Author Topic: VCE Methods Question Thread! (Read 5743956 times) Tweet Share

Stick · « **Reply #2040 on:** June 04, 2013, 08:47:32 pm »

THANK YOU!

emchun · « **Reply #2041 on:** June 05, 2013, 02:52:01 pm »

Hello, I did a practise SAC today and I have my real one tomorrow. I had quite a few questions I didn't understand from the practise one. It'll be great if someone can help! Thanks.

1. Find the expression of y in terms of x

$9+\frac{1}{3}$ $\log_{e} x$ = $\log_{e} y^3$

2. Find the exact solution

$e^{6+x}$ = $\frac{7e^\frac{1x}{2}}{4}$

3. Find the equation and asymptopes of a logarithm with (base e), it is being reflected on the x-axis, and has a horizontal translation of 4 units to the right.

4. $y=Ae^{bx}$
y(0)=60 and y(1)=10
Find A and b.

5a. $v(t)=3sin(\frac{\pi t}{12}) +\frac{1}{2}$
What is the volume when t=44?
What is the rate of change, and what is the volume when t=44.

I had another that had to do with matrices, I think it asked me to find $T\begin{matrix} x \\ y \end{matrix}$
They gave me two equations:
$p(x)=x^\frac{7}{3}$ and $q(x)=6+\frac{1}{8} (7-x)^\frac{7}{3}$
I don't really remember this question so maybe the information is not enough to work it out.

Phy124 · « **Reply #2042 on:** June 05, 2013, 03:30:27 pm »

Use laws

$\log_e(b^a) =a \log_e(b) ,b>0$

$\log_e(e^a)=a$

$e^{\log_e(a)}=a, a>0$

$\log_e(a)+\log_e(b)=\log_e(ab), a,b>0$

$\frac{x^a}{x^b}=x^{a-b}$

Quote from: emchun on June 05, 2013, 02:52:01 pm

Hello, I did a practise SAC today and I have my real one tomorrow. I had quite a few questions I didn't understand from the practise one. It'll be great if someone can help! Thanks.

1. Find the expression of y in terms of x

$9+\frac{1}{3}$ $\log_{e} x$ = $\log_{e} y^3$

$\Leftrightarrow \log_e(e^9)+\log_e(x^{\frac{1}{3}})=\log_e(y^3)$

$\Leftrightarrow \log_e(e^9x^{\frac{1}{3}})=\log_e(y^3)$

$\Leftrightarrow e^9x^{\frac{1}{3}}=y^3$

$\Leftrightarrow y = \sqrt[3]{e^9x^{\frac{1}{3}}}$

$\Rightarrow y = e^3\sqrt[3]{x^{\frac{1}{3}}}$

Quote from: emchun on June 05, 2013, 02:52:01 pm

2. Find the exact solution

$e^{6+x}$ = $\frac{7e^\frac{1x}{2}}{4}$

$\Leftrightarrow 1 = \frac{7e^\frac{x}{2}}{4e^{6+x}}$

$\Leftrightarrow 1 = \frac{7e^{\frac{x}{2}-(6+x)}}{4}$

$\Leftrightarrow \frac{4}{7}=e^{-\frac{x}{2}-6}$

$\Leftrightarrow -\frac{x}{2}-6 = \log_e\left(\frac{4}{7}\right)$

$\Leftrightarrow -\frac{x}{2}=\log_e\left(\frac{4}{7}\right)+6$

$\Rightarrow x = -2(\log_e\left(\frac{4}{7}\right)+6)=2(\log_e\left(\frac{7}{4}\right)-6)$

I'm going back to my own studies for a little while and I'll do the rest later (if someone else doesn't chose to do so in the mean time)

shadows · « **Reply #2043 on:** June 05, 2013, 09:10:59 pm »

Quote from: Alwin on June 04, 2013, 08:33:38 pm

MERRY CHIRSTMAS STICK:

Re: Methods [3/4] Question Thread!

I uploaded solutions there ^^

Considering just uploading the whole booklet now lol.

haha needed this doc as i was stuck on this question as well..

Part A: Why is it in Radians in the first place. isnt initial Angle is given in degrees??

Part B: Isn't it suppose to be square root (3)/15 km/ s rather than km/hr?

Alwin · « **Reply #2044 on:** June 05, 2013, 09:49:58 pm »

Quote from: shadows on June 05, 2013, 09:10:59 pm

haha needed this doc as i was stuck on this question as well..

Part A: Why is it in Radians in the first place. isnt initial Angle is given in degrees??

Part B: Isn't it suppose to be square root (3)/15 km/ s rather than km/hr?

