Author Topic: VCE Specialist 3/4 Question Thread! (Read 2544796 times) Tweet Share

Will T · « **Reply #1170 on:** January 21, 2013, 08:53:29 pm »

Chapter review essentials chapter 4 short answer question 16.
I am majorly in struggle town.

Will T · « **Reply #1171 on:** January 21, 2013, 09:11:13 pm »

Calling out question 20 just after as a typo on the editor's part.
It says re(z)=0 then displays the line im(z)=0.
Can I get confirmation?

b^3 · « **Reply #1172 on:** January 21, 2013, 09:43:29 pm »

Quote from: Will T on January 21, 2013, 08:53:29 pm

Chapter review essentials chapter 4 short answer question 16.
I am majorly in struggle town.

We can use the factor theorem here, that is if P(ki)=0, then Z-ki is a factor,

Spoiler

$ \begin{alignedat}{1}P\left(ki\right) & =0 \\ \left(ki\right)^{3}+\left(2+i\right)\left(ki\right)^{2}+\left(2+2i\right)\left(ki\right)+4 & =0 \\ k^{3}i^{3}+2k^{2}i^{2}+k^{2}i^{3}+2ki+2ki^{2}+4 & =0 \\ -k^{3}i-2k^{2}-k^{2}i+2ki-2k+4 & =0 \\ i\left(-k^{3}-k^{2}+2k\right)-\left(2k^{2}+2k-4\right) & =0+0i \\ -k\left(k^{2}+k-2\right)+\left(2k-2\right)\left(k+2\right) & =0+0i \\ k\left(k+2\right)\left(k-1\right)=0 &\text{ or } \left(2k-2\right)\left(k+2\right)=0 \\ k=-2,\:1,\:0\: &\text{ or } k=-2,\:1 \end{alignedat} $
Note that the solutions have to satisfy both the real and complex parts

Then that means that the following two are factors
$\begin{alignedat}{1}z=-2i & \:\text{and }z=i \\ \left(z+2i\right)\left(z-i\right) & =z^{2}-zi+2zi-2i^{2} \\ & =z^{2}+2+zi \end{alignedat} $
Now if you use long division on the original expression and divide it by the above, you will get z+2, so that means the roots are -2i, i, -2

Homer · « **Reply #1173 on:** January 21, 2013, 09:44:12 pm »

q20, i think it should be im(z)=0

b^3 · « **Reply #1174 on:** January 21, 2013, 09:47:11 pm »

Quote from: Will T on January 21, 2013, 09:11:13 pm

Calling out question 20 just after as a typo on the editor's part.
It says re(z)=0 then displays the line im(z)=0.
Can I get confirmation?

Should be $\text{Im}(z)=0$

Spoiler

$\begin{alignedat}{1}|z+i| & =|z-i| \\ \text{Let\:}z=x+iy \\ |x+i\left(y+1\right)| & =|x+i\left(y-1\right)| \\ \sqrt{x^{2}+\left(y+1\right)^{2}} & =\sqrt{x^{2}+\left(y-1\right)^{2}} \\ x^{2}+y^{2}+2y+1 & =x^{2}+y^{2}-2y+1 \\ 2y & =-2y \\ 4y & =0 \\ y & =0 \\ \text{Hence\: Im}\left(z\right) & =0 \end{alignedat} $

Will T · « **Reply #1175 on:** January 21, 2013, 10:18:02 pm »

Cheers b3. Can't believe I didn't try for that for. Q16.

dossie58 · « **Reply #1176 on:** January 22, 2013, 08:59:26 pm »

Hey guys,
When sketching inequations of ellipses, hyperbolas and circles, how do we know where to colour in, for example: x^2/4 +y^2 is equal to or less than 1, and the equation: y is equal to or less than x?

Thanks!

brightsky · « **Reply #1177 on:** January 22, 2013, 09:06:48 pm »

you always have two regions from which to choose. to determine which corresponds to the given inequation, test a point in either of the two regions. if the point satisfies the inequation, then colour the region that contains the point. if the point does not satisfy the inequation, then colour the other region. for example, consider x^2/4 + y^2 =< 1 (equal to or less than 1). test a random point in either of the two regions available. let's choose (0,0). now we know that 0^2/4 + 0^2 =< 1, so (0,0) satisfies the inequation. therefore we colour in the region that contains (0,0), i.e. the inside of the circle.

Will T · « **Reply #1178 on:** January 22, 2013, 09:07:56 pm »

Ellipses are straightforward. Pick a point inside the ellipse, usually the origin unless that's a point on the ellipse. Substitute it into the inequation, if that inequation becomes true after substitution, then inside the ellipse is the required region, so shade accordingly.

For hyperbolas, it's the same idea.

Edit: what bright sky said.

Will T · « **Reply #1179 on:** January 23, 2013, 10:51:41 am »

Essentials question 9 extended response chapter review chapter 4.
Majorly struggling, also I typed it into my CAS and it said it was false

brightsky · « **Reply #1180 on:** January 23, 2013, 11:35:01 am »

9 a)

so we have z = cost + i*sint (i shall use t instead of theta for convenience).
(1+z)/(1-z) = [(1+cost) + i*sint]/[(1-cost) - i*sint]
now convert the numerator and the denominator into polar form.
for the numerator:
modulus = sqrt((1+cost)^2 + sin^t) = sqrt(1 + 2cost + cos^2 t + sin^2t) = sqrt(2 + 2cost)
argument = arctan(sint/(1+cost)) = arctan[(2sin(t/2)cos(t/2))/(1+2cos^2 (t/2) - 1)] = arctan[(2sin(t/2)cos(t/2)/(2cos^2(t/2))] = arctan(tant/2) = t/2
so polar form is sqrt(2+2cost) cis (t/2)
for the denominator:
modulus = sqrt((1-cost)^2 + sin^2t) = sqrt(1 - 2cost + cos^2(t) + sin^2(t)) = sqrt(2-2cost)
argument = arctan(-sint/(1-cost)) = -arctan(sint/(1-cost)) = -arctan(2sin(t/2)cos(t/2)/(1-(1-2sin^2(t/2)) = -arctan(2sin(t/2)cos(t/2)/2sin^2(t/2)) = -arctan(cot(t/2)) = -arctan(tan(pi/2 - t/2)) = -(pi/2 - t/2) = t/2 - pi/2
so polar form is sqrt(2-2cost) cis (t/2-pi/2)
now everything is straightforward:
[sqrt(2+2cost)cis(t/2)]/[sqrt(2-2cost) cis(t/2-pi/2)] = sqrt((2+2cost)/(2-2cost)) cis(pi/2) = sqrt((1+cost)/(1-cost))* i = i*sqrt((1+2cos^2(t/2) - 1)/(1 - 1 + 2sin^2(t/2)) = i* sqrt((2cos^2(t/2))/2sin^2(t/2)) = i*sqrt(cot^2(t/2)) = i*cot(t/2), as req

is that all you needed help with?

brightsky · « **Reply #1181 on:** January 23, 2013, 11:38:22 am »

hint for 9 c, use vectors. positions vectors are pretty much exactly the same thing as cartesian complex numbers. so all you need to show is that vector OQ . vector OP = 0. and you know how to find magnitudes already.

dossie58 · « **Reply #1182 on:** January 23, 2013, 01:24:08 pm »

How do we figure out the range and domain of a parametric hyperbola? The examples of 1H off Essentials are confusing!

Also, could someone show how to do 2f off 1H Essentials?

Thanks.

polar · « **Reply #1183 on:** January 23, 2013, 03:12:14 pm »

Quote from: dossie58 on January 23, 2013, 01:24:08 pm

How do we figure out the range and domain of a parametric hyperbola? The examples of 1H off Essentials are confusing!

Also, could someone show how to do 2f off 1H Essentials?

Thanks.

the domain is set of x values a relation takes, if I substitute all the values of t into x(t), I get all the values that x can take. this equal to the domain. the range can be worked out the same way. so for example 33
$x=3\sec(t), y=3\tan(t), t\in\left(\frac{\pi}{2},\frac{3\pi}{2}\right)$
the domain of the hyperbola is all the values $x$ can take, this is dependent on the values that $t$ can take. so to find the domain, draw the graph of $x=3\sec(t), t\in\left(\frac{\pi}{2},\frac{3\pi}{2}\right)$ and you'll see that $x \in (-\infty, -3]$ which means the domain is $(-\infty,-3]$
for the range, you do the same thing, draw the graph of $y=3\tan(t), t\in\left(\frac{\pi}{2},\frac{3\pi}{2}\right)$ which has range $R$ , hence, $y \in R$

so, for 2(f)
$\begin{aligned} x = 1- \sec(2t), \sec(2t) &= 1-x \\ y = 1+ \tan(2t), \tan(2t) &= y-1 \\ \text{Since \;} \sec^2(2t) - \tan^2(2t) &= 1 \\ (1-x)^2 - (y-1)^2 &= 1 \\ (x-1)^2 - (y-1)^2 &= 1 \end{aligned}$

to work out the domain, draw the graph of $x= 1- \sec(2t)$ on a CAS over the domain $t\in\left(\frac{\pi}{4},\frac{3\pi}{4}\right)$ which will give a range of $x \in [2,\infty]$

Jaswinder · « **Reply #1184 on:** January 24, 2013, 10:38:04 am »

$\int \frac {2x^2-9x+7} {x^2-6x+9} dx$

I keep getting $2x+3ln \left | x-3 \right | +c$

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