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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: cltf on May 27, 2010, 08:54:33 pm

Title: Units 1 Questions thread
Post by: cltf on May 27, 2010, 08:54:33 pm: I am in dire need of help, simply the matter is I seem to have lost my ability to answer simple questions:

1. For $-4\left ( x+2 \right )^{2}+5$ find the largest domain so that the INVERSE function does exist, then state the inverse function.
My answers: $\left ( - \infty,-2 \right ]$ and $y= \sqrt{\frac{x-5}{-4}}-2$ ???

2. a can has a volume of $500 cm^{3}$ the height is "h" the radius is "r"
express "h" in terms of "r"
Then, Show that the total surface area, $A cm^{2}$ of the can can be given by $A=\frac{1000}{r}-2\pi r^{2}$

3. An open tank is made from a sheet of metal 84cm by 40cm, by cutting squares of X length of the corners, express the volume of the tank in terms of X, hence state the maximal do mail of your function for V, the volume of the tank.

4. Make "a" the subject in $\frac{1}{b}-\frac{1}{a}=\frac{2}{c}$
Title: Re: Units 1 Questions thread
Post by: the.watchman on May 27, 2010, 09:06:25 pm: Here you go! (if you need more explanation, I'll tell you at school tomorrow)

(1) For this the 'largest' domain is actually $[-2,\infty)$ , because it is slightly 'bigger' than what you had (sounds weird I know but the extra four units makes a big difference)

So for $f^{-1}$ : Let $f(y)=x$

$-4(y+2)^2 + 5 = x$

$(y+2)^2 = \frac{x-5}{-4}$

$y= \pm \sqrt{\frac{x-5}{-4}} - 2$

BUT if the domain of f is [-2,infinity), then the range of f^-1 is also [-2,infinity)

So $y= - \sqrt{\frac{x-5}{-4}} - 2$ is rejected

$f^{-1}(x)= \sqrt{\frac{x-5}{-4}} - 2$

(2) $V=\pi r^2 \cdot h = 500$

$h = \frac{500}{\pi r^2}$

So $A = 2\pi rh + 2\pi r^2$

$A = 2\pi r(\frac{500}{\pi r^2}) + 2\pi r^2$

$A = \frac{1000}{r} + 2\pi r^2$ (i think you put a minus in accidentally?)

(3) Visualising the sheet (as a net for the tank),

Length is (84-2X)cm and width is (40-2X)cm and height is Xcm

so $V=X(84-2X)(40-2X)$ , 0<X<20 (because if $X \geq 20$ , then the width is zero, or negative)

(4) x abc: $ac-bc=2ab$

$ac-2ab=bc$

$a(c-2b)=bc$

$a=\frac{bc}{c-2b}$
Title: Re: Units 1 Questions thread
Post by: cltf on May 27, 2010, 09:10:47 pm: Ahhh that new domain would explain so much! It explains my Y intercepts when I tried to sketch it, I should have known.

ah cool, i was pretty close to most of them, that's good to know. thanks once again!

one more think, just in reference to question $x \left (84-2x \right) \left (40-2x \right) =v$ what is the maximal domain? How do I find it?
Title: Re: Units 1 Questions thread
Post by: TrueTears on May 27, 2010, 10:40:48 pm: x(84-2x)(40-2x)=v

the domain is R. I am assuming you are not relating this question to any specifics. If so then you must apply restrictions, i haven't read any of the above posts.

ALL polynomial functions have domain of R (try prove this :D very fun proof! think about the converse, what if they didn't have a domain of R?)
Title: Re: Units 1 Questions thread
Post by: the.watchman on May 28, 2010, 06:16:57 am: Sorry, but for the question scenario, it is 0<X<20

For the tank to exist, the height, length and width all must be greater than zero

So, for height: X>0
For length: 84-2X>0, X<42
For width: 40-2X>0, X<20

For all these to be true, the maximal domain must be 0<X<20
Title: Re: Units 1 Questions thread
Post by: cltf on May 28, 2010, 09:38:55 pm: if $f\left ( x \right )=-4\left ( x+2 \right )^{2}+5$ and $f^{-1}\left ( x \right )=\sqrt{\frac{x-5}{-4}}-2$ how do i find the intersection point?

also what is I meant to do here?
The value(s) of "m" that will give the equation $mx^{2}-6mx-3=0$ two real roots lie in the following
a) $\left ( -\infty ,\frac{-1}{3} \right )$
b) $\left ( 0,\infty \right )$
c) $\left ( \frac{1}{3},\infty \right )$
d) $\left ( -\infty ,0 \right )$
e) $\left ( -\infty ,\frac{-1}{3} \right )\cup \left ( 0,\infty \right )$
Title: Re: Units 1 Questions thread
Post by: moekamo on May 28, 2010, 10:37:19 pm: 1 let f(x) = x and solve for x, this is the x and y value of the intersection point/s

2.

for 2 real roots the discriminant must be greater than 0:
$\Delta = b^2-4ac = 36m^2-4(m)(-3)$

so $36m^2 + 12m > 0$

$\implies 3m^2 + m >0 \implies m(3m+1)>0$

by doing it graphically we can see that $m \in \left (-\infty, -\frac{1}{3} \right ) \cup (0,\infty)$
Title: Re: Units 1 Questions thread
Post by: cltf on May 30, 2010, 07:20:32 am: Quote from: moekamo on May 28, 2010, 10:37:19 pm
1 let f(x) = x and solve for x, this is the x and y value of the intersection point/s

2.
so $36m^2 + 12m > 0$

$\implies 3m^2 + m >0 \implies m(3m+1)>0$

so when
$36m^2 + 12m > 0$

$\implies 3m^2 + m >0 \implies m(3m+1)>0$

we can just factorize and ignore the 12?

and i dont get the intersection point part.
Title: Re: Units 1 Questions thread
Post by: the.watchman on May 30, 2010, 10:00:44 am: (1)
As you should know, the graph of the inverse is a reflection in the line y=x
So the graphs of the inverse and the original function should intersect on the line y=x

Therefore, instead of solving $f(x)=f^{-1}(x)$ , it will suffice to just solve $f(x)=x$
Also, the y-values are the same as the x-values (because y=x), which is what moekamo said

(2)
It's not 'ignoring' the 12, it's dividing both sides by 12 :)
Title: Re: Units 1 Questions thread
Post by: m@tty on May 30, 2010, 09:20:45 pm: Quote from: the.watchman on May 27, 2010, 09:06:25 pm
The 'largest' domain is actually $[-2,\infty)$ , because it is slightly 'bigger' than what you had (sounds weird I know but the extra four units makes a big difference)

If asked for the maximal domain would the other domain(which is equally infinite) be marked wrong?
Title: Re: Units 1 Questions thread
Post by: the.watchman on May 30, 2010, 10:10:31 pm: Quote from: m@tty on May 30, 2010, 09:20:45 pm
Quote from: the.watchman on May 27, 2010, 09:06:25 pm
The 'largest' domain is actually $[-2,\infty)$ , because it is slightly 'bigger' than what you had (sounds weird I know but the extra four units makes a big difference)

If asked for the maximal domain would the other domain(which is equally infinite) be marked wrong?

I'm not sure, but $[-2,\infty)$ seems 'more right', eh? :P
I have no idea though...
Title: Re: Units 1 Questions thread
Post by: cltf on May 31, 2010, 04:45:40 pm: Quote from: the.watchman on May 30, 2010, 10:10:31 pm
Quote from: m@tty on May 30, 2010, 09:20:45 pm
Quote from: the.watchman on May 27, 2010, 09:06:25 pm
The 'largest' domain is actually $[-2,\infty)$ , because it is slightly 'bigger' than what you had (sounds weird I know but the extra four units makes a big difference)

If asked for the maximal domain would the other domain(which is equally infinite) be marked wrong?

I'm not sure, but $[-2,\infty)$ seems 'more right', eh? :P
I have no idea though...

If you sketch it $[-2,\infty)$ is more correct.
Title: Re: Units 1 Questions thread
Post by: the.watchman on May 31, 2010, 05:37:56 pm: Quote from: cltf on May 31, 2010, 04:45:40 pm
Quote from: the.watchman on May 30, 2010, 10:10:31 pm
Quote from: m@tty on May 30, 2010, 09:20:45 pm
Quote from: the.watchman on May 27, 2010, 09:06:25 pm
The 'largest' domain is actually $[-2,\infty)$ , because it is slightly 'bigger' than what you had (sounds weird I know but the extra four units makes a big difference)

If asked for the maximal domain would the other domain(which is equally infinite) be marked wrong?

I'm not sure, but $[-2,\infty)$ seems 'more right', eh? :P
I have no idea though...

If you sketch it $[-2,\infty)$ is more correct.

That's what I thought, but I'm not sure if the other one would be marked wrong.
Clarification anyone?
Title: Re: Units 1 Questions thread
Post by: m@tty on May 31, 2010, 08:22:25 pm: Quote from: cltf on May 31, 2010, 04:45:40 pm
Quote from: the.watchman on May 30, 2010, 10:10:31 pm
Quote from: m@tty on May 30, 2010, 09:20:45 pm
Quote from: the.watchman on May 27, 2010, 09:06:25 pm
The 'largest' domain is actually $[-2,\infty)$ , because it is slightly 'bigger' than what you had (sounds weird I know but the extra four units makes a big difference)

If asked for the maximal domain would the other domain(which is equally infinite) be marked wrong?

I'm not sure, but $[-2,\infty)$ seems 'more right', eh? :P
I have no idea though...

If you sketch it $[-2,\infty)$ is more correct.

It *appears* more correct.

But against the enormity of infinity 4 units is infinitesimal, even to the point of becoming irrelevant.

Technically they're equal, but logically they are different. I suppose they should be considered of different sizes, though I have never seen this addressed...
Title: Re: Units 1 Questions thread
Post by: cltf on June 01, 2010, 08:44:17 pm: got a dilemma.
In the methods unit 1 semester exam, no doubt there will be a cubic graph, how do i find the maximum and minimums?
Because my cas either keeps giving me a not accurate enough pinpoint, or not enough decimal places. Is there any other method CAS aside i can use to find the max and min? cause my friend mentioned something about differentiation...
Title: Re: Units 1 Questions thread
Post by: superflya on June 01, 2010, 08:46:38 pm: yes u differentiate and let it equal zero. to find ur y coordinate sub the x value/s back into ur original equation. to test for the nature of the point use the box thingy, dont think u wouldve learnt second derivatives.
Title: Re: Units 1 Questions thread
Post by: cltf on June 01, 2010, 08:51:56 pm: Quote from: superflya on June 01, 2010, 08:46:38 pm
yes u differentiate and let it equal zero. to find ur y coordinate sub the x value/s back into ur original equation. to test for the nature of the point use the box thingy, dont think u wouldve learnt second derivatives.

haven't learnt it yet.

so let $ax^{3}+bx^{2}+cx+d=0$ ? then.. but how would you find the y coordinate?
Title: Re: Units 1 Questions thread
Post by: davidle_10 on June 01, 2010, 09:26:42 pm: So say you have the equation y=ax^3+bx^2+cx+d (cubic)
Firstly you have to differentiate so: dy/dx =3ax^2 + 2bx + c (differentiation of polynomials)
Then when the derivative=0 (stationary point), solve for x
After you have found the x value(s) you can sub them back into the original equation,
y=ax^3+bx^2+cx+d, to find the corresponding y coordinate.
Title: Re: Units 1 Questions thread
Post by: cltf on June 01, 2010, 09:32:38 pm: so $y=ax^{3}+bx^{2}+cx+d => dy/dx=3ax^{2} + {2}bx + c$
or does dy/dx have an actual rule?
Title: Re: Units 1 Questions thread
Post by: davidle_10 on June 01, 2010, 09:55:54 pm: dy/dx (called the derivative or gradient function) is essentially the gradient of an equation at any given point. There are rules in determining the derivative of a function. For a polynomial, x^p, the derivative is px^p-1. In the case of y=ax^3+bx^2+cx+d, dy/dx= 3ax^(3-2)+2bx^(2-1)+cx^(1-1)
= 3ax^2+2bx+c
But unless your class hasn't started doing differentiation yet, I would just use the calculator to find the point. If you are using the TI-nspire then you will need to press menu,trace,graph trace. Then just move the cursor to the location of the maximum and minimum turning points and the coordinates will show up.
Title: Re: Units 1 Questions thread
Post by: cltf on June 01, 2010, 10:03:15 pm: Quote from: davidle_10 on June 01, 2010, 09:55:54 pm
dy/dx (called the derivative or gradient function) is essentially the gradient of an equation at any given point. There are rules in determining the derivative of a function. For a polynomial, x^p, the derivative is px^p-1. In the case of y=ax^3+bx^2+cx+d, dy/dx= 3ax^(3-2)+2bx^(2-1)+cx^(1-1)
= 3ax^2+2bx+c
But unless your class hasn't started doing differentiation yet, I would just use the calculator to find the point. If you are using the TI-nspire then you will need to press menu,trace,graph trace. Then just move the cursor to the location of the maximum and minimum turning points and the coordinates will show up.

if reference to the bottom part, the problem im facing is that the points don't go to 2dp. or at not actual enough. so what do i do?
Title: Re: Units 1 Questions thread
Post by: the.watchman on June 01, 2010, 10:05:42 pm: Erm ... guys, this is MM unit 1
And anyway cltf, none of the (CGS) 1/2 classes have done differentiation yet, dont worry :P

For this, you would need to use a calculator for finding local max and min :)
Title: Re: Units 1 Questions thread
Post by: the.watchman on June 01, 2010, 10:07:08 pm: Sorry for the double post, but you could use the fmin() and fmax() functions on your Ti-Nspire, if you restrict the domain
Title: Re: Units 1 Questions thread
Post by: the.watchman on June 01, 2010, 10:21:22 pm: Check it in the catalog, but it's fmin(FUNC., VARIABLE)
So for example:

$fmin((x-2)^2 (x+2)^2,x)|-3 \leq x \leq -1$ gives $x=-2$ :)
Title: Re: Units 1 Questions thread
Post by: the.watchman on June 02, 2010, 02:54:50 pm: That's because the min of $(x-2)^2 (x+2)^2$ is 0, when x = -2 or 2, check the graph :)

Oh, and yes that's the symbol, it means "given that"
Title: Re: Units 1 Questions thread
Post by: cltf on June 03, 2010, 04:12:49 pm: does any one have MME u1&2 semester one - calculator trial exams?
Title: Re: Units 1 Questions thread
Post by: cltf on June 04, 2010, 05:33:52 pm: if a turning point touches the x axis is that considered an intercept? if so why?
Title: Re: Units 1 Questions thread
Post by: luken93 on June 04, 2010, 06:51:18 pm: Quote from: cherylim23 on June 04, 2010, 06:04:14 pm
yes, discriminant = 0

edit: anything that touches/cut the x-axis is a root.
In other words,
$b^2 - 4ac > 0 =$ Two solutions/intercepts (Cutting the line)
$b^2 - 4ac = 0 =$ One Solution/Intercept (Touching the line, aka tangent to the line)
$b^2 - 4ac < 0 =$ No Solutions/Intercepts (Does not touch the line)
Title: Re: Units 1 Questions thread
Post by: TrueTears on June 07, 2010, 07:27:24 pm: Quote from: cltf on June 04, 2010, 05:33:52 pm
if a turning point touches the x axis is that considered an intercept? if so why?
it's a matter of definition which is not too important in mathematics, just define what u need but anyways for the sake for completeness, a quote from wikipedia:

"a root (or a zero) of a real-, complex- or generally vector-valued function ƒ is a member x of the domain of ƒ such that ƒ(x) vanishes at x, an alternative name for the root in this context is the x-intercept."

thus if a tp touches the x axis it is an intercept.
Title: Re: Units 1 Questions thread
Post by: luken93 on June 08, 2010, 06:56:57 pm: okay, heres one, it was on our exam today, and i'm not sure if i got it, so if anyone wants to have a go...

A circle has its centre at (5 , -1) and a radius of 5
a) Show that ( 8 , 3 ) is a point on the circle

y= mx is tangent to the circle
b) Find the value(s) of m which makes the line y = mx tangent to the circle

i) Find the value(s) of m which crosses the circle twice

ii) Find the value(s) of m which does not touch the circle at all
Title: Re: Units 1 Questions thread
Post by: the.watchman on June 08, 2010, 09:26:24 pm: a) Eqn: $(x-5)^2 + (y+1)^2 = 25$

sub x=8 into above: $(8-5)^2 + (y+1)^2 = 25$

$(y+1)^2=16$

$y+1=\pm 4$

$y=-5,3$

So (8,3) is on the circle

b) $(x-5)^2 + (y+1)^2 = 25$

Let $y=mx$ :

$(x-5)^2 + (mx+1)^2 = 25$

$(m^2+1)x^2 + (2m-10)x +26=25$

$(m^2+1)x^2 + (2m-10)x +1=0$

Because $\triangle =0$ for one solution (thus tangent):

$\triangle=(2m-10)^2 - 4(m^2+1)\cdot 1=0$

$96-40m=0$

$m=\frac{12}{5}$

Follow the same process as above to find the other two answers (discriminant greater than zero, less than zero)
Title: Re: Units 1 Questions thread
Post by: luken93 on June 08, 2010, 09:38:07 pm: When its a show/prove that queestion, should you just sub in one value in hope that if finds the other one, or can you sub in x and y and find the answer to be 25?

Yay i got $\frac{12}{5}$ , I was worried though because it said values but I was pretty sure it was only 1, because the circle would touch the y-int?

For the other two, it would be $0 < m < \frac{12}{5}$ and $\frac{12}{5} < m < 0$ ???

Thanks
Title: Re: Units 1 Questions thread
Post by: the.watchman on June 08, 2010, 10:02:15 pm: You can sub in one value, or show that LHS=RHS, whatever you prefer

Well your answer for the other two is not quite right:

For $\triangle=96-40m>0$ , $m<\frac{12}{5}$ (m can be less than zero :P)

And your second answer is wild, how can m be greater than 12/5 and less than zero?? :D

For $\triangle=96-40m<0$ , $m>\frac{12}{5}$
Title: Re: Units 1 Questions thread
Post by: luken93 on June 08, 2010, 10:21:42 pm: Quote from: the.watchman on June 08, 2010, 10:02:15 pm
You can sub in one value, or show that LHS=RHS, whatever you prefer

Well your answer for the other two is not quite right:

For $\triangle=96-40m>0$ , $m<\frac{12}{5}$ (m can be less than zero :P)

And your second answer is wild, how can m be greater than 12/5 and less than zero?? :D

For $\triangle=96-40m<0$ , $m>\frac{12}{5}$
oh woops, what was i saying
dw, i was thinking on another tangent hahah
now i cant remember what i wrote though.....