Part A: When you diff or antidiff, it always has to be in radians for trig functions. eg try diff on CAS in degrees mode and it'll start giving you pi and 180s. to be consistent I probably should have used theta equals pi / 6 radians.

Part B: You're completely right!! haha thanks man, I'll go back and change it. You're the first one to notice, feel special

shadows · « **Reply #2045 on:** June 05, 2013, 09:53:13 pm »

Haha np!

I think that it's great what you are doing!

Your notes are fantastic as well

emchun · « **Reply #2046 on:** June 05, 2013, 10:45:08 pm »

Thank you for the help 2/cos(c)!

shadows · « **Reply #2047 on:** June 07, 2013, 04:39:23 pm »

If f(x)= (x-1)(x-a)^2 a is bigger than 1

X val of s.ps occur at a and (2+a)/3

Can you find nature by table method?

Lasercookie · « **Reply #2048 on:** June 07, 2013, 05:11:34 pm »

Quote from: shadows on June 07, 2013, 04:39:23 pm

If f(x)= (x-1)(x-a)^2 a is bigger than 1

X val of s.ps occur at a and (2+a)/3

Can you find nature by table method?

$f(x)=(x-1)(x-a)^2$ , which means $f'(x) = 2(x-a)(x-1) + (x-a)^2$ by the product rule.

We have a stationary point at $x = a$ , so $f'(a) = 0$ and at $x = \frac{2+a}{3}$ , so $f'(\frac{2+a}{3})=0$

You can use the table method. Let's check the points $\pm 1$ of each of those x-values
$f'(a+1) = 2(1)(a) + 1 = 2a + 1$
$f'(a-1) = 2(-1)(a-2) + 1 = -2a+5$

So we can then see that if $2a + 1 > 0 \implies a > \frac{-1}{2}$ and $-2a+5>0 \implies a < \frac{5}{2}$ .

So we have f'(a+1) and f'(a-1) positive if $\frac{-1}{2} < a < \frac{5}{2}$ . Hence x = a is a stationary point of inflection if this condition holds.

Then we supposedly have f'(a+1) and f'(a-1) negative if $a < \frac{-1}{2}$ and $a > \frac{5}{2}$ . This is clearly a contradiction, so we can't have f'(a+1) and f'(a-1) both negative for the same value of a.

What this does mean is if $a < \frac{-1}{2}$ then we have f'(a+1) negative and f'(a-1) positive. That is a local maximum at x =a

It also means that if $a > \frac{5}{2}$ then we have f'(a+1) positive and f'(a-1) negative. That is a local minimum at x = a

You can do the same thing for the other stationary point. We're basically just looking at what different kind of stationary points we get depending on what values of a we have.

edit: fixed my mistake

b^3 · « **Reply #2049 on:** June 07, 2013, 05:25:56 pm »

~~EDIT: I overcomplicated it a little, stick with Lazyred's method above.~~

Little bit tricky, first you need to check which turning point is bigger, this is where the $a>1$ comes in.
$\begin{alignedat}{1}a>1 \\ \implies2<2a \\ \implies a+2<2a \\ \implies\frac{a+2}{3}<a \end{alignedat}$
That is the left most turning point is at $x=\frac{a+2}{3}$ .
$\begin{alignedat}{1}x<\frac{a+2}{3} \\ \implies x<a \\ \implies\left(x-a\right)<0 \\ \left(3x-2-a\right)<0 \\ \implies f'\left(x\right)>0 & \:\:\text{-ve\ensuremath{\times}-ve=+ve} \\ \\ \frac{a+2}{3}<x<a \\ \implies\left(x-a\right)<0 \\ \left(3x-2-a\right)>0 \\ \implies f'\left(x\right)<0 & \:\:\text{-ve\ensuremath{\times}+ve=-ve} \\ \\ x>a \\ \implies x>\frac{a+2}{3} \\ \left(x-a\right)>0 \\ \left(3x-2-a\right)>0 \\ \implies f'\left(x\right)>0 & \:\:\text{+ve\ensuremath{\times}+ve=+ve} \end{alignedat} $
Now you can draw up the gradient table.

Which means that the turning point at $x=\frac{a+2}{3}$ is a local maximum and the turning point at $x=a$ is a local minimum.

Alwin · « **Reply #2050 on:** June 07, 2013, 05:32:41 pm »

Quote from: shadows on June 07, 2013, 04:39:23 pm

If f(x)= (x-1)(x-a)^2 a is bigger than 1

X val of s.ps occur at a and (2+a)/3

Can you find nature by table method?

Yes, you can (and probably should since this is a method's question). You need 5 values, one $x<\frac{2+a}{3}$ , $x=\frac{2+a}{3}$ , one $\frac{2+a}{3}<x<a$ , $x=a$ , and one $x>a$ . Note, $\frac{2+a}{3}<a$ , since $a>1$ . You can prove this for yourself.

