Author Topic: VCE Specialist 3/4 Question Thread! (Read 2547350 times) Tweet Share

zvezda · « **Reply #1125 on:** January 15, 2013, 11:07:13 am »

Quote from: Jenny_2108 on January 15, 2013, 03:54:49 am

you already found a, so just sub value of a back into equation $a^2 -b^2=1$ to find b value

$2 ^ { \frac {1} {4} } cis( \frac {\pi } {8}) = \sqrt { \frac {1+ \sqrt {2} } {2}} + \sqrt { \frac {-1+ \sqrt {2} } {2} } i$

$\implies 2 ^ { \frac {1} {4} } cos( \frac {\pi } {8}) = \sqrt { \frac {1+ \sqrt {2} } {2}}$

$\implies cos( \frac {\pi } {8}) = \frac{({ \frac {1+ \sqrt {2} } {2}})^\frac{1}{2}}{2 ^ {\frac {1} {4}} }$

$\implies cos( \frac {\pi } {8}) = (\frac{1+ \sqrt {2} } {2\sqrt{2}})^{\frac{1}{2}}$

$\implies cos( \frac {\pi } {8}) = (\frac{(1+ \sqrt {2})\times \sqrt{2} } {4}) ^{\frac{1}{2}}$

$\implies cos( \frac {\pi } {8}) =\frac{\sqrt {{2+\sqrt{2}}}}{2}$

To find sin(pi/8), you do similarly. I'm so lazy to type LaTex again

ok, when i use that equation to find the b value it works. Cos in fact, initially, i used 2ab=1.

sorry, but is it all right if you explain the thinking process between steps 2 and 3 for part b? I think you've converted 2^1/4 to sqrt 2^1/2 and then multiplied the bases with the same power (2^1/2 and sqrt 2^1/2) to get 2sqrt2 all to the power of 1/2), but thats not mathematically correct is it?

Homer · « **Reply #1126 on:** January 15, 2013, 11:26:24 am »

After Step two square both sides
$\implies cos ^ 2 ( \frac {\pi } {8}) = \frac {({ \frac {1+ \sqrt {2} } {2}})}{2 ^ { \frac {1} {2}} }$

then

$\implies cos ^ 2( \frac {\pi } {8}) = (\frac {1+ \sqrt {2} } {2\sqrt{2}})$

which then comes to,

$\implies cos( \frac {\pi } {8}) = (\frac{1+ \sqrt {2} } {2\sqrt{2}})^{\frac{1}{2}}$

then as jenny_2108 said,

$\implies cos ( \frac {\pi } {8}) = ( \frac{(1+ \sqrt {2}) ({2 \sqrt{2}}) } {2 \sqrt{2} ({2 \sqrt{2}})}) ^ { \frac {1} {2}}$

$\implies cos ( \frac {\pi } {8})= ( \frac {1+ \sqrt {2}({2 \sqrt {2}}) } {8}) ^ { \frac {1} {2}}$

$\implies cos( \frac {\pi } {8}) = (\frac{(1+ \sqrt {2})\times \sqrt{2} } {4}) ^{\frac{1}{2}}$

$\implies cos( \frac {\pi } {8}) =\frac{\sqrt {{2+\sqrt{2}}}}{2}$

zvezda · « **Reply #1127 on:** January 15, 2013, 01:51:46 pm »

Quote from: Homer on January 15, 2013, 11:26:24 am

After Step two square both sides
$\implies cos ^ 2 ( \frac {\pi } {8}) = \frac {({ \frac {1+ \sqrt {2} } {2}})}{2 ^ { \frac {1} {2}} }$

then

$\implies cos ^ 2( \frac {\pi } {8}) = (\frac {1+ \sqrt {2} } {2\sqrt{2}})$

which then comes to,

$\implies cos( \frac {\pi } {8}) = (\frac{1+ \sqrt {2} } {2\sqrt{2}})^{\frac{1}{2}}$

then as jenny_2108 said,

$\implies cos ( \frac {\pi } {8}) = ( \frac{(1+ \sqrt {2}) ({2 \sqrt{2}}) } {2 \sqrt{2} ({2 \sqrt{2}})}) ^ { \frac {1} {2}}$

$\implies cos ( \frac {\pi } {8})= ( \frac {1+ \sqrt {2}({2 \sqrt {2}}) } {8}) ^ { \frac {1} {2}}$

$\implies cos( \frac {\pi } {8}) = (\frac{(1+ \sqrt {2})\times \sqrt{2} } {4}) ^{\frac{1}{2}}$

$\implies cos( \frac {\pi } {8}) =\frac{\sqrt {{2+\sqrt{2}}}}{2}$

thanks homer, 100% clear. cheers. thanks to you too jenny

Homer · « **Reply #1128 on:** January 15, 2013, 03:22:25 pm »

$\int \frac{ln(x)}{x}dx=\frac{[ln(x)]^2}{2}+c$
could someone show me their method. thanks

e^1 · « **Reply #1129 on:** January 15, 2013, 03:27:58 pm »

Quote from: Homer on January 15, 2013, 03:22:25 pm

$\int \frac{ln(x)}{x}dx=\frac{[ln(x)]^2}{2}+c$
could someone show me their method. thanks

$\begin{align*}\text{Let } u = log_{e}(x) \\ \therefore \frac{du}{dx} &= \frac{1}{x} \\ \therefore du &= \frac{1}{x} dx \end{align*}$

$\begin{align*} \therefore \int \frac{ln(x)}{x}dx &= \int u du \\ &= \frac{u^2}{2} + C \\ &= \frac{[ln(x)]^2}{2}+C \end{align*}$

zvezda · « **Reply #1130 on:** January 15, 2013, 11:15:03 pm »

Hey, another question, modulus of (z-1)/z=1, how would i find the equation. Ive reached a dead end with modulus of (x^2-2x+2yi+y^2)/(x^2+y^2)=1.
thanks in advance

brightsky · « **Reply #1131 on:** January 15, 2013, 11:46:54 pm »

move the x^2 + y^2 to the RHS (basically multiply both sides by (x^2 + y^2) and then equate real part and imaginary part.

zvezda · « **Reply #1132 on:** January 15, 2013, 11:48:30 pm »

Quote from: brightsky on January 15, 2013, 11:46:54 pm

move the x^2 + y^2 to the RHS (basically multiply both sides by (x^2 + y^2) and then equate real part and imaginary part.

It's in modulus form though?

brightsky · « **Reply #1133 on:** January 16, 2013, 12:14:27 am »

abs((z-1)/z)=1
abs(x+yi - 1)/(x+yi) = 1
abs((x-1) + yi)/(x+yi)) = 1
abs(((x-1) + yi)(x-yi)/(x^2+y^2)) = 1
abs(((x^2-x+y^2)+yi))/(x^2+y^2)) = 1
sqrt(x^2-2x+y^2 +1)/sqrt(x^2+y^2) = 1
x^2 -2x + y^2 + 1 = x^2 + y^2
-2x+1=0
x=1/2

should be a more elegant way but yeah

Homer · « **Reply #1134 on:** January 16, 2013, 10:22:52 am »

zvezda · « **Reply #1135 on:** January 16, 2013, 10:52:02 am »

Quote from: Homer on January 16, 2013, 10:22:52 am

$\frac { \left | z-1 \right |}{ \left | z \right |} =1$

${ \left | z-1 \right |} ={\left | z \right |}$

${ \left | (x-1)+yi \right |} ={ \left | x+yi \right |}$

$\sqrt {(x-1)^2+y^2}= \sqrt {x^2+y^2}$

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

$x ^2 - 2 x + 1 + y ^2= x ^2+ y ^2$

$-2x + 1 =0$

$x= \frac {1}{2}$

Oh wow.... why didnt i see that...
thanks for that.
and brightsky, i was following your method up until when you sqrt. you didnt seem to square the terms before sqrt as you would do - as far as i know - with modulus

zvezda · « **Reply #1136 on:** January 16, 2013, 01:56:22 pm »

2 more questions lol.
1: If, with an Argand diagram with origin O, the point P represents z and Q represents 1/(conj z), prove that O, P and Q are collinear and find the ratio OP:OQ in terms of modulus z.

I know how to do the first part, the second part i thought i knew how to do until i came across a really messy equation and i just want to see what you guys would do.

2: Let w=2z. Describe the locus w if z describes a circle with centre (1,2) and radius 3.

I know that the answer is a circle with centre (2,4) and radius 6, but how would i show that? ive tried some possible ways to work it out on paper rather than in my head, but none have worked so far, most of them returning back to the original circle equation.

brightsky · « **Reply #1137 on:** January 16, 2013, 03:02:43 pm »

let z = rcis theta. OP has length r and OQ has length 1/r. so OP/OQ = r/(1/r) = r^2 = square of mod z.

i'm quite unfamiliar with the notation used in q2...where did you source this question?

zvezda · « **Reply #1138 on:** January 16, 2013, 03:29:30 pm »

Quote from: brightsky on January 16, 2013, 03:02:43 pm

let z = rcis theta. OP has length r and OQ has length 1/r. so OP/OQ = r/(1/r) = r^2 = square of mod z.

i'm quite unfamiliar with the notation used in q2...where did you source this question?

of course. cheers. It's from the essentials text book. Ex4H q 15

polar · « **Reply #1139 on:** January 16, 2013, 04:18:28 pm »

let z=x+yi, all points that lie on z are in the form (3cos(t)+1,3sin(t)+2)

if w=2z, then w=2(x+yi) = 2x+2yi, thus, double both the real and imaginary components to get (6cos(t)+2,6sin(t)+4)
now, x=6cos(t)+2 and y=6sin(t)+4,

solve for cos(t) and sin(t)
cos(t) = (x-2)/6 and sin(t) = (y-4)/6

since cos^2(t) + sin^2(t) =1
((x-2)/6)^2 + ((y-4)/6)^2 = 1
(x-2)^2/36 + (y-4)^2/36 = 1
(x-2)^2 + (y-4)^2 = 36

hence, w is a circle with centre (2,4) and radius 6 units.

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