Author Topic: VCE Specialist 3/4 Question Thread! (Read 2580584 times) Tweet Share

BubbleWrapMan · « **Reply #1935 on:** July 06, 2013, 09:39:30 am »

I would say so.

Nato · « **Reply #1936 on:** July 08, 2013, 03:35:50 pm »

this is a quick question from GMA

regard the ratio of vectors when i have something like $\frac{3}{5}\vec{AP}=\vec{AB}$ why does the ratio between two
of them equal $5:3$

btw i am expecting it to be $3:5$ as i moved the 5 making $3\vec{AP}=5\vec{AB}$

thanks

SocialRhubarb · « **Reply #1937 on:** July 08, 2013, 03:50:28 pm »

If you actually think about it, if $\frac{3}{5}\vec{AP}=\vec{AB}$ , then this implies that vector AP has a greater magnitude, since vector AB is three fifths of AP, so AP must be longer.

It makes sense then, that the ratio $\vec{AP}:\vec{AB}$ is $5:3$ , since the magnitude of vector AP must be greater than the magnitude of vector AB.

jono88 · « **Reply #1938 on:** July 08, 2013, 06:45:26 pm »

In question 4iii)

Is y=5.2 a mistake? My answer is y=2(-6.4)+16=3.2

zvezda · « **Reply #1939 on:** July 08, 2013, 06:51:20 pm »

A spherical ball with radius 24 mm is immersed in an acid bath. The volume of the ball decreases at a (time) rate proportional to the surface area at that instant. If the ball is completely dissolved in 8 minutes, find:
i) the formula connecting radius with time

The answer is apparently, r=24-3t. What gives this away? I can't just assume that this relationship is linear; it could be exponential?
Help appreciated

lzxnl · « **Reply #1940 on:** July 08, 2013, 06:59:13 pm »

So surface area = $4\pi r^2$
$\frac{dV}{dt} = -4k\pi r^2$
$\frac{dV}{dr} = 4\pi r^2$
using chain rule
$\frac{dr}{dt} = \frac{\frac{dV}{dt}}{\frac{dV}{dr}} = -k$
$r = -kt + c$

As t = 0, r = 24, c = 24
Also, t = 8, r = 0
So r = -3t + 24

if r is in mm and t is in min

b^3 · « **Reply #1941 on:** July 08, 2013, 07:02:37 pm »

Probably seems like a long way of doing it, but this shows why it will be linear.
$\begin{alignedat}{1}V & =\frac{4}{3}\pi r^{3} \\ \frac{dV}{dt} & \propto4\pi r^{2}\:\left(\text{surface area of a sphere}\right) \\ \frac{dV}{dt} & =4k\pi r^{2},\: k\in\mathbb{R} \\ \frac{dr}{dt} & =\frac{dr}{dV}\frac{dV}{dt} \\ \frac{dV}{dr} & =4\pi r^{2} \\ \frac{dr}{dt} & =\frac{1}{4\pi r^{2}}\times4k\pi r^{2} \\ \frac{dr}{dt} & =k \\ r\left(t\right) & =\int kdt \\ & =kt+c \\ r\left(0\right) & =24 \\ 24 & =0+c \\ \implies c & =24 \\ r\left(8\right) & =0 \\ 0 & =8k+24 \\ \implies k & =-3 \\ \therefore r\left(t\right) & =24-3t,\: t\geq0 \end{alignedat}$

EDIT: Beaten.

aestheticatar · « **Reply #1942 on:** July 09, 2013, 04:43:37 pm »

Hey guys,
Is there a way to do this by hand and hence find an exact answer?
(Despite the question asking for 3dp)

Thanks in advance.
Have a great day!

zvezda · « **Reply #1943 on:** July 09, 2013, 04:56:52 pm »

Ah yeah I see it now.
Cheers nlieu and b^3

Alwin · « **Reply #1944 on:** July 09, 2013, 05:32:54 pm »

Quote from: aestheticatar on July 09, 2013, 04:43:37 pm

Hey guys,
Is there a way to do this by hand and hence find an exact answer?

So we have:
$V = \pi \int_2^3{(log_e{x})^2} dx$
CAS could give you an exact answer, but you asked for 'by hand' method. I'm sorry, but this is outside of the course, pretty much anything involving integration by parts (eg antidiff of ln(x)) or other calculus methods to find volume (edit: eg shell method which brightsky suggested, but removed his post)

You could, however use your CAS or program such as Wolfram Alpha to find the integral of (ln(x))^2:

from Wolfram Alpha

From here, you can solve normally by substitution of terminals, etc etc.
But again: its outside of the course to find such an integral without aid of CAS

Lasercookie · « **Reply #1945 on:** July 09, 2013, 06:07:45 pm »

Quote from: Alwin on July 09, 2013, 05:32:54 pm

But again: its outside of the course to find such an integral without aid of CAS

Yeah, it's pretty much outside of the course, but I guess you could rephrase it as a multiparted integration by recognition question, which would make it possible to evaluate using Year 12 knowledge. You've probably seen a fair bit of integration by recognition in methods, it's comes up in spesh too, right? (I don't remember, I think it does).

You've probably figured out what the integral of $log_e(x)$ is using integration by recognition before. The idea is something along the lines of we want some function that we can differentiate, that will allow us to separate log_e(x) e.g. $xlog_e(x)$ being the usual example. That question could be worded something like find the derivative of $xlog_e(x)$ , and hence antidifferentiate $log_e(x)$ .

Spoiler

$\frac{d}{dx}(xlog_e(x)) = log_e(x) + 1 \implies \int log_e(x) \ dx = xlog_e(x) - \int 1 \ dx = xlog_e(x) - x + c$

So what about $\int log_e(x)^2 \ dx$ ? What's something that we can differentiate that will allow us to separate log_e(x)^2? $xlog_e(x)^2$ , right? Once we know what the antiderivative of that is, we can evaluate this integral: $V = \pi \int_2^3(log_e(x))^2 \ dx$

So figure out what the integral of log_e(x)^2 is, by diffing xlog_e(x)^2 first

Spoiler

$\frac{d}{dx}(xlog_e(x)^2) = log_e(x)^2 + 2log_e(x) \implies \int log_e(x)^2 \ dx = xlog_e(x)^2 - 2\int log_e(x) \ dx$

and simplify

Spoiler

We figured out what the integral of log_e(x) was before, so just substitute that in and simplify.
$\int log_e(x)^2 \ dx = xlog_e(x)^2 - 2(xlog_e(x) - x) + c = xlog_e(x)^2 - 2xlog_e(x) + 2x + c$

Then evaluate our definite integral

Spoiler

Hence $V = \pi \int_2^3(log_e(x))^2 \ dx = \pi \big [ xlog_e(x)^2 -2xlog_e(x) + 2x \big ]_2^3$

and simplify

Spoiler

$=\pi \big(3\log_e(3)^2 - 6\log_e(3) + 6 - 2\log_e(2)^2 + 4\log_e(2) - 4 \big )$ and simplify some more, and there you have your exact value expression

I wouldn't expect that to be on a Year 12 exam, but hopefully that's a way that you might find do-able by hand.

barydos · « **Reply #1946 on:** July 09, 2013, 08:48:13 pm »

Quote from: b^3 on July 02, 2013, 11:19:31 pm

10b. Find an equation for the circle, then try rotating around the $y$ axis, restricting the part you rotating by using terminals in terms of $r$
Spoiler
$\begin{alignedat}{1}x^{2}+y^{2} & =r^{2} \\ x^{2} & =r^{2}-y^{2} \\ V & =\pi\int_{\frac{3}{4}r}^{r}x^{2}dy \\ & =\pi\int_{\frac{3}{4}r}^{r}\left(r^{2}-y^{2}\right)dy \\ & =\pi\left[r^{2}y-\frac{1}{3}y^{3}\right]_{\frac{3}{4}r}^{r} \\ & =\pi\left(r^{3}-\frac{1}{3}r^{3}-\left(\frac{3}{4}r^{3}-\frac{1}{3}\frac{27}{64}r^{3}\right)\right) \\ & =\pi\left(\frac{2}{3}r^{3}-\frac{48-9}{64}r^{3}\right) \\ & =\pi\left(\frac{128-117}{192}r^{3}\right) \\ & =\frac{11\pi r^{3}}{192} \end{alignedat}$

Part a is asking for an expression for the volume if you filled any shape of height H up fully. While part b asks for the volume when you only half fill it up. They are different because in this case half filling a container of this shape that is 10 cm high is not the same as completely filling a container of this shape that is 5 cm high.

What you need to do for part b is similar to the previous problem, set up the volume of revolution integral, but with the lower terminal as zero, and the upper terminal as $\frac{H}{2}$ .

Sorry I'm a bit late in responding, but I was a bit busy. Thank you for your help!
Umm, in terms of question 15 b), while doing all the algebra and what not, I managed to simplify the thing down to
$\frac {\pi H}{24(b-a)} * (b^3 + 3b^2a + 3a^2b - 7a^3)$
Is there a trick somewhere in dividing the (b-a) ? The only thing I could do was long division, just wondering if there was a shortcut here somewhere. THANKS AGAIN!

b^3 · « **Reply #1947 on:** July 09, 2013, 09:21:37 pm »

Quote from: Anonymiza on July 09, 2013, 08:48:13 pm

Sorry I'm a bit late in responding, but I was a bit busy. Thank you for your help!
Umm, in terms of question 15 b), while doing all the algebra and what not, I managed to simplify the thing down to
$\frac {\pi H}{24(b-a)} * (b^3 + 3b^2a + 3a^2b - 7a^3)$
Is there a trick somewhere in dividing the (b-a) ? The only thing I could do was long division, just wondering if there was a shortcut here somewhere. THANKS AGAIN!

I'm a little bit confused as to how you got $b-a$ in the denominator? This is how I would have simplified it.

$\begin{alignedat}{1}m & =\frac{H-0}{b-a} \\ m & =\frac{H}{b-a} \\ y & =\left(\frac{H}{b-a}\right)\left(x-a\right) \\ x-a & =\left(\frac{b-a}{H}\right)y \\ x & =\left(\frac{b-a}{H}\right)y+a \\ V & =\pi\int_{0}^{\frac{H}{2}}\left(\left(\frac{b-a}{H}\right)y+a\right)^{2}dy \\ & =\pi\int_{0}^{\frac{H}{2}}\left(\left(\frac{b-a}{H}\right)^{2}y^{2}+2a\left(\frac{b-a}{H}\right)y+a^{2}\right)dy \\ & =\pi\left[\frac{1}{3}\left(\frac{b-a}{H}\right)^{2}y^{3}+a\left(\frac{b-a}{H}\right)y^{2}+a^{2}y\right]_{0}^{\frac{H}{2}} \\ & =\pi\left(\frac{1}{3}\frac{\left(b-a\right)^{2}}{H^{2}}\left(\frac{H^{3}}{8}\right)+a\left(\frac{b-a}{H}\right)\left(\frac{H^{2}}{4}\right)+a^{2}\left(\frac{H}{2}\right)\right) \\ & =\pi\left(\frac{\left(b-a\right)^{2}H}{24}+\frac{Ha\left(b-a\right)}{4}+\frac{Ha^{2}}{2}\right) \\ & =\frac{\pi H}{24}\left(b^{2}-2ab+a^{2}+6ab-6a^{2}+12a^{2}\right) \\ & =\frac{\pi H}{24}\left(7a^{2}+4ab+b^{2}\right)\text{ cm}^{3} \end{alignedat}$

barydos · « **Reply #1948 on:** July 10, 2013, 12:08:19 pm »

Quote from: b^3 on July 09, 2013, 09:21:37 pm

I'm a little bit confused as to how you got $b-a$ in the denominator? This is how I would have simplified it.
$\begin{alignedat}{1}m & =\frac{H-0}{b-a} \\ m & =\frac{H}{b-a} \\ y & =\left(\frac{H}{b-a}\right)\left(x-a\right) \\ x-a & =\left(\frac{b-a}{H}\right)y \\ x & =\left(\frac{b-a}{H}\right)y+a \\ V & =\pi\int_{0}^{\frac{H}{2}}\left(\left(\frac{b-a}{H}\right)y+a\right)^{2}dy \\ & =\pi\int_{0}^{\frac{H}{2}}\left(\left(\frac{b-a}{H}\right)^{2}y^{2}+2a\left(\frac{b-a}{H}\right)y+a^{2}\right)dy \\ & =\pi\left[\frac{1}{3}\left(\frac{b-a}{H}\right)^{2}y^{3}+a\left(\frac{b-a}{H}\right)y^{2}+a^{2}y\right]_{0}^{\frac{H}{2}} \\ & =\pi\left(\frac{1}{3}\frac{\left(b-a\right)^{2}}{H^{2}}\left(\frac{H^{3}}{8}\right)+a\left(\frac{b-a}{H}\right)\left(\frac{H^{2}}{4}\right)+a^{2}\left(\frac{H}{2}\right)\right) \\ & =\pi\left(\frac{\left(b-a\right)^{2}H}{24}+\frac{Ha\left(b-a\right)}{4}+\frac{Ha^{2}}{2}\right) \\ & =\frac{\pi H}{24}\left(b^{2}-2ab+a^{2}+6ab-6a^{2}+12a^{2}\right) \\ & =\frac{\pi H}{24}\left(7a^{2}+4ab+b^{2}\right)\text{ cm}^{3} \end{alignedat}$

Ah, on the line of integration, I didn't expand the $((\frac{b-a}{H})y +a )^2$ when I integrated so I got the $\frac{H}{3(b-a)}$ as a coefficient of a cubic, so then I had to perform long division in the very end to get rid of $(b-a)$
So yeah, I think I like your method better, I should've expanded haha, thanks again!

Alwin · « **Reply #1949 on:** July 10, 2013, 08:28:08 pm »

Hey guys

I had a q about assumed knowledge. This is from the Heinemann textbook, chapter review (and yes, before you ask, I do do questions from other textbooks.. just for the sake of it)

My (geometric) approach was:
$\begin{alignedat}{1}\overrightarrow{OM}& = \mathbf{a}+\frac{1}{2}\mathbf{c} \\ \text{Since P is the median of the triangle OAB, } |\overrightarrow{OP}|& = \frac{2}{3} |\overrightarrow{OM}| \\ \Rightarrow\overrightarrow{OP} & = \frac{2}{3} \left( \mathbf{a}+\frac{1}{2}\mathbf{c} \right) \\ \therefore \overrightarrow{OP} & = \frac{2}{3} \mathbf{a}+\frac{1}{3}\mathbf{c} \end{alignedat} $

I'm just not sure if this is in the spesh course, my teacher never mentioned it either... I suppose a spesh method would be to find the intersection using vector methods like scale factors or something.

Does someone know if that is an acceptable method?? thanks!

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Author Topic: VCE Specialist 3/4 Question Thread! (Read 2580584 times) Tweet Share

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