 2012, 11:24:53 pm »

Should be $-48x^2$ , and you didn't subtract $128x$ properly

Homer · « **Reply #1084 on:** January 03, 2013, 10:08:23 am »

If i write {z: Im(z)Re(z-2)=-1+2Re(z-2)} instead of {z:Im(z-2)Re(z-2)=-1} would it mean the same if i was to convert it into Cartesian form?

BubbleWrapMan · « **Reply #1085 on:** January 03, 2013, 11:28:48 am »

Assuming you meant $\left\{z:\mathrm{Im}(z-2i)\mathrm{Re}(z-2)=-1\right\}$ , yes

sin0001 · « **Reply #1086 on:** January 03, 2013, 11:46:48 pm »

Came across this exam question from heinemann:
Find an anti derivative of (2+6x)/sqrt(4-x^2)
Thanks!

b^3 · « **Reply #1087 on:** January 03, 2013, 11:49:00 pm »

Hint, split it up into $\frac{2}{\sqrt{4-x^{2}}}+\frac{6x}{\sqrt{4-x^{2}}}$ then the first part will become a sin inverse, and try a u substitution for the second part.

sin0001 · « **Reply #1088 on:** January 04, 2013, 12:04:11 am »

Omg never thought of that. Ty!

Homer · « **Reply #1089 on:** January 06, 2013, 04:32:15 pm »

not sure how to do dis $:-\$

Hancock · « **Reply #1090 on:** January 06, 2013, 04:38:58 pm »

$\frac{A}{x-c} + \frac{B}{2x+c}$

$=\frac{A(2x+c)+B(x-c)}{(x-c)(2x+c)}$

$=\frac{2Ax+Ac+Bx-Bc}{(x-c)(2x+c)}$

$=\frac{x(2A+B)+c(A-B)}{(x-c)(2x+c)}$

Let $a = 2A+B$ and $b = c(A-B)$

$=\frac{ax+b}{(x-c)(2x+c)}$

So, answer b if I'm not mistaken.

Homer · « **Reply #1091 on:** January 06, 2013, 04:48:10 pm »

thanks man, i got upto here $=\frac{x(2A+B)+c(A-B)}{(x-c)(2x+c)}$ and then didn't know what to do :/ btw, what if you cant identify that a = 2A+B and b = c(A-B)

Hancock · « **Reply #1092 on:** January 06, 2013, 04:49:18 pm »

You should be able to identify that a = 2A + B and so on because 2A+B is just a scalar number. It makes sense to introduce a new variable 'a' to simplify it

Homer · « **Reply #1093 on:** January 07, 2013, 02:01:34 pm »

We have to find the second derivative of the question attached. The answer i get is slightly different to the answer provided. I have upload my workings (http://sphotos-c.ak.fbcdn.net/hphotos-ak-snc6/254695_390068854410942_1240762991_n.jpg) it'd be great if someone could pin point my mistakes. Thanks

b^3 · « **Reply #1094 on:** January 07, 2013, 02:51:41 pm »

The answer is equivalent to what the calculator gives, its just in a different form, so you haven't made a mistake, just not got it in the same form. So it's correct, I'll post up in a couple of minutes to show that they are the same.

$\begin{alignedat}{1}\frac{\frac{15}{4}\sin^{2}\left(2x\right)x^{\frac{1}{2}}+4x^{\frac{5}{2}}\sin^{2}\left(2x\right)-10\sin\left(2x\right)\cos\left(2x\right)x^{\frac{3}{2}}+8x^{\frac{5}{2}}\cos^{2}\left(2x\right)}{\sin^{3}\left(2x\right)} & =\frac{\frac{15}{4}\left(1-\cos^{2}\left(2x\right)\right)x^{\frac{1}{2}}+4x^{\frac{5}{2}}\left(1-\cos^{2}\left(2x\right)\right)-10\sin\left(2x\right)\cos\left(2x\right)x^{\frac{3}{2}}+8x^{\frac{5}{2}}\cos^{2}\left(2x\right)}{\sin^{3}\left(2x\right)}\:\:\:\left(\mathrm{using\:}\sin^{2}\left(2x\right)+\cos^{2}\left(2x\right)=1\right) \\ & =\frac{x^{\frac{1}{2}}\left(15-15\cos^{2}\left(2x\right)+16x^{2}-16x^{2}\cos^{2}\left(2x\right)-40x\sin\left(2x\right)\cos\left(2x\right)+32x^{2}\cos\left(2x\right)\right)}{4\sin^{3}\left(2x\right)} \\ & =\frac{x^{\frac{1}{2}}\left(\left(32x^{2}-16x^{2}-15\right)\cos^{2}\left(2x\right)-40x\sin\left(2x\right)\cos\left(2x\right)+16x^{2}+15\right)}{} \\ & =\frac{x^{\frac{1}{2}}\left(\left(16x^{2}-15\right)\cos^{2}\left(2x\right)-40x\sin\left(2x\right)\cos\left(2x\right)+16x^{2}+15\right)}{4\sin^{3}\left(2x\right)} \end{alignedat} $

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