March 28, 2017, 08:55:30 pm

0 Members and 1 Guest are viewing this topic.

#### anotherworld2b

• Forum Obsessive
• Posts: 483
• Respect: 0
##### Re: 3U Maths Question Thread
« Reply #1185 on: January 11, 2017, 11:12:11 pm »
0
I was doing this question but i got the wrong answer. I am not sure what i did wrong

#### RuiAce

• Moderator
• ATAR Notes Legend
• Posts: 3595
• It will be ok
• Respect: +95
##### Re: 3U Maths Question Thread
« Reply #1186 on: January 11, 2017, 11:13:47 pm »
0
I was doing this question but i got the wrong answer. I am not sure what i did wrong
Line 3: Error in differentiation

dy/dx = -(...)^-2 . ( 2 + 6x )
Interested in Mathematics Tutoring? Message for details!

ATAR: 98.60
Currently studying: Bachelor of Actuarial Studies/Bachelor of Science (Advanced Maths) @ UNSW

#### anotherworld2b

• Forum Obsessive
• Posts: 483
• Respect: 0
##### Re: 3U Maths Question Thread
« Reply #1187 on: January 11, 2017, 11:25:22 pm »
0
oh I see now Thank you for the help RuiAce

Line 3: Error in differentiation

dy/dx = -(...)^-2 . ( 2 + 6x )

#### anotherworld2b

• Forum Obsessive
• Posts: 483
• Respect: 0
##### Re: 3U Maths Question Thread
« Reply #1188 on: January 12, 2017, 12:42:44 am »
0
For this question I'm not sure what I have to do

#### jamonwindeyer

• ATAR Notes HSC Lