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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: GerrySly on January 27, 2009, 10:37:15 pm

Title: A few Specialist Problems
Post by: GerrySly on January 27, 2009, 10:37:15 pm: Questions I have had trouble with (or need a little push to get started) :)
Title: Re: A few Specialist Problems
Post by: fredrick on January 28, 2009, 12:16:06 pm: all those questions can be/must be done by calc. one way is you can sketch both equations and find(x-value) where they intercept. Dont forget the domain
Title: Re: A few Specialist Problems
Post by: Mao on January 28, 2009, 02:43:55 pm: question 3: http://vcenotes.com/forum/index.php/topic,2171.0.html

question 5: substitute values for A and r then solve using calculator.
Title: Re: A few Specialist Problems
Post by: GerrySly on February 10, 2009, 08:53:34 pm: Find correct to two decimal places, the angle between the line $y = \frac{1}{2}x+1$ and the tangent to the graph of $x=y^2-y$ at the point of intersection $(2,2)$
Title: Re: A few Specialist Problems
Post by: Mao on February 10, 2009, 09:26:46 pm: it is difficult to deal with $x=y^2-y$ , hence, we take the inverse of both functions [which does not change the angle between the two slopes]

$y_1^{-1}: y=2x-2$

$y_2^{-1}: y=x^2-x$

point of intersection is (2,2) [take the inverse of the point]

slope of first graph: $\frac{dy}{dx}=2\implies \tan (\theta_1)=2 \implies \theta_1 = \tan^{-1} (2)$

slope of second graph: $\frac{dy}{dx} \right| _{x=2} = 2\times(2) - 1 = 3\implies \theta_2 = \tan^{-1} (3)$

angle in between: $|\theta_2 - \theta_1| \approx 0.142^c = 8.13^o$
Title: Re: A few Specialist Problems
Post by: GerrySly on February 11, 2009, 09:25:17 pm: $\frac {d}{dx} -4\cos^{-1} (\frac{5x-3}{2})$

I came up with the following answer $\frac {4}{\sqrt {-(5x-4)(5x-2)}}$ , but the answer is actually $\frac {20}{\sqrt {-5(5x^2 - 6x + 1)}}$

I seem to not be able to do the questions with a constant in front of x, when there is no constant I am able to differentiate. This is the following equation I used to differentiate $cos^{-1}$ with no constant in front of x.

$\frac {d}{dx} \cos^{-1} (\frac{x}{a}) = \frac {-1}{\sqrt {a^2 - x^2}}$

There seems to be something I am missing

Thanks
Title: Re: A few Specialist Problems
Post by: Mao on February 11, 2009, 10:02:44 pm: instead, try using the chain rule:

$\frac{d}{dx} \left(-4\cos^{-1}\left(\frac{5x-3}{2}\right)\right) = -4\frac{d}{du}\cos^{-1}(u)\cdot \frac{d}{dx}\left(\frac{5x-3}{2}\right) = \frac{4}{\sqrt{1-u^2}}\cdot \frac{5}{2} = \frac{10}{\sqrt{1-\frac{(5x-3)^2}{4}}} = \frac{20}{\sqrt{4-(5x-3)^2}}$

expand that --> simply would give you the correct answer
Title: Re: A few Specialist Problems
Post by: GerrySly on March 03, 2009, 07:46:57 pm: $\int \cot^2{(x)} dx$

$\int \frac{\cos^2 (x)}{\sin^2 (x)} dx$

$\int \frac{1 - \sin^2(x)}{\sin^2 (x)} dx$

$\int \left [ \frac{1}{\sin^2 (x)} - \frac{\sin^2(x)}{\sin^2(x)} \right ] dx$

$\int \left [ \sin^{-2} {(x)} - 1 \right ] dx$

$-\cot (x) - x + c$

Alright, I managed to $\LaTeX$ hack my way through most of the equation but how do I integrate $\sin^{-2} {(x)}$ ?
Title: Re: A few Specialist Problems
Post by: Mao on March 03, 2009, 08:26:23 pm: firstly, $\frac{1}{\sin x} = \csc x \neq \sin^{-1} x$ , I was a little confused reading your post :P

to integrate $\csc^2(x)$ and $\sec^2(x)$ , the only way [that I know of] is by recognition:

$\frac{d}{dx}\tan (x) = \sec^2(x) + C$

$\frac{d}{dx}\cot (X) = -\csc^2 (x) + C$
Title: Re: A few Specialist Problems
Post by: GerrySly on March 06, 2009, 10:34:02 pm: $\int cosec^2 \left ( x - \frac{\pi}{2} \right )$

and

$\int \sin^2 (x) \cos^4 (x)$

And if any help can be given on $\int \cot^2 (x)$ , our teacher says he's not giving any help on how to integrate it lol
Title: Re: A few Specialist Problems
Post by: Mao on March 07, 2009, 05:40:53 pm: $\int \csc^2 \left( x-\frac{\pi}{2}\right) \; dx = -\cot \left( x-\frac{\pi}{2}\right) + C$

$\begin{align*} \int \sin^2(x)\cos^4(x)\; dx &= \int \cos^4(x) - \cos^6(x)\; dx \\ &= \int \left(\frac{\cos(2x)+1}{2}\right)^2 - \left(\frac{\cos(2x)+1}{2}\right)^3\; dx\\ &= \int \frac{\cos^2(2x) + 2\cos(2x) + 1}{4} - \frac{\cos^3(2x) +3\cos^2(2x) + 3\cos(2x) +1}{8}\; dx\\ &= \frac{1}{4}\int \frac{\cos(4x)+1}{2}+ 2\cos(2x) + 1\; dx - \frac{1}{8}\int (1-\sin^2(2x))\cos(2x) + 3\frac{\cos(4x)+1}{2} + 3\cos(2x) + 1\; dx\\ &= \frac{\sin(4x)}{32}+ \frac{x}{8} + \frac{\sin(2x)}{4} + \frac{x}{4} - \frac{1}{16}\left(\sin(2x) - \frac{\sin^3(2x)}{3}\right) - \frac{3}{64}\sin(4x) -\frac{3}{16}x - \frac{3}{16}\sin(2x) - \frac{x}{16} + C \\ &= -\frac{\sin(4x)}{64} + \frac{\sin^3(2x)}{48} + \frac{x}{8} + C\\ \end{align*}$
(checked)

$\int \cot^2(x)\; dx = \int \csc^2 (x) - 1\; dx = -\cot(x) - x + C$
Title: Re: A few Specialist Problems
Post by: GerrySly on March 23, 2009, 04:02:40 pm: A few more that I came across...

Find the asymptotes of the graph with equation $\frac{x^2+x+2}{x}$

$\int_{0}^{\sqrt{3}}\frac{2x + 3}{x^2 + 9}\, dx$
Title: Re: A few Specialist Problems
Post by: GerrySly on March 23, 2009, 06:30:42 pm: Oh also I wanted to know how do I write up a multi choice SAC? Our teacher told us to get the correct answer and then explain why the others are incorrect. How do I write that though?
Title: Re: A few Specialist Problems
Post by: GerrySly on March 23, 2009, 07:19:18 pm: Quote from: GerrySly on March 23, 2009, 04:02:40 pm
A few more that I came across...

Find the asymptotes of the graph with equation $\frac{x^2+x+2}{x}$

$\int_{0}^{\sqrt{3}}\frac{2x + 3}{x^2 + 9}\, dx$
Just made attempt number 3 on the second problem and got this

$\begin{align*} \int_{0}^{\sqrt{3}}\frac{2x + 3}{x^2 + 9}\, dx &= \int_{0}^{\sqrt{3}}\frac{2x}{x^2 + 9} + \frac{3}{x^2 + 9}\, dx \\ &= \begin{bmatrix} \log_e{(x^2 + 9)} + \tan^{-1} {(x/3)} \end{bmatrix}_{0}^{\sqrt{3}} \\ &= \left [ \log_e {(12)} + \frac{\pi}{6} \right ] - \left [ \log_e {(9)} \right ]\\ &= \log_e { \left ( \frac{4}{3} \right )} - \frac{\pi}{6}\\ &\approx 0.811 \end{align*}$

Gosh I love how LaTeX makes me see things more clearer, now just to find out how I find asymptotes, I am sure I have learnt it before but I just can't figure out what it is heh
Title: Re: A few Specialist Problems
Post by: Mao on March 23, 2009, 07:42:31 pm: Quote from: GerrySly on March 23, 2009, 04:02:40 pm
Find the asymptotes of the graph with equation $\frac{x^2+x+2}{x}$

dividing through by x to get
$\implies (x + 1) + \frac{2}{x}$

as $x\to -\infty$ , $\frac{2}{x}\to 0^-$ , so on the left, the graph becomes $(x+1) + 0^- = x+1$ . Do the same thing on the right to obtain $(x+1)+0^+ = x+1$ . Hence $y=x+1$ is a slant asymptote.

as $x\to 0^-$ (approach from the left), $\frac{2}{x}\to -\infty$ , $(x+1)+\frac{2}{x}\to -\infty$ . as $x\to 0^+$ , $\frac{2}{x}\to \infty$ , $(x+1)+\frac{2}{x}\to \infty$ , so there is a verticle asymptote $x=0$ .
Title: Re: A few Specialist Problems
Post by: GerrySly on March 23, 2009, 07:53:47 pm: Quote from: Mao on March 23, 2009, 07:42:31 pm
Quote from: GerrySly on March 23, 2009, 04:02:40 pm
Find the asymptotes of the graph with equation $\frac{x^2+x+2}{x}$

dividing through by x to get
$\implies (x + 1) + \frac{2}{x}$

as $x\to -\infty$ , $\frac{2}{x}\to 0^-$ , so on the left, the graph becomes $(x+1) + 0^- = x+1$ . Do the same thing on the right to obtain $(x+1)+0^+ = x+1$ . Hence $y=x+1$ is a slant asymptote.

Ah ok, yeah that is one of my problems heh, asymptotes that are equations, thanks for that Mao. It seems you are the only one who can help me lol
Title: Re: A few Specialist Problems
Post by: GerrySly on April 19, 2009, 03:57:14 pm: An inverted cone with depth 50 cm and radius 25cm is intially full. Water drains out at 0.5 litres per minute. The depth of water in the cone is h cm at time t minutes.(Find an expression for $\frac{dh}{dt}$ )
Title: Re: A few Specialist Problems
Post by: kamil9876 on April 19, 2009, 04:14:10 pm: $V=\frac{1}{3}h\pi r^2$

$\frac{dV}{dt}=-500$

The slope of the line on the side is a constant $\frac{50}{25}=\frac{h}{r}$
Use that expression to find r in terms of h. Then you can sub that into the equation for V to get V in terms of h. This yields an easy situation.
Title: Re: A few Specialist Problems
Post by: GerrySly on April 19, 2009, 05:13:32 pm: Quote from: kamil9876 on April 19, 2009, 04:14:10 pm
$V=\frac{1}{3}h\pi r^2$

$\frac{dV}{dt}=-500$

The slope of the line on the side is a constant $\frac{50}{25}=\frac{h}{r}$
Use that expression to find r in terms of h. Then you can sub that into the equation for V to get V in terms of h. This yields an easy situation.
$\frac{dV}{dt}=-500$ , why is that -500 and not -1/2?

I attempted the equation and ended with the answer $-\frac{1}{72\pi h^2}$ , I think I am not fully understanding something.

2 weeks of no maths at all has really hit me hard now :P
Title: Re: A few Specialist Problems
Post by: dcc on April 19, 2009, 05:17:02 pm: Quote from: GerrySly on April 19, 2009, 05:13:32 pm
$\frac{dV}{dt}=-500$ , why is that -500 and not -1/2?

1 litre = 1000 cm^3. Since all the other measurements are in cm, it makes sense to get volume in cm as well.
Title: Re: A few Specialist Problems
Post by: kamil9876 on April 19, 2009, 05:45:28 pm: $r=0.5h$
$V=\frac{1}{12}\pi h^3$
$\frac{dV}{dh}=\frac{1}{4}\pi h^2$
$\frac{dh}{dV}=\frac{4}{\pi h^2}$

and now chain rule it to get -2000 on the numerator for dh/dt.
Title: Re: A few Specialist Problems
Post by: GerrySly on April 19, 2009, 08:43:21 pm: Thanks for that guys, appreciate the help, starting to get back into the swing of things. Got another though that I can't seem to do

(http://img8.imageshack.us/img8/3573/question9e3.png)

I know it has something to do with using the relationship between $A m^2$ and $x$ but I am not sure how to do that as x is constantly changing so it can't be the radius (which I could then use $A=6x$ , to solve)
Title: Re: A few Specialist Problems
Post by: Mao on April 19, 2009, 09:22:54 pm: the circular cross section can be modelled by a circle of radius 2 with a center of (2,0): $(x-2)^2 + y^2 = 4$
[think of this as the tank is laying on its side with water flowing out to the left (-x direction)]

Then, at any given x value between 0 and 4, the width is 2y

$|y| = \sqrt{4-(x-2)^2}$

$\implies 2y = 2\sqrt{4-(x-2)^2}$

The surface area A can then be expressed as $A = (2y)\times 6 = 12\sqrt{4-(x-2)^2} = 12\sqrt{4x-x^2}$ [this area is the area of water that is exposed to atmosphere, ambiguous wording...]

hence, $\frac{dx}{dt} = \frac{-0.025\sqrt{x}}{12\sqrt{4x-x^2}} = -\frac{1}{480\sqrt{4-x}}$

(if I remember correctly, that is)
Title: Re: A few Specialist Problems
Post by: GerrySly on May 03, 2009, 04:14:01 pm: (http://img131.imageshack.us/img131/7710/kinematics.jpg)

Having some troubles with those two questions. I thought I had an idea of how to do 10 a) but I have trouble working it out without 3 bits of information. I thought that I just subbed it into the equation $v^2=u^2+2as$ , and then solved but with two unknowns I am not sure where to go from there

$\begin{align*}v^2&=(1.2)^2 + 2(a)(3.2)\\&=(6.4)a+1.44\end{align*}$

Number 11 I know I have to split it into two parts and solve, i.e. forwards and backwards, but no idea where to go from there.

Help appreciated :)
Title: Re: A few Specialist Problems
Post by: Flaming_Arrow on May 03, 2009, 04:37:28 pm: 10.

a = a
u = 1.2
d = 3.2
v = 0

$0 = 1.44 + 2 * a * 3.2$

$0 = 1.44 + 6.4a$

$-1.44= 6.4a$

$a= - 0.225$
Title: Re: A few Specialist Problems
Post by: GerrySly on May 03, 2009, 06:13:52 pm: Quote from: Flaming_Arrow on May 03, 2009, 04:37:28 pm
10.

a = a
u = 1.2
d = 3.2
v = 0

$0 = 1.44 + 2 * a * 3.2$

$0 = 1.44 + 6.4a$

$-1.44= 6.4a$

$a= - 0.225$

Oh wow, probably should have figured that out heh, thanks

$\begin{align*} t&=\frac{v-u}{a}\\ &=\frac{0-1.2}{-0.225}\\ &=\frac{-\frac{12}{10}}{-\frac{9}{40}}\\ &=\frac{16}{3} s \end{align*}$
Title: Re: A few Specialist Problems
Post by: Flaming_Arrow on May 03, 2009, 06:30:38 pm: 11.

d = ut
d = 12 * 4 = 48m

u = -4 a= ? t= 20 d=48

d= -80 + 0.5at^2

48 = -80 + 0.5 * a * 400

a = 0.64 ms^-2

i don't get what b is asking
Title: Re: A few Specialist Problems
Post by: GerrySly on May 03, 2009, 07:04:06 pm: Thanks for that, I thought B was asking for the time in which the final velocity of the first section = 0. So when the particle slows to a stop and begins going backwards

The way I thought to do that was the following

$\begin{align*} v&=\int -0.64 dt\\ &=-(0.64)t+c v(12)&=4\\ 4&=-(0.64)(12)+c\\ c&=11.68\\ \therefore v(t)&=-(0.64)t+(11.68) \end{align*}$

Then solve that for when $v(t)=0$ , but I keep getting $\frac{73}{4}$ when the answer is $\frac{55}{4}$
Title: Re: A few Specialist Problems
Post by: Mao on May 03, 2009, 10:47:34 pm: 11. Setting $t_0$ at 12 seconds, the displacement x is +48m, velocity is +4. to get back to origin, it must have a displacement of -48.

$s = ut + \frac{1}{2}at^2$

$\implies -48 = 80 + 200a \implies a = -\frac{16}{25}$

The time the particle has a negative velocity:

$v = u + at \implies 0 = 4 - \frac{16}{25}\cdot t \implies t = 6.25$ (this is the point where the particle momentarily stops)

Hence it spends the rest of the time, 13.75 seconds, going backwards.
Title: Re: A few Specialist Problems
Post by: GerrySly on May 31, 2009, 06:32:23 pm: Show that

$\begin{align*} \sin{(\theta)}+\cos{(\theta)}i&=cis \left ( \frac{\pi}{2} - \theta \right )\\ \cos{(\theta)}-\sin{(\theta)}i&=cis \left ( -\theta \right )\\ \sin{(\theta)}-\cos{(\theta)}i&=cis \left ( \theta - \frac{\pi}{2} \right )\\ \end{align*}$

Just showing me how to do the first one should be helpful, thanks
Title: Re: A few Specialist Problems
Post by: d0minicz on May 31, 2009, 06:46:14 pm: did u mean cis on the right hand side ?
Title: Re: A few Specialist Problems
Post by: kamil9876 on May 31, 2009, 07:43:16 pm: Quote from: GerrySly on May 31, 2009, 06:32:23 pm
Show that

$\begin{align*} \sin{(\theta)}+\cos{(\theta)}i&=cis \left ( \frac{\pi}{2} - \theta \right )\\ \cos{(\theta)}-\sin{(\theta)}i&=cis \left ( -\theta \right )\\ \sin{(\theta)}-\cos{(\theta)}i&=cis \left ( \theta - \frac{\pi}{2} \right )\\ \end{align*}$

Just showing me how to do the first one should be helpful, thanks
expand the RHS:
$=cis(-\theta)(cis\frac{\pi}{2})$
$=(cos(-\theta)+isin(-\theta))i$
$=(cos(\theta)-isin(\theta))i$
$=icos(\theta)-i^2sin(\theta)$
$=icos(\theta)+sin(\theta)$

which should render once latex gets working...
Title: Re: A few Specialist Problems
Post by: evaporade on May 31, 2009, 08:07:40 pm: Quicker this way:

cis(pi/2-@) = cos(pi/2 - @) + isin(pi/2 - @) = sin@ + icos@
Title: Re: A few Specialist Problems
Post by: GerrySly on June 16, 2009, 06:33:30 pm: (http://img198.imageshack.us/img198/3859/vectorquestion.png)

Now for the first part I am pretty sure how to get that answer. $\overleftarrow{X Y}$ appears to be exactly in between $\overrightarrow{A B}$ $\overrightarrow{B C}$ , hence $\frac{1}{2}(\mathbf{a}+\mathbf{b})$ . But can you just deduce that from the graphic or is there something else you can use?

Now for part b I have no idea how to begin... just began working with vectors today
Title: Re: A few Specialist Problems
Post by: Mao on June 16, 2009, 06:57:05 pm: The easiest way to think about this is to have two trapezium make a parallelogram (flip the second one horizontally). In this case, you will have

top edge = b + a
bottom edge = a + b
middle line is exactly in the centre since the line is from X to X', both are midpoints. since the line is XYX', the length XY is also half of top or bottom edge, i.e. $\frac{1}{2}(a + b)$

To prove it is parallel, since DC is parallel to AB, any sum of a and b will be parallel. hence XY is parallel to AB.

[But this is a badly written question.]
Title: Re: A few Specialist Problems
Post by: kamil9876 on June 16, 2009, 07:02:25 pm: $let \vec{AD}=c, \vec{CB}=d$

Let's travel from X to D to C to Y. This gives vector XY:

$\vec{XY}=\frac{1}{2}c+b+\frac{1}{2}d$

However, traveling from A to X to Y to B gives AB=a:

$a=\frac{1}{2}c+\vec{XY}+\frac{1}{2}d$

Subtract the two equations to get rid of $c$ and $d$
Title: Re: A few Specialist Problems
Post by: GerrySly on June 16, 2009, 07:39:55 pm: (http://img188.imageshack.us/img188/386/vectorquestion2.png)

Ok just a clarification question for this question, I can't remember how our teacher showed us to show they are linearly dependent / independent, but wikipedia says I can arrange them in a 3x3 matrix and find the determinent and if it equals 0 then it is linearly independent. Is this the way we are supposed to show it? I have been ridiculed for using methods found in wikipedia before...
Title: Re: A few Specialist Problems
Post by: dcc on June 16, 2009, 08:02:33 pm: Essentially, you want to find out the nature of the solutions to the equation

$\alpha_1 \textbf{a} + \alpha_2 \textbf{b} + \alpha_3 \textbf{c} = 0$

which can also be expressed as:

$\left[ \begin{matrix} \textbf{a} & \textbf{b} & \textbf{c} \end{matrix} \right] \left[ \begin{matrix} \alpha_1 \\ \alpha_2 \\ \alpha_3\\ \end{matrix} \right] = \textbf{0}$

Now, if the matrix $\left[ \begin{matrix} \textbf{a} & \textbf{b} & \textbf{c} \end{matrix} \right]$ is INVERTIBLE (that is, the determinant is non-zero), then you can immediately see that our solution will be of the form:

$\left[ \begin{matrix} \alpha_1 \\ \alpha_2 \\ \alpha_3 \\ \end{matrix} \right] = \left[ \begin{matrix} \textbf{a} & \textbf{b} & \textbf{c} \end{matrix} \right]^{-1} \textbf{0} = \textbf{0}$ , that is, the matrices are linearly independent.

If the matrix is not invertible, then you can do ninja shit and they should be linearly dependent.

Applying this to the examples you have provided:

a) Our 'big' matrix is $A = \left[ \begin{matrix} 4 & 2 & -4\\ 1 & -1 & 2\\ 3 & 3 & 6\\ \end{matrix} \right]$

Notice that $\det(A) \neq 0$ , so these vectors are linearly indepedent.

I'm sure the rest are doable in much the same manner (unless you can see straight off the bat that one of the provided vectors is a scalar multiple of another)
Title: Re: A few Specialist Problems
Post by: Mao on June 16, 2009, 11:19:54 pm: for the first case only:

Your teacher's method:

$a \textbf{a} + b \textbf{b} = \textbf{c}$ will give you three simultaneous equations

4a + 2b = -4
a - b = 2
3a + 3b = 6

if the above has a solution $a,b \in \mathbb{R}$ , then they are linearly dependant, otherwise independant.

The proper method used in uni is as dcc has shown above.
Title: Re: A few Specialist Problems
Post by: GerrySly on June 21, 2009, 02:50:09 pm: (http://img198.imageshack.us/img198/6586/vectorsproblem.png)

Ok, 6.a.i and ii are easy enough (6.a.ii $\overrightarrow{A B}=(3\mathbf{i}-2\mathbf{j}-\mathbf{k})-(\mathbf{i}+\mathbf{j}-5\mathbf{k})=2\mathbf{i}-3\mathbf{j}+4\mathbf{k}$ and 6.a.ii is just $\frac{4}{5}$ of that according to the ratio) but I always seem to have trouble with $\mathbf{O}$ . I thought $\mathbf{O}$ was just $0\mathbf{i}-0\mathbf{j}+0\mathbf{k}$ , but using that position vector gives me something weird.

Just helping me understand what the are asking for $\overrightarrow{O M}$ should be enough, I should be able to work it out from there.

BTW the answer is $\frac{1}{5}(13\mathbf{i}-7\mathbf{j}-9\mathbf{k})$
Title: Re: A few Specialist Problems
Post by: TrueTears on June 21, 2009, 03:15:42 pm: $\overrightarrow{OM}$ is just the position vector of M

$\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM}$

or

$\overrightarrow{OM} = \overrightarrow{OB} + \overrightarrow{BM}$

Same thing.
Title: Re: A few Specialist Problems
Post by: GerrySly on June 21, 2009, 05:21:27 pm: Thanks TrueTears, vectors are kicking my ass here...

(http://img291.imageshack.us/img291/2530/vectorprobleme.png)

The answer is just out of my reach, I know it had something to do with using $\mathbf{c.d}=|\mathbf{c}||\mathbf{d}|\cos{(\theta)}$ but I am not sure how to relate that to $\overrightarrow{C D}$
Title: Re: A few Specialist Problems
Post by: TrueTears on June 21, 2009, 05:31:35 pm: Find $cos(\theta)$ using the information provided.

Now pretend you have a triangle namely OCD.

Using the cosine rule to work out length of CD.

You know OC = 5, OD = 7.
Title: Re: A few Specialist Problems
Post by: GerrySly on July 09, 2009, 08:18:33 pm: (http://img265.imageshack.us/img265/4438/chapter11question8.png)

I always have problems with these "solution" questions, not sure how to approach them. I have problems visualising it, I can visualise the other formula questions where there is an easy exit and entry amount.

By the way that is Chapter 11 Question 8 Cambridge Essential
Title: Re: A few Specialist Problems
Post by: TrueTears on July 09, 2009, 08:24:34 pm: http://vcenotes.com/forum/index.php/topic,9192.msg121038.html#msg121038

That would help immensely.
Title: Re: A few Specialist Problems
Post by: GerrySly on July 09, 2009, 08:45:47 pm: Quote from: TrueTears on July 09, 2009, 08:24:34 pm
http://vcenotes.com/forum/index.php/topic,9192.msg121038.html#msg121038

That would help immensely.
Thanks for that TrueTears, helped partially, got some clarifications now :)

The answer is $\frac{dQ}{dt}=6-\frac{Q}{10}$ and I am able to get the 6 as $Inflow=3 g/L \cdot 2 L/min=6 g/min$ but I am having trouble getting the Outflow. By my understanding of kj_'s post the outflow is $\frac{mass}{volume}$ and the volume exits and enters the container at the same amount therefore there is no $t$ variable.

This means that the answer should be $\frac{dQ}{dt}=6-\frac{Q}{20}$ right? Not sure how they got it over 10.
Title: Re: A few Specialist Problems
Post by: Damo17 on July 09, 2009, 09:03:48 pm: Quote from: GerrySly on July 09, 2009, 08:45:47 pm
Quote from: TrueTears on July 09, 2009, 08:24:34 pm
http://vcenotes.com/forum/index.php/topic,9192.msg121038.html#msg121038

That would help immensely.
Thanks for that TrueTears, helped partially, got some clarifications now :)

The answer is $\frac{dQ}{dt}=6-\frac{Q}{10}$ and I am able to get the 6 as $Inflow=3 g/L \cdot 2 L/min=6 g/min$ but I am having trouble getting the Outflow. By my understanding of kj_'s post the outflow is $\frac{mass}{volume}$ and the volume exits and enters the container at the same amount therefore there is no $t$ variable.

This means that the answer should be $\frac{dQ}{dt}=6-\frac{Q}{20}$ right? Not sure how they got it over 10.

$Concentration = \frac{mass}{volume}$

$Outflow= \frac{mass}{volume} \times outflow \ rate$

$Outflow= \frac{Q}{20} \times 2 = \frac{Q}{10}$
Title: Re: A few Specialist Problems
Post by: GerrySly on July 14, 2009, 12:09:27 pm: (http://img20.imageshack.us/img20/111/chapter11question21.png)

Just a clarification question, is this normal notation? The umlat on the top of the y? Seems a tad strange to me...
Title: Re: A few Specialist Problems
Post by: GerrySly on July 14, 2009, 12:30:55 pm: (http://img160.imageshack.us/img160/1124/chapter11question24.png)

Another simple question haha

I was able to get the equation for displacement (x) as $x=5(9-t^2)$ , which allowed me to find the time at which the rock hits the ground, $t=3$ . Now when I subbed it into the equation for velocity, $v=-10t$ I get $-30 m/s$ . I am not sure why I am getting a negative but they are getting a positive?
Title: Re: A few Specialist Problems
Post by: kamil9876 on July 14, 2009, 12:55:25 pm: Yes, Issac Newton used this notation.

It's because the word 'speed' refers to magnitude.
Btw if you havn't already seen then there is a formula:

$v^2=u^2+2as$

Which is derived from using the method that you used. ie: sub in $t=\frac{v-u}{a}$ into $s=\frac{1}{2}at^2+ut$ and you get the formula.

Anyways, the formula i provided saves time for this problem.
Title: Re: A few Specialist Problems
Post by: GerrySly on July 14, 2009, 01:31:13 pm: Quote from: kamil9876 on July 14, 2009, 12:55:25 pm
Yes, Issac Newton used this notation.

It's because the word 'speed' refers to magnitude.
Btw if you havn't already seen then there is a formula:

$v^2=u^2+2as$

Which is derived from using the method that you used. ie: sub in $t=\frac{v-u}{a}$ into $s=\frac{1}{2}at^2+ut$ and you get the formula.

Anyways, the formula i provided saves time for this problem.
Ah ok, yeah I was going to use that formula but decided against for some stupid reason. Thanks for that
Title: Re: A few Specialist Problems
Post by: GerrySly on July 14, 2009, 02:46:51 pm: Alright just another simple question, I managed to get $\int_0^{\frac{\pi}{6}} \sin^{n}{(x)}\cos{(x)}.dx=\frac{1}{64}$ down to the following equation $\frac{1}{64}=\frac{2^{-n}}{2(n+1)}$ , and was wondering is there anyway to solve that by hand? I used nSolve on the calculator and got 3.
Title: Re: A few Specialist Problems
Post by: GerrySly on July 14, 2009, 04:07:02 pm: (http://img16.imageshack.us/img16/6934/chapter11question72.png)

No idea where to even begin with this one, most differential equations I solve I am given $y=$ and then I differentiate as many times as needed and sub in to solve.
Title: Re: A few Specialist Problems
Post by: /0 on July 14, 2009, 04:13:58 pm: $\frac{dy}{dx} = 1-y$

Now we flip so you get

$\frac{dx}{dy} =\frac{1}{1-y}$

Now you can solve for x in terms of y by integrating both sides. Later you can solve for y again.
Title: Re: A few Specialist Problems
Post by: dcc on July 14, 2009, 06:11:05 pm: But does $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}$ ?
Title: Re: A few Specialist Problems
Post by: /0 on July 14, 2009, 06:28:30 pm: er.... does it?
Title: Re: A few Specialist Problems
Post by: dcc on July 14, 2009, 06:31:03 pm: Quote from: /0 on July 14, 2009, 06:28:30 pm
er.... does it?

I'm willing to say sometimes. Other times it would be nonsense.
Title: Re: A few Specialist Problems
Post by: kamil9876 on July 14, 2009, 06:58:06 pm: /0, your username provides one idea of a problematic scenario.

how about $y=x^2$ . what is $\frac{d}{dy}(x)|_{y=1}$ ? is it 0.5 or -0.5? But for one-to-one function this will not arise such as the logarithms and stuff that you ussually deal with, or application problems where you stick to some domain.
Title: Re: A few Specialist Problems
Post by: GerrySly on July 16, 2009, 05:34:33 pm: A bowl can be described as the solid of revolution formed by rotating the graph of $y=\frac{1}{4}x^2$ around the y-axis for $0 \le y \le 25$

Find the volume of the bowl $V=1250\pi$

The bowl is filled with water and then, at time $t=0$ , the water begins to run out of a small hole in the bottom. The rate at which the water runs out is proportional to the depth, $h$ , of the water at time $t$ . Let $V$ denote that volume of water at time $t$ .

Show that $\frac{dh}{dt}=\frac{-k}{4\pi}, k > 0$

Now this is the part that I am having trouble with. I know I must do the whole $\frac{dh}{dt}\propto$ to something, then introduce a constant of proportionality but I am not sure where to go with this.

Any help is appreciated :)
Title: Re: A few Specialist Problems
Post by: TrueTears on July 16, 2009, 05:36:34 pm: I thought it was $\frac{dV}{dt} \propto h$

So $\frac{dV}{dt} = -kh$ where $k > 0$

Now find $\frac{dV}{dh}$ [Let y = h, integrate and find V as a function of h] then use chain rule.
Title: Re: A few Specialist Problems
Post by: GerrySly on July 21, 2009, 02:36:39 pm: A container intially contains 500L of pure water. Salt water is being poured in at 5L/min with 10kg salt concentrate. It is leaving the container at 10L/min. Assume m kg is the amount of salt in the container at time t hours, find $\frac{dm}{dt}$

That was a paraphrased question from our analysis task today (don't have the actual test with me)

Here is how I solved it

$\begin{align*} \frac{dm}{dt}&=Inflow - Outflow\\ &=(10\cdot 5) - \left ( \frac{m}{500} \cdot 10 \right )\\ &=50-\frac{m}{50} (kg/min) \end{align*}$

I am not sure if I went correct or not, because I had another equation to begin with but I scrapped it for this approach.

Note: My test is finished so you are not actually helping me complete a test, just helping me see my mistakes
Title: Re: A few Specialist Problems
Post by: kamil9876 on July 21, 2009, 02:43:18 pm: Quote
10kg salt concentrate

if 10kg/L then your inflow is correct.

Your outflow is wrong however:

$ouflow=C * 10$
$=\frac{m}{V} *10$

You assumed V is a constant 500, but really it changes over time: V=500+5t-10t=V=500-5t
Title: Re: A few Specialist Problems
Post by: GerrySly on July 21, 2009, 03:05:29 pm: Quote from: kamil9876 on July 21, 2009, 02:43:18 pm
Quote
10kg salt concentrate

if 10kg/L then your inflow is correct.

Your outflow is wrong however:

$ouflow=C * 10$
$=\frac{m}{V} *10$

You assumed V is a constant 500, but really it changes over time: V=500+5t-10t=V=500-5t
Hmmm, that sucks, thanks kamil
Title: Re: A few Specialist Problems
Post by: GerrySly on July 22, 2009, 05:02:08 pm: A particle travels in a path such that the position vector $\mathbf{r}(t)$ at time $t$ is given by $\mathbf{r}(t)=3\cos{(t)}\mathbf{i}+2\sin{(t)}\mathbf{j}, t\ge 0$

If the positive y axis points north and the positive x axis point east, find correct to two decimal places, the bearing of point $P$ , the position of the particle at $t=\frac{3\pi}{4}$ from:

The origin

Alright I was able to find this one out fairly easily, I found the point and the quadrant the point was in, 4th quadrant point $\left ( \frac{3}{\sqrt{2}} , \sqrt{2} \right )$ . Then I knew that the angle I find needs to add $270^o$ to it, so I went..

$\begin{align*} \tan{(\theta)}&=\frac{\sqrt{2}}{\frac{3}{\sqrt{2}}}\\ \theta&=\tan^{-1}{\left ( \frac{2}{3} \right )}\\ &=33.69^o \end{align*}$

I then added $270^o$ to it and I got $303.69^o$ which is the answer.

Now here is the part I was having trouble with

The initial position

The initial position vector was $\mathbf{r}(0)=3\mathbf{i}$ which I then converted to cartesian form to find the bearing and I got $(3,0)$ . Now my assumption was because it was on the x-axis I just subtract $90^o$ from my previous answer to get the bearing but that gave me a different answer.

Appreciate any help :)
Title: Re: A few Specialist Problems
Post by: dcc on July 22, 2009, 06:56:30 pm: $3 \cos \left(\frac{3 \pi}{4} \right) \neq \frac{3}{\sqrt{2}}$ ?
Title: Re: A few Specialist Problems
Post by: GerrySly on September 02, 2009, 09:35:14 pm: Wow it's been a while since I've hit the Spec books heh, started to get into revision mode and hit a problem straight up with Vectors heh

(http://img215.imageshack.us/img215/1108/question.png)
Title: Re: A few Specialist Problems
Post by: kamil9876 on September 02, 2009, 10:18:42 pm: tricky one. Use the cosine rule:

(i will denote dp as that dot product in the equation given).

Verify that the following two equations are true by drawing a diagram and applying the cosine rule twice:

$|AO|^2=|AB|^2 + |OB|^2 - 2dp$

$|OB|^2=|AB|^2 + |AO|^2 - 2dp$

$\implies |AO|^2-|OB|^2=|OB|^2-|AO|^2$
$\implies 2|AO|^2=2|OB|^2$
$\implies |AO|=|OB|$

hence two sides are of equal length, but the other pair is not, hence isosceles.
Title: Re: A few Specialist Problems
Post by: GerrySly on September 18, 2009, 11:11:34 am: (http://img42.imageshack.us/img42/1000/questionj.png)

Ok I am fine with doing everything up to part c. I am just not sure about the wording. What does "The Spider decides to continue along AB to join MN" mean mathematically? Scalar multiples, perpendicular? Just not sure...
Title: Re: A few Specialist Problems
Post by: GerrySly on September 18, 2009, 06:27:38 pm: Also another question, consider the following shape...

(http://img195.imageshack.us/img195/8567/questionc.png)

If we were asked to find $\overrightarrow{A C}$ is it technically wrong if we expanded like so? $\overrightarrow{A C}=\overrightarrow{A C} + \overrightarrow{C D}$ because I keep going a different way to the book and get a different answer only because I take a different router (the book expands $\overrightarrow{A C}=\overrightarrow{A B} + \overrightarrow{B D}$ ) I realise why it is silly to do it my way in this case (we are told to find $\overrightarrow{B D}$ earlier) but in an exam if we have enough information for both ways will the examiners have two answers or only the one and if we go the wrong way they will mark it incorrect?
Title: Re: A few Specialist Problems
Post by: TrueTears on September 18, 2009, 06:38:41 pm: I don't think $\overrightarrow{AC} = \overrightarrow{AC} + \overrightarrow{CD}$ .

I think book is wrong as well $\overrightarrow{AC} \neq \overrightarrow{AB} + \overrightarrow{BD}$

In any case even if you take a different route the final answer should always be the same.

Say you did $\overrightarrow{AC} = \overrightarrow{AD} + \overrightarrow{DC}$

This would yield the exact same answer if you did $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$
Title: Re: A few Specialist Problems
Post by: biggzee on September 18, 2009, 07:15:30 pm: Quote from: GerrySly on September 18, 2009, 11:11:34 am
(http://img42.imageshack.us/img42/1000/questionj.png)

Ok I am fine with doing everything up to part c. I am just not sure about the wording. What does "The Spider decides to continue along AB to join MN" mean mathematically? Scalar multiples, perpendicular? Just not sure...

allow "OA +mAB = OM + nMN" where m is any value of R, and n is between 0 and 1. solve in 3 dimensions i, j, and k
Title: Re: A few Specialist Problems
Post by: GerrySly on September 18, 2009, 07:43:19 pm: Quote from: TrueTears on September 18, 2009, 06:38:41 pm
I don't think $\overrightarrow{AC} = \overrightarrow{AC} + \overrightarrow{CD}$ .

I think book is wrong as well $\overrightarrow{AC} \neq \overrightarrow{AB} + \overrightarrow{BD}$

In any case even if you take a different route the final answer should always be the same.

Say you did $\overrightarrow{AC} = \overrightarrow{AD} + \overrightarrow{DC}$

This would yield the exact same answer if you did $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$
Ok that makes sense now, I should be getting the answer no matter what (I think I typed it up wrong but you gave me a good explanation anyway thanks :))

(http://img156.imageshack.us/img156/917/questionn.png)

Another one I am having trouble with, question a.ii in particular

Ok I can get as far as this...

$\begin{align*} |\overrightarrow{A E}| &= |\overrightarrow{A B} + \overrightarrow{B E}|\\ &=|m\mathbf{u}+n\mathbf{v}|\\ &=\sqrt{(m\mathbf{u}+n\mathbf{v})^2}\\ &=\sqrt{m^2\cdot |\mathbf{u}|^2 + 2mn\mathbf{u}\cdot \mathbf{v} + n^2|\mathbf{v}|^2}\\ &=\sqrt{m^2 + 2mn\mathbf{u}\cdot \mathbf{v}+n^2}\\ &=\sqrt{m^2+mv+n^2} \end{align*}$

Now there answer is $\sqrt{m^2-mv+n^2}$ but I not sure how they got the minus, aren't all angles within an equilateral triangle 60 degrees? How are they getting 120? Wouldn't that be saying it isn't an equilateral triangle?
Title: Re: A few Specialist Problems
Post by: bem9 on September 18, 2009, 07:50:08 pm: angle has to be tail to tail vectors so the angle is 120 degrees
Title: Re: A few Specialist Problems
Post by: GerrySly on September 18, 2009, 07:53:53 pm: Quote from: bem9 on September 18, 2009, 07:50:08 pm
angle has to be tail to tail vectors so the angle is 120 degrees
I could kiss you right now, sitting here studying for 11 hours straight makes things seem that much harder :P
Title: Re: A few Specialist Problems
Post by: GerrySly on September 20, 2009, 02:37:47 pm: (http://img27.imageshack.us/img27/2510/questionp.png)
Title: Re: A few Specialist Problems
Post by: /0 on September 20, 2009, 02:45:11 pm: Position vector is $r(t) = 6ti+(t^2+4)j$

$\Rightarrow v(t)=6i+2tj$ [v(t) is always tangent to r(t)]

When $t=4$ , $v(t) = 6i+8j$ .

The unit vector is $\frac{1}{10}(6i+8j)=\frac{1}{5}(3i+4j)$
Title: Re: A few Specialist Problems
Post by: Flaming_Arrow on September 20, 2009, 05:23:06 pm: r(t) =6ti + (t^2+4)j
r'(t) = 6i + 2tj
r'(4) = 6i + 8j
|r'(4)| = rt[36 + 64]
= 10

unit vector = r'(4) / |r'(4)|

(6i + 8j) / 10
Title: Re: A few Specialist Problems
Post by: GerrySly on September 20, 2009, 07:05:13 pm: Thanks for that guys, appreciate the help :)

(http://img21.imageshack.us/img21/6264/questionk.png)

I'm having trouble with part c. I have no idea how to find out the time when they are "nearest" each other.
Title: Re: A few Specialist Problems
Post by: TrueTears on September 20, 2009, 07:10:35 pm: There's 2 ways depending on whatever floats your boat.

1. The magnitude of |PQ| can be used. Find the minimum under the square root and thus you found the minimum t at which this occurs. [This is utilizing the result from b) ii)]

2. You can also dot the velocity vectors of OP and PQ and find when it equals 0, you must dot the velocity vectors and not the displacement because they give the direction of motion. [Since two particles are closest when they are perpendicular to each other, ie a straight line, notice this would only work if the path they travel in is a straight line.] But for this question you don't know the velocity vector for time t thus you should go with option 1.
Title: Re: A few Specialist Problems
Post by: Mao on September 21, 2009, 12:04:57 am: Quote from: TrueTears on September 20, 2009, 07:10:35 pm
There's 2 ways depending on whatever floats your boat.

1. The magnitude of |PQ| can be used. Find the minimum under the square root and thus you found the minimum t at which this occurs. [This is utilizing the result from b) ii)]

2. You can also dot the velocity vectors of OP and PQ and find when it equals 0, you must dot the velocity vectors and not the displacement because they give the direction of motion. [Since two particles are closest when they are perpendicular to each other, ie a straight line, notice this would only work if the path they travel in is a straight line.] But for this question you don't know the velocity vector for time t thus you should go with option 1.

the first method is valid.

However, I'm not sure how you got the second method. Two objects do not have to have perpendicular velocity at the closest point, this question included. [The velocities here are constant, not dependent on time]
Title: Re: A few Specialist Problems
Post by: GerrySly on September 21, 2009, 08:47:08 am: (http://img44.imageshack.us/img44/9400/questionpt.png)

I'm having trouble with part e. I know we have to use the dot product and solve for the cosine then arccos it, but I'm not sure what two vectors to use. Do I take a vector one second after being hit and the position of the hole, 100i, just not sure.

Cheers
Title: Re: A few Specialist Problems
Post by: kamil9876 on September 21, 2009, 12:19:30 pm: Quote from: Mao on September 21, 2009, 12:04:57 am
Quote from: TrueTears on September 20, 2009, 07:10:35 pm
There's 2 ways depending on whatever floats your boat.

1. The magnitude of |PQ| can be used. Find the minimum under the square root and thus you found the minimum t at which this occurs. [This is utilizing the result from b) ii)]

2. You can also dot the velocity vectors of OP and PQ and find when it equals 0, you must dot the velocity vectors and not the displacement because they give the direction of motion. [Since two particles are closest when they are perpendicular to each other, ie a straight line, notice this would only work if the path they travel in is a straight line.] But for this question you don't know the velocity vector for time t thus you should go with option 1.

the first method is valid.

However, I'm not sure how you got the second method. Two objects do not have to have perpendicular velocity at the closest point, this question included. [The velocities here are constant, not dependent on time]

The second method is reminiscent of method used when you have a path(possible straight line) and want to find the minimum distance between a FIXED point and the path, except you would take the dot product of velocity and the vector connecting the 'vehicle' and fixed point. (although there possibly could be a lesser distance at the end points of the path, or you could have several of these perpendicular cases(but this is mostly rare). Completely analogous to finding minimum by setting gradient to zero, in fact it's the same if you think about it geometrically).

edit: actually this method can sometimes give you maximum distance. By drawing a graph you can see whether it is minimum, max or something else (again, completely analogous to setting derivative to zero when finding min/max of y=f(x)). So I'll just always go with method 1 unless it's something obvious like a striaght line or curve bending away from (convex to??) FIXED point in question.
Title: Re: A few Specialist Problems
Post by: GerrySly on September 21, 2009, 08:51:02 pm: (http://img84.imageshack.us/img84/9840/questionx.png)
Title: Re: A few Specialist Problems
Post by: kamil9876 on September 21, 2009, 10:01:04 pm: turning these into vectors:

3i+1j and ai+bj.

The angle between these must be $60^o$ and the magnitude of both must be equal. Using the dot product:
$3a+b=(\sqrt{3^2+1})(\sqrt{3^2+1})cos60^o$
$3a+b=5$

but $ 10=a^2+b^2$ (from equating magnitudes).
$ \implies 10=a^2+(5-3a)^2$
$10=a^2+9a^2-30a+25$
$0=10a^2-30a+15$
$0=2a^2-6a+3$

and use the the info that a and b are positive.
Title: Re: A few Specialist Problems
Post by: Mao on September 22, 2009, 08:46:46 am: Another method:

OP = 3i + j
OQ = ai + bj
PQ = (a-3)i + (b-1)j
$|OP|^2 = |OQ|^2 = |PQ|^2$

hence $10 = a^2 + b^2$ (1)

$10 = (a-3)^2 + (b-1)^2$ (2)

equating them gives $a^2 + b^2 = a^2 - 6a + b^2 - 2b + 10 \implies 3a + b = 5$ , substituting back into (1) will yield the same answer as kamil
Title: Re: A few Specialist Problems
Post by: Mao on September 22, 2009, 08:54:50 am: Quote from: kamil9876 on September 21, 2009, 12:19:30 pm
Quote from: Mao on September 21, 2009, 12:04:57 am
Quote from: TrueTears on September 20, 2009, 07:10:35 pm
There's 2 ways depending on whatever floats your boat.

1. The magnitude of |PQ| can be used. Find the minimum under the square root and thus you found the minimum t at which this occurs. [This is utilizing the result from b) ii)]

2. You can also dot the velocity vectors of OP and PQ and find when it equals 0, you must dot the velocity vectors and not the displacement because they give the direction of motion. [Since two particles are closest when they are perpendicular to each other, ie a straight line, notice this would only work if the path they travel in is a straight line.] But for this question you don't know the velocity vector for time t thus you should go with option 1.

the first method is valid.

However, I'm not sure how you got the second method. Two objects do not have to have perpendicular velocity at the closest point, this question included. [The velocities here are constant, not dependent on time]

The second method is reminiscent of method used when you have a path(possible straight line) and want to find the minimum distance between a FIXED point and the path, except you would take the dot product of velocity and the vector connecting the 'vehicle' and fixed point. (although there possibly could be a lesser distance at the end points of the path, or you could have several of these perpendicular cases(but this is mostly rare). Completely analogous to finding minimum by setting gradient to zero, in fact it's the same if you think about it geometrically).

edit: actually this method can sometimes give you maximum distance. By drawing a graph you can see whether it is minimum, max or something else (again, completely analogous to setting derivative to zero when finding min/max of y=f(x)). So I'll just always go with method 1 unless it's something obvious like a striaght line or curve bending away from (convex to??) FIXED point in question.
yes yes yes, I see now. For moving objects P and Q, the dot product is $\vec{PQ}\cdot \dot{r}(P) = \vec{PQ}\cdot \dot{r}(Q) = 0$ , where the velocity must be non-zero, and either P and Q are travelling parallel to each other, or they intersect.
Title: Re: A few Specialist Problems
Post by: bigtick on September 22, 2009, 03:44:16 pm: OR in polar form.
p=rt(10)cis(artan1/3)
q=rt(10)cis(artan1/3 + pi/3)

.:a=rt(10)cos(artan1/3 + pi/3) and b=rt(10)sin(artan1/3 + pi/3).
Use compound angle formulae to evaluate.
Title: Re: A few Specialist Problems
Post by: GerrySly on September 22, 2009, 06:51:27 pm: Thanks guys :) Always good to see multiple approaches at questions

(http://img153.imageshack.us/img153/1108/question.png)

I'm not sure how the book got $2000 N$ . I keep getting $7200 N$ instead of what they want. The only way I could incorporate the speeds was to find the acceleration and then to find F=ma and solve.
Title: Re: A few Specialist Problems
Post by: Flaming_Arrow on September 22, 2009, 06:53:45 pm: u= 50/3 v = 20/3 t = 5

v = u + at
20/3 = 50/3 + 5a
a= -2

f = 1000 * 2
= 2000N
Title: Re: A few Specialist Problems
Post by: /0 on September 22, 2009, 06:54:09 pm: did you convert to m/s :O
Title: Re: A few Specialist Problems
Post by: GerrySly on October 04, 2009, 07:01:30 pm: Alright got a couple questions after doing a couple practice exams...

I am constantly getting implicit differentiation incorrect. It seems to be the correct equation but not in their form, how flexible are they with this? E.g.

(http://img22.imageshack.us/img22/1497/questionr.png)

I did this and came up with $\frac{dy}{dx}=\frac{y^2-x^2y}{2yx-x^3}$ and yet there answer is $\frac{dy}{dx}=\frac{2(1-xy)}{x^2+2y}$ . This has happened more often than not but for the rest of the question I get the correct gradient so it's the correct form. I am just not sure how I can go further than what I have.

Also in reading time what are allowed to use? Calculators, reference material?

Cheers :)
Title: Re: A few Specialist Problems
Post by: d0minicz on October 04, 2009, 07:05:17 pm: i got the same as you lol
used quotient rule instead of their one?
Title: Re: A few Specialist Problems
Post by: TrueTears on October 04, 2009, 08:12:32 pm: I remember a similar problem /0 brought up [actually I think it's the same]

They should accept both forms, but to get their form use the product rule.
Title: Re: A few Specialist Problems
Post by: GerrySly on October 04, 2009, 08:22:29 pm: Quote from: TrueTears on October 04, 2009, 08:12:32 pm
I remember a similar problem /0 brought up [actually I think it's the same]

They should accept both forms, but to get their form use the product rule.
Heh righto, I used the product rule to get that ( $\frac{y^2}{x}=y^2x^{-1}$ )
Title: Re: A few Specialist Problems
Post by: Mao on October 04, 2009, 09:37:50 pm: Quote from: GerrySly on October 04, 2009, 07:01:30 pm
(http://img22.imageshack.us/img22/1497/questionr.png)

I did this and came up with $\frac{dy}{dx} =\frac{y^2-x^2y}{2yx-x^3}$ and yet there answer is $\frac{dy}{dx}=\frac{2(1-xy)}{x^2+2y}$

$\begin{align*} \frac{dy}{dx} & =\frac{y^2-x^2y}{2yx-x^3} \\ & = \frac{x\left( \frac{y^2}{x} - xy\right)}{x(2y - x^3)} \\ & = \frac{\frac{y^2}{x} + xy - 2xy}{2y -x^3},\; x \neq 0 \\ & = \frac{2 - 2xy}{2y - x^3} \\ & = \frac{2(1-xy)}{2y - x^3}\\ \end{align*}$
Title: Re: A few Specialist Problems
Post by: GerrySly on October 18, 2009, 08:56:14 pm: 2007 VCAA EXAM 2 MC Q 19
A particle is initially travelling at 5 m/s. A constant force of 18 newtons is applied to the particle, in the direction of motion, for 4 seconds. The particle accelerates at $1.5 \frac{m}{s^2}$ .
The momentum of the particle after 4 seconds is...

2007 VCAA EXAM 2 MC Q 21
A 12kg mass moves in a straight line under the action of a variable force F, so that is velocity $v \frac{m}{s}$ when is is x metres from the origin is given by $v=\sqrt{3x^2-x^3+16}$ .
The force F acting on the mass is given by...

2007 VCAA EXAM 2 SA Q 4c
(http://img30.imageshack.us/img30/7196/questiond.png)

Just did 2007 VCAA Exam 2 as you can see, those were the questions I couldn't solve even after looking at the assessor's report. I always seem to have trouble with the style of questions as the last one (landing and angles) so if you could explain it generally it would be helpful :)

Thanks a bunch :)
Title: Re: A few Specialist Problems
Post by: TonyHem on October 18, 2009, 09:05:20 pm: Question 19)
F = mA
18 = M(1.5)
M = 12 kg

P = mv
Find v at t =4
u = 5, v =? t = 4, a = 1.5
v = u + at
v = 5 + 6
v = 11
p = 12*11 = 132

21) v^2 = 3x^2-x^3+16
(1/2)v^2 = 1/2(3x^2-x^3+16)
Derive it, to get a = whatever
then F = MA
so times both sides by m = 12

as for 4C, I asked that question myself so I can't help :|
Title: Re: A few Specialist Problems
Post by: GerrySly on October 18, 2009, 09:07:37 pm: Ah right thanks for that TonyHem, I knew I had to do something with other forms of acceleration for 21 but didn't know how, thanks for clearing that up :)
Title: Re: A few Specialist Problems
Post by: kamil9876 on October 18, 2009, 10:11:00 pm: Quote
I always seem to have trouble with the style of questions as the last one (landing and angles) so if you could explain it generally it would be helpful

There's a good discussion about it starting from this post, and goes till the next page too. If unsure by the details after reading this then feel free to ask specific things.

http://vcenotes.com/forum/index.php/topic,11962.msg174393.html#msg174393
Title: Re: A few Specialist Problems
Post by: Mao on October 19, 2009, 01:28:03 am: An alternative solution to Q19)

$F = \frac{dP}{dt}$ (Newton's Second Law, see Essentials Textbook)

$\implies \Delta P = F \cdot \Delta t = 72 Ns$

$F = ma \implies m = 12 kg$

$P_i = 12\times 5 = 60\implies P_f = 132\; Ns$
Title: Re: A few Specialist Problems
Post by: Mao on October 19, 2009, 01:47:56 am: Q4c)

The angle of landing deals with the angle the vector path meets with the horizontal plane. The angle is the angle between the vector tangent to the path and the horizontal, i.e. between the velocity vector and the horizontal.

The horizontal vector must point in the same direction as velocity vector, hence $\tilde{h} = 30\tilde{i} - 40\tilde{j}$

$\implies \tilde{h}\cdot \tilde{v} = |\tilde{h}||\tilde{v}|\cos \theta$ , the rest is fairly trivial.
Title: Re: A few Specialist Problems
Post by: GerrySly on October 19, 2009, 08:05:31 am: Quote from: Mao on October 19, 2009, 01:47:56 am
Q4c)

The angle of landing deals with the angle the vector path meets with the horizontal plane. The angle is the angle between the vector tangent to the path and the horizontal, i.e. between the velocity vector and the horizontal.

The horizontal vector must point in the same direction as velocity vector, hence $\tilde{h} = 30\tilde{i} - 40\tilde{j}$

$\implies \tilde{h}\cdot \tilde{v} = |\tilde{h}||\tilde{v}|\cos \theta$ , the rest is fairly trivial.

Ah right, it's the velocity vector. The vector, $\tilde{h} = 30\tilde{i} - 40\tilde{j}$ , you just used the velocity vector but removed the k component yeah? Can I do that for every question like these? (Velocity vector).(velocity vector minus k component) etc. ?
Title: Re: A few Specialist Problems
Post by: kamil9876 on October 19, 2009, 05:28:46 pm: That's assuming the "ground" is the x-y plane. What if the ground was the x-y plane tilted say $30^o$ ? :coolsmiley:
Title: Re: A few Specialist Problems
Post by: GerrySly on October 19, 2009, 05:35:03 pm: Quote from: kamil9876 on October 19, 2009, 05:28:46 pm
That's assuming the "ground" is the x-y plane. What if the ground was the x-y plane tilted say $30^o$ ? :coolsmiley:
Interesting, I assume you could just right angle triangle that shit and go j component is [tex]40\cdot \sin{(30^o)}=40\cdot \frac{1}{2}=20/tex]?
Title: Re: A few Specialist Problems
Post by: kamil9876 on October 19, 2009, 06:55:40 pm: Ahh just realised the solution may not be as simple as I thought. Kinda post without thinking :/

Another kool way of doing the original version is to find the angle from taking $v.(-k)$ . This angle is the angle between velocity and $-k$ vector. A little geometry tells you that the required angle is $90^o$ minus this. In the new example, the vector taken from tilting $k$ 30 degrees acts like the new $k$ . Yeah... as i said not that as simple as i thought it could be :/. But at least you get the picture as to why the dot product is taken and not simply the tangent of the magnitudes of the i and k component.
Title: Re: A few Specialist Problems
Post by: GerrySly on October 29, 2009, 08:36:31 am: (http://img262.imageshack.us/img262/2504/tssm.jpg)

Not quite sure about the solutions, if you could shed some light in it it'd be appreciated :)
Title: Re: A few Specialist Problems
Post by: Mao on October 29, 2009, 11:34:31 am: Oh wow, that's pretty challenging

$A_{shaded} = A_{OCD\; sector} - A_{triangle}$

$\implies A_s = \frac{\theta}{2\pi} \pi r^2 - \frac{r^2 \sin\theta}{2}$

$V = 8 \times A_s = 4\cdot 1.5^2 \theta - 4 \cdot 1.5^2 \sin \theta = 9 \theta - 9 \sin \theta = 9(\theta - \sin \theta)$
Title: Re: A few Specialist Problems
Post by: brenny on October 29, 2009, 11:41:33 am: just a quick question how would you integrate x= -800v/(10,000+v^2) to find x in terms of v
Title: Re: A few Specialist Problems
Post by: GerrySly on October 29, 2009, 02:34:42 pm: Quote from: Mao on October 29, 2009, 11:34:31 am
Oh wow, that's pretty challenging

$A_{shaded} = A_{OCD\; sector} - A_{triangle}$

$\implies A_s = \frac{\theta}{2\pi} \pi r^2 - \frac{r^2 \sin\theta}{2}$

$V = 8 \times A_s = 4\cdot 1.5^2 \theta - 4 \cdot 1.5^2 \sin \theta = 9 \theta - 9 \sin \theta = 9(\theta - \sin \theta)$

Is that formula of a sector bit required knowledge? Never seen it before :S
Title: Re: A few Specialist Problems
Post by: arthurk on October 29, 2009, 02:41:47 pm: Quote from: brenny on October 29, 2009, 11:41:33 am
just a quick question how would you integrate x= -800v/(10,000+v^2) to find x in terms of v

let u = 10000 + v^2 and u should end up with a log by the looks of it
Title: Re: A few Specialist Problems
Post by: TonyHem on October 29, 2009, 03:11:19 pm: Just curious, where is this question from? I've browsed through heaps of exams n haven't seen this q.
Title: Re: A few Specialist Problems
Post by: GerrySly on October 29, 2009, 03:15:23 pm: Quote from: TonyHem on October 29, 2009, 03:11:19 pm
Just curious, where is this question from? I've browsed through heaps of exams n haven't seen this q.
Mine? From TSSM '09
Title: Re: A few Specialist Problems
Post by: GerrySly on November 02, 2009, 05:52:13 pm: Well I survived Specialist maths lol

Thanks go out to TrueTears, Mao, kamil9876, dcc, Flaming_Arrow, evaporade, Damo17, /0, bem9, biggzee, bigtick and TonyHem for all the help over the year, truly appreciate it :)