We need to test a value between $\frac{2+a}{3}$ and $a$ , but we shouldn't just use $a-1$ imo. eg if $a=2$ , $\frac{2+a}{3}=\frac{4}{3}$ . Clearly $a-1=1<\frac{4}{3}$ . So, we would be testing the gradient on the other side of the other turning point

Without loss of generality, I let the 'middle' x value be the 'middle' of $\frac{2+a}{3}$ and $a$ , guaranteeing that it's between $\frac{2+a}{3}$ and a no matter what.
Hence the middle value is $\frac{a+\frac{2+a}{3}}{2}=\frac{2 a+1}{3}$
(see image)

EDIT beaten, probably spent a bit too much time on my pretty pic

But, the only bit I don't like about Lazyred's method is that he does a+1 and a-1. I prefer using the mid point

Lasercookie · « **Reply #2051 on:** June 07, 2013, 05:43:20 pm »

Quote from: Alwin on June 07, 2013, 05:32:41 pm

Note, (2+a)/3<a, since a>1. You can prove this for yourself.

Quote from: b^3 on June 07, 2013, 05:25:56 pm

Little bit tricky, first you need to check which turning point is bigger, this is where the $a>1$ comes in.

Oops I read the question wrong, I missed the $a > 1$ and ended up treating $a$ as a real number. I read the question as $f(x) = (x-1)(x-a)^2 > 1$ (which I kind of ignored in my working anway, that inequality is clearly not consistent with the conclusions I made anyway). -_-

Anyway in case it comes up, that's why my answers differ from b^3's and Alwin's.

shadows · « **Reply #2052 on:** June 07, 2013, 06:16:54 pm »

Wow thanks.

Ugh it would be soo much easier just to find the double derivative...

shadows · « **Reply #2053 on:** June 07, 2013, 06:25:46 pm »

Quote from: b^3 on June 07, 2013, 05:25:56 pm

~~EDIT: I overcomplicated it a little, stick with Lazyred's method above.~~

Little bit tricky, first you need to check which turning point is bigger, this is where the $a>1$ comes in.
$\begin{alignedat}{1}a>1 \\ \implies2<2a \\ \implies a+2<2a \\ \implies\frac{a+2}{3}<a \end{alignedat}$
That is the left most turning point is at $x=\frac{a+2}{3}$ .
$\begin{alignedat}{1}x<\frac{a+2}{3} \\ \implies x<a \\ \implies\left(x-a\right)<0 \\ \left(3x-2-a\right)<0 \\ \implies f'\left(x\right)>0 & \:\:\text{-ve\ensuremath{\times}-ve=+ve} \\ \\ \frac{a+2}{3}<x<a \\ \implies\left(x-a\right)<0 \\ \left(3x-2-a\right)>0 \\ \implies f'\left(x\right)<0 & \:\:\text{-ve\ensuremath{\times}+ve=-ve} \\ \\ x>a \\ \implies x>\frac{a+2}{3} \\ \left(x-a\right)>0 \\ \left(3x-2-a\right)>0 \\ \implies f'\left(x\right)>0 & \:\:\text{+ve\ensuremath{\times}+ve=+ve} \end{alignedat} $
Now you can draw up the gradient table.
Which means that the turning point at $x=\frac{a+2}{3}$ is a local maximum and the turning point at $x=a$ is a local minimum.

Arent we always required to state the vales if the derivative when using table method?

b^3 · « **Reply #2054 on:** June 07, 2013, 06:44:30 pm »

I don't think you have to have to exactly, as long as you've shown that in that interval the gradient is negative or positive (the essentials textbook does it this way). If you do have the values then it's best to put them in. (I've messaged Timmeh to double check on this though).

Search the forums now!

We have moved!

Author Topic: VCE Methods Question Thread! (Read 5743956 times) Tweet Share

Stick

Re: VCE Methods Question Thread!

emchun

Re: VCE Methods Question Thread!

Phy124

Re: VCE Methods Question Thread!

shadows

Re: VCE Methods Question Thread!

Alwin

Re: VCE Methods Question Thread!

shadows

Re: VCE Methods Question Thread!

emchun

Re: VCE Methods Question Thread!

shadows

Re: VCE Methods Question Thread!

Lasercookie

Re: VCE Methods Question Thread!

b^3

Re: VCE Methods Question Thread!

Alwin

Re: VCE Methods Question Thread!

Lasercookie

Re: VCE Methods Question Thread!

shadows

Re: VCE Methods Question Thread!

shadows

Re: VCE Methods Question Thread!

b^3

Re: VCE Methods Question Thread!

Recent Posts

User:
Password: