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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: olly_s15 on February 04, 2010, 07:47:59 pm

Title: Olly_s15 Question Thread
Post by: olly_s15 on February 04, 2010, 07:47:59 pm: Just need some help with the following:

If $z-3i$ is a factor of $2z^4-4z^3+21z^2-36z+27$ , find the remaining factors.

I know that the conjugate is also a factor but cannot find the remaining.
Title: Re: Olly_s15 Question Thread
Post by: the.watchman on February 04, 2010, 08:02:03 pm: So the first two factors are (z-3i) and (z+3i)

This is probably a bad way but, using quad. formula:

$z=\frac{4\pm\sqrt{-8}}{4}$

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

From this, you can get the other factors, which are:

$(z-1+\frac{\sqrt{2}}{2}i)$ AND $(z-1-\frac{\sqrt{2}}{2}i)$
Title: Re: Olly_s15 Question Thread
Post by: GerrySly on February 04, 2010, 08:07:45 pm: You just expand $(z-3i)(z+3i)$ then divide $2z^4-4z^3+21z^2-36z+27$ by that. That'll give you another quadratic which you can then factorise the way you want it
Title: Re: Olly_s15 Question Thread
Post by: the.watchman on February 04, 2010, 08:09:39 pm: Oops, I forgot to show the first few steps:

Divide the expression by $(z-3i)(z+3i)$ to get $2z^2-4z+3$ ,

Then solve (as in my previous post)

Sorry about that :)
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on February 04, 2010, 08:15:28 pm: Im having troubles dividing the expression because there is no z term when i make the expansion
Title: Re: Olly_s15 Question Thread
Post by: the.watchman on February 04, 2010, 08:17:54 pm: Quote from: olly_s15 on February 04, 2010, 08:15:28 pm
Im having troubles dividing the expression because there is no z term when i make the expansion

If you're using long division (or even other methods) just put a 'ghost' term 0z into the divisor
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on February 04, 2010, 08:22:12 pm: Omg, I see. Brain not working. Thanks ;)
Title: Re: Olly_s15 Question Thread
Post by: the.watchman on February 04, 2010, 08:23:34 pm: Quote from: olly_s15 on February 04, 2010, 08:22:12 pm
Omg, I see. Brain not working. Thanks ;)

No prob.
Somehow, my pea-sized knowledge of complex numbers got me through that one (phew! :P)
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on February 09, 2010, 10:55:43 pm: Find the square root of $1+\sqrt{3i}$ in cartesian form.
Title: Re: Olly_s15 Question Thread
Post by: superflya on February 09, 2010, 11:31:47 pm: let $z=rcis\theta$

$r^{2}=1^{2}+(\sqrt{3})^{2}$ hence $\Rightarrow r=2$

now to find ur $\theta$

$\theta =\tan^{-1}(\frac{\sqrt{3}}{1})$
$\theta =\frac{\pi }{3}$

$2\theta =\frac{\pi }{3}+2k\pi$

and now combine

$2\theta =\frac{\pi }{3}+2k\pi$

$\therefore 2cis(\frac{\pi}{6}+\pi k)$
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on February 09, 2010, 11:35:04 pm: zzzz i made a mistake.. thought inverse tan of root 3 was pi/4. stupid
Title: Re: Olly_s15 Question Thread
Post by: superflya on February 09, 2010, 11:36:10 pm: Quote from: olly_s15 on February 09, 2010, 11:35:04 pm
zzzz i made a mistake.. thought inverse tan of root 3 was pi/4. stupid

lol it happens.
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on February 09, 2010, 11:46:21 pm: thanks though :)
Title: Re: Olly_s15 Question Thread
Post by: superflya on February 09, 2010, 11:48:05 pm: not a prob, its good practice for me anyway :)
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on February 13, 2010, 09:56:39 pm: Hey guys, need help with question 2f of this one.

See attached file. Thanks.
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on February 13, 2010, 09:59:31 pm: Answer attached here.
Title: Re: Olly_s15 Question Thread
Post by: Mao on February 14, 2010, 10:50:33 pm: By inspection, I am fairly sure [not 100%] that the answers are wrong, |z-z1|<|z-z2| should give the line Im(z)=-Re(z) [or y=-x], with the bottom left shaded.

Edit: confirmed, the answers are wrong
Title: Re: Olly_s15 Question Thread
Post by: Mao on February 14, 2010, 10:58:38 pm: How to do this question geometrically:

Interpretations:
|z-z1| - distance from z1
|z-z2| - distance from z2

|z-z1| = |z-z2| - the perpendicular bisector of z1 and z2. Note at any point on the perpendicular bisector, distance from the two points are the same.

|z-z1| < |z-z2| - distance from z1 is less than distance from z2. To illustrate this, draw the perpendicular bisector, the side including z1 will satisfy the condition that it is closer to z1 than z2

|z-z1| + |z-z2| = 4 - this is the geometric definition of an ellipse: sum of distances from two points are constant [compare to the definition of a circle: distance from one point is constant]

You can also do this algebraically to convince yourself, but that is too tedious and I can't be bothered.
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on February 22, 2010, 07:42:13 pm: What is the best method to express something like $\sqrt{3+4i}$ in $x+yi$ form?
Can someone please show their working. Thanks.
Title: Re: Olly_s15 Question Thread
Post by: brightsky on February 22, 2010, 07:45:07 pm: A useful way would be to express it as simply: $(3 + 4i)^{\frac{1}{2}}$ , which helps a lot when your working in cis form. But I'm not 100% sure if this answers your question. Hope this helps. :)
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on February 22, 2010, 08:08:35 pm: I think i find keeping it in cartesian form and equating real and imaginary parts then solving simultaneously works better.
I think i just answered my own question. Thanks anyway.
Title: Re: Olly_s15 Question Thread
Post by: GerrySly on February 22, 2010, 08:10:26 pm: $\begin{align*} \text{Let } z&=3+4i\\ z&=5 \text{cis}(\tan^{-1}{(\frac{4}{3})})\\ \therefore \sqrt{z}&=5 \text{cis}(\tan^{-1}{(\frac{4}{3})})\\ z&=\sqrt{5}\text{cis}(\frac{1}{2}\tan^{-1}{(\frac{4}{3})})\\ &=\sqrt{5}\left ( \cos{(\frac{1}{2}\tan^{-1}{(\frac{4}{3})})}+i\sin{(\frac{1}{2}\tan^{-1}{(\frac{4}{3})})}\right )\\ &=\sqrt{5}\left ( \frac{2}{\sqrt{5}}+i\frac{1}{\sqrt{5}}\right )\\ &=2+i \end{align*}$

That's how I did it anyway, allowed me to have it in cis form just in case further parts of the question asked for it. I'm sure theres an easier way I just found it more helpful when I did it this way
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on February 22, 2010, 08:13:09 pm: That is true. Although the question was just simply "express in x + yi form".
Title: Re: Olly_s15 Question Thread
Post by: the.watchman on February 22, 2010, 08:17:57 pm: How about:

Let $(a+bi)^2 = 3+4i$

$a^2 - b^2 + 2abi = 3+4i$

So $2ab = 4, b = \frac{2}{a}$ AND $(a+b)(a-b) = 3$

$(a+\frac{2}{a})(a-\frac{2}{a}) = 3$

$a^2 - \frac{4}{a^2} = 3$

$a^4 - 3a^2 - 4 = 0$

$a = \pm 2$

Sub in for b

Is there a shorter way?
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on February 22, 2010, 08:49:33 pm: Well I've been using the same method as the.watchman. But you could use polar form too.

I don't think there is a shorter algebraic way.
Title: Re: Olly_s15 Question Thread
Post by: Mao on February 23, 2010, 10:28:28 am: Quote from: olly_s15 on February 22, 2010, 08:49:33 pm
Well I've been using the same method as the.watchman. But you could use polar form too.

I don't think there is a shorter algebraic way.

In this case, you'll have to somehow evaluate $\sin (1/2 \tan^{-1} (4/3))$ , which would involve a calculator, or a LOT of trig identities (not undoable, but hard).

You'll also have to take into account the multiple solutions by adding the right increment to the argument. So the algebraic way is the easiest (unless you love trig).

Also, on a side note, $\sqrt{z}$ denotes the primary square root. The primary root will have the same sign for the imaginary part as original complex number (you can prove it using polar form).
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on May 16, 2010, 10:23:51 pm: new question.. it's been a while

$\int_{-1}^{1} \frac{-1}{4-(x-1)^2}\, dx$
Title: Re: Olly_s15 Question Thread
Post by: Mao on May 16, 2010, 10:49:08 pm: $\begin{align*} I & = \int_{-1}^{1} \frac{1}{(x-1)^2 - 4}\; dx \\ & = \int_{-1}^{1} \frac{1}{(x-3)(x+1)}\; dx \\ \end{align*}$

Now, partial fractions. :)
$\begin{align*} I & = \frac{1}{4}\int_{-1}^{1}\frac{1}{x-3} - \frac{1}{x+1}\; dx \\ & = \frac{1}{4} \left[ \log_e \left| \frac{x-3}{x+1} \right| \right]_{-1}^1 \\ & = \frac{1}{4} (\log_e (1) - \log_e (4/0)) ....\\ \end{align*}$

AHA I see your dilemma. Let me get my thinking cap on.
Title: Re: Olly_s15 Question Thread
Post by: Mao on May 16, 2010, 11:03:50 pm: The integrand is not continuous on the closed interval [-1,1], thus the integral cannot be evaluated directly. Using a limit approach:

$\begin{align*} I & = \int_{-1}^{1} \frac{1}{(x-3)(x+1)}\; dx \\ & = \lim_{t \to -1} \int_{t}^{1} \frac{1}{(x-3)(x+1)}\; dx \\ & = \lim_{t \to -1} \left( \frac{1}{4} \left(\log_e(1) - \log_e \left| \frac{t-3}{t+1} \right| \right) \right) \\ & = -\frac{1}{4} \lim_{t \to -1} (\log_e |t - 3| - \log_e | t + 1|) \\ & = -\frac{1}{4} (\log_e(4) - \lim_{t\to -1} \log_e | t + 1| ) \\ & = -\frac{1}{4} (\log_e(4) - \lim_{t \to 0} \log_e (t) ) \\ & = -\frac{1}{4} (\log_e(4) - (-\infty)) \\ & = -\infty \\ \end{align*}$

So informally, that integral evaluates to negative infinity. Formally, it is divergent (and thus has no integral). It really depends on if you want to take the 'area-under-the-graph' interpretation or not.
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on May 16, 2010, 11:10:11 pm: Quote from: Mao on May 16, 2010, 11:03:50 pm
The integrand is not continuous on the closed interval [-1,1], thus the integral cannot be evaluated directly. Using a limit approach:

$\begin{align*} I & = \int_{-1}^{1} \frac{1}{(x-3)(x+1)}\; dx \\ & = \lim_{t \to -1} \int_{t}^{1} \frac{1}{(x-3)(x+1)}\; dx \\ & = \lim_{t \to -1} \left( \frac{1}{4} \left(\log_e(1) - \log_e \left| \frac{t-3}{t+1} \right| \right) \right) \\ & = -\frac{1}{4} \lim_{t \to -1} (\log_e |t - 3| - \log_e | t + 1|) \\ & = -\frac{1}{4} (\log_e(4) - \lim_{t\to -1} \log_e | t + 1| ) \\ & = -\frac{1}{4} (\log_e(4) - \lim_{t \to 0} \log_e (t) ) \\ & = -\frac{1}{4} (\log_e(4) - (-\infty)) \\ & = -\infty \\ \end{align*}$

So informally, that integral evaluates to negative infinity. Formally, it is divergent (and thus has no integral). It really depends on if you want to take the 'area-under-the-graph' interpretation or not.

Yes this question is quite obscure. The book provides an answer of -pi/2 which is quite strange.

I did what you had done up until you undertook a limit approach and was no good. I think the book has made a typo and a question of this style shouldn't really be in there but meh, when you pay for the book you pay for their mistakes too.

Thanks Mao....
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on May 16, 2010, 11:11:22 pm: just looking at it, i think they meant to enclose the denominator in a square root which would provide an circular function anti derivative hence obtaining something like -pi/2.
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on June 15, 2010, 10:16:07 pm: ok having trouble with making y the subject while solving for this DE

$y'(x)=8-\frac{y}{5}$
Title: Re: Olly_s15 Question Thread
Post by: kamil9876 on June 15, 2010, 11:01:37 pm: $\frac{dy}{dx}=8- \frac{y}{5}$

$\frac{dx}{dy}=\frac{5}{40-y}=\frac{-5}{y-40}$
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on June 16, 2010, 09:20:55 pm: Quote from: kamil9876 on June 15, 2010, 11:01:37 pm
$\frac{dy}{dx}=8- \frac{y}{5}$

$\frac{dx}{dy}=\frac{5}{40-y}=\frac{-5}{y-40}$

not really what I meant but it actually helped eliminate the problem i encountered - thanks

new question:

A solid sphere of radius 6 cm has a cylindrical hole of radius 1 cm bored through it's center. What is the volume of water in the bowl?
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on July 12, 2010, 10:40:42 pm: A large container of coffee holds 10 liters of boiling water in which coffee is dissolved so that the concentration of the coffee is 40g/L. If hot water is added at a rate of 0.5L/min while the coffee solution is run off at the same rate, find the concentration of coffee in the container after 5 minutes. Assume the mixture remains uniform.

Answer at back of book:31.15g/L

my answer is different :S
Title: Re: Olly_s15 Question Thread
Post by: Mao on July 12, 2010, 11:30:11 pm: The book is correct.

The DE is $\frac{dQ}{dt} = \frac{-Q\times 0.5}{10}$
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on July 13, 2010, 09:30:27 pm: Quote from: Mao on July 12, 2010, 11:30:11 pm
The book is correct.

The DE is $\frac{dQ}{dt} = \frac{-Q\times 0.5}{10}$

ok cheers for that i'll re-attempt
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on August 01, 2010, 08:23:12 pm: An object moves with a position vector $\tilde{r}(t)=(t^3-t)\tilde{i}+2t^{-1}\tilde{j}$ . At what time is the object travelling parallel to the unit vector $\tilde{j}$
Title: Re: Olly_s15 Question Thread
Post by: fady_22 on August 01, 2010, 08:58:59 pm: Quote from: olly_s15 on August 01, 2010, 08:23:12 pm
An object moves with a position vector $\tilde{r}(t)=(t^3-t)\tilde{i}+2t^{-1}\tilde{j}$ . At what time is the object travelling parallel to the unit vector $\tilde{j}$
$\tilde{r}(t)=(t^3-t)\tilde{i}+2t^{-1}\tilde{j}=k\tilde{j}$

Therefore: $(t^3-t)=0$
t=1, 0. (reject negative solution)

t also cannot equal 0 as the j component will be undefined.

Therefore, it occurs at t=1.
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on August 01, 2010, 09:38:07 pm: Quote from: fady_22 on August 01, 2010, 08:58:59 pm
Quote from: olly_s15 on August 01, 2010, 08:23:12 pm
An object moves with a position vector $\tilde{r}(t)=(t^3-t)\tilde{i}+2t^{-1}\tilde{j}$ . At what time is the object travelling parallel to the unit vector $\tilde{j}$
$\tilde{r}(t)=(t^3-t)\tilde{i}+2t^{-1}\tilde{j}=k\tilde{j}$

Therefore: $(t^3-t)=0$
t=1, 0. (reject negative solution)

t also cannot equal 0 as the j component will be undefined.

Therefore, it occurs at t=1.

oh okay i see thanks for that
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on August 31, 2010, 10:22:29 pm: New one!

A 1.0-kg mass is just in equilibrium on a 12degree inclined plane. The angle of inclination is increased to 24degrees. What is the size of an applied force on the mass parallel to the plane so that is accelerates upwards at 1.0m/s^2?
Title: Re: Olly_s15 Question Thread
Post by: Mao on August 31, 2010, 10:36:37 pm: Tricky.

Assumption: There is static friction force. Assume the coefficient for static friction is the same as the coefficient for kinetic friction. [In real life, this is NOT true]

Thus, initially at equilibrium, friction act upwards wile gravity acts downwards.

$\mu N = mg \sin 12$

Now, accelerating upwards at 1 m/s² means friction acts downwards, as well as gravity.

$F - (\mu N + mg \sin 24) = ma \implies F = ma + mg \sin 12 + mg \sin 24$

EDIT: I can't read <.<
Title: Re: Olly_s15 Question Thread
Post by: olly_s15 on August 31, 2010, 10:48:52 pm: nvm i worked out the correct answer

letting A denote applied force, F denote friction
$A-(F+mgsin24)=ma \implies A= ma + \mu N +mgsin24 \implies A= 1 + (tan12*9.8cos24)+9.8sin24$