Author Topic: VCE Methods Question Thread! (Read 5649273 times) Tweet Share

~T · « **Reply #1860 on:** April 23, 2013, 10:49:53 pm »

Do we seriously need to be able to do left endpoint/right endpoint estimate for integration? As in, will it ever be on an exam? I see absolutely no reason for its existence, except perhaps for understanding/intuition of what integration is... Or if it's a graph without a function...

I've just gotten up to chapter 12, and some of these questions require little thought, but soooooo much time!

~T · « **Reply #1861 on:** April 23, 2013, 11:07:32 pm »

Quote from: Essendon2013 on April 23, 2013, 10:49:22 pm

Hey guys, need some help for a question.

The simultaneous linear equations
mx + 12y = 24
3x + my = m

Find the values for which m give
-> A unique solution
-> Infinitely many solutions

Help would be greatly appreciated!

If we were to write it as a matrix equation, then the determinant of the matrix of coefficients would be:
$Det = m*m - 12*3$
$Det = m^{2} - 36$

For a unique solution, the determinant will not equal zero.
For infinitely many solutions, the determinant will equal zero AND the two graphs will represent the same line.
If they do not represent the same line, but the determinant still equals zero (y=4x+3 compared to y=4x+7) then there are no solutions.

So, firstly we solve for Det = 0:
$m^{2} - 36 = 0$
$m^{2} = 36$
$m \in \left\{-6 , 6 \right\}$

For a unique solution, the determinant does not equal zero. So $m \in R \setminus \left\{-6 , 6 \right\}$

For infinitely many, it could be either -6 or 6. So let's inspect:
When m = -6 the two equations are:
$-6x + 12y = 24$
$3x - 6y = -6$ OR $-6x + 12y = 12$ (multiplying by -2)
These equations are different and hence represent equations with no solutions.

When m = 6 the two equations are:
$6x + 12y = 24$
$3x + 6y = 6$ OR $6x + 12y = 12$ (multiplying by 2)
These equations are different and hence represent equations with no solutions.

Hence, there are no values of m which give infinitely many solutions

Essendon2013 · « **Reply #1862 on:** April 24, 2013, 11:49:12 am »

Quote from: Tim...blahhh on April 23, 2013, 11:07:32 pm

If we were to write it as a matrix equation, then the determinant of the matrix of coefficients would be:
$Det = m*m - 12*3$
$Det = m^{2} - 36$

For a unique solution, the determinant will not equal zero.
For infinitely many solutions, the determinant will equal zero AND the two graphs will represent the same line.
If they do not represent the same line, but the determinant still equals zero (y=4x+3 compared to y=4x+7) then there are no solutions.

So, firstly we solve for Det = 0:
$m^{2} - 36 = 0$
$m^{2} = 36$
$m \in \left\{-6 , 6 \right\}$

For a unique solution, the determinant does not equal zero. So $m \in R \setminus \left\{-6 , 6 \right\}$

For infinitely many, it could be either -6 or 6. So let's inspect:
When m = -6 the two equations are:
$-6x + 12y = 24$
$3x - 6y = -6$ OR $-6x + 12y = 12$ (multiplying by -2)
These equations are different and hence represent equations with no solutions.

When m = 6 the two equations are:
$6x + 12y = 24$
$3x + 6y = 6$ OR $6x + 12y = 12$ (multiplying by 2)
These equations are different and hence represent equations with no solutions.

Hence, there are no values of m which give infinitely many solutions

Thank you! Definately understand it now!

shadows · « **Reply #1863 on:** April 25, 2013, 05:41:04 pm »

Hey guys, I haven't done indices in a long time so my skills might be little rusty.

This is 5C from essentials book

Q1 f
$\frac{15(x^5y^{-2})^4} {3(x^4y)^{-2}}$

my answer is $\frac{25x^{28}}{y^6}$

but apparently thats not the answer D:?

another question

how do you solve this?

1 i
$\frac{x^2+y^2}{x^{-2}+y^{-2}}$

thank you guys. Learnt latex as well.. took soo long to write the equations
but makes it so much easier to read for you guys

Daenerys Targaryen · « **Reply #1864 on:** April 25, 2013, 05:51:09 pm »

Quote from: shadows on April 25, 2013, 05:41:04 pm

Hey guys, I haven't done indices in a long time so my skills might be little rusty.

This is 5C from essentials book

Q1 f
$\frac{15(x^5y^{-2})^4} {3(x^4y)^{-2}}$

my answer is $\frac{25x^{28}}{y^6}$

but apparently thats not the answer D:?

another question

how do you solve this?

1 i
$\frac{x^2+y^2}{x^{-2}+y^{-2}}$

thank you guys. Learnt latex as well.. took soo long to write the equations
but makes it so much easier to read for you guys

$\frac{15(x^{20}y^{-8}}{3(x^{-8}y^{-2}}$

$5(x^{28}y^{-6})$

$\frac{5x^{28}}{y^6}$

CAS came up with same

Scooby · « **Reply #1865 on:** April 25, 2013, 05:54:58 pm »

Quote from: shadows on April 25, 2013, 05:41:04 pm

Hey guys, I haven't done indices in a long time so my skills might be little rusty.

This is 5C from essentials book

Q1 f
$\frac{15(x^5y^{-2})^4} {3(x^4y)^{-2}}$

my answer is $\frac{25x^{28}}{y^6}$

but apparently thats not the answer D:?

$\frac{15(x^5y^{-2})^4} {3(x^4y)^{-2}}$

= $\frac{15(x^{20}y^{-8})} {3(x^{-8}y^{-2})}$

= ${5x^{28}y^{-6}}$

= $\frac{5x^{28}} {y^6}$

Phy124 · « **Reply #1866 on:** April 25, 2013, 05:55:48 pm »

Quote from: shadows on April 25, 2013, 05:41:04 pm

$\frac{15(x^5y^{-2})^4} {3(x^4y)^{-2}}$

How come you have done $\frac{15}{3} = 25$ ?

(The $\frac{x^{28}}{y^6}$ part is right)

Quote from: shadows on April 25, 2013, 05:41:04 pm

how do you solve this?

1 i
$\frac{x^2+y^2}{x^{-2}+y^{-2}}$

I assume by solve you mean simplify, in which case:

$\frac{x^2+y^2}{x^{-2}+y^{-2}}= \frac{x^2+y^2}{\frac{1}{x^{2}}+\frac{1}{y^{2}}}=\frac{x^2+y^2}{\frac{x^2+y^2}{x^2y^{2}}}=x^2y^2$

Quote from: shadows on April 25, 2013, 05:41:04 pm

thank you guys. Learnt latex as well.. took soo long to write the equations
but makes it so much easier to read for you guys

Thanks, we appreciate it

shadows · « **Reply #1867 on:** April 25, 2013, 05:59:53 pm »

15/3 is 5

doesn't 5 have to multiply both x and y?

and y moves down... leaving the extra 5 to times with the other 5?/

i havent done this in a year... so please help me out xD

Daenerys Targaryen · « **Reply #1868 on:** April 25, 2013, 06:01:23 pm »

The 5 would be times by x and y if: $a(b+c)=ab+ac \text { where as in this case }a(bc)=abc$

shadows · « **Reply #1869 on:** April 25, 2013, 06:04:25 pm »

oh right!

can't believe i'm getting these concepts muddled up D:

math doesn't occur to me naturally lol..

shadows · « **Reply #1870 on:** April 25, 2013, 06:18:59 pm »

if you get more than 1 result for exponential/ logs

do you always need to sub back in to equation to ensure that it works?

Daenerys Targaryen · « **Reply #1871 on:** April 25, 2013, 06:24:41 pm »

Quote from: shadows on April 25, 2013, 06:18:59 pm

if you get more than 1 result for exponential/ logs

do you always need to sub back in to equation to ensure that it works?

yup.

tae · « **Reply #1872 on:** April 25, 2013, 08:37:46 pm »

Ehh, I don't really understand how to do this question

Find exact solutions of $4sin2(x-\frac{2\pi}{3})+5=7$ over the domain [ $0, \pi\$ ]

Alwin · « **Reply #1873 on:** April 25, 2013, 08:55:41 pm »

Quote from: tae on April 25, 2013, 08:37:46 pm

Ehh, I don't really understand how to do this question

Find exact solutions of $4sin2(x-\frac{2\pi}{3})+5=7$ over the domain [ $0, \pi\$ ]

You rearrange for $sin(2(x-\frac{2\pi}{3}))$ and then find the basic angle etc.

Full solution:

$4sin(2(x-\frac{2\pi}{3}))+5=7$
$sin(2(x-\frac{2\pi}{3}))=\frac{1}{2}$
$\text{basic angle: } sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$
$\text{Domain: }x \in [0, \pi] \text{ hence } 2(x-\frac{2\pi}{3}) \in [-\frac{2\pi}{3}, \frac{4\pi}{3}]$
$2(x-\frac{2\pi}{3})=\frac{-7\pi}{6}, \frac{\pi}{6}, \frac{5\pi}{6}$
$x=\frac{\pi}{12}, x=\frac{3\pi}{4}, x=\frac{13\pi}{12}$

Clearly the last solution is not in the specified domain, hence
$x=\frac{\pi}{12}, x=\frac{3\pi}{4}$

Hope that helps!

shadows · « **Reply #1874 on:** April 27, 2013, 02:54:29 pm »

another question!
$10=ae^{-6(\frac{1}{3}loge(5))$

How to get a solution for a?

Is it ok if you can show me step by step? I get confused with log laws....

Thank you.

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~T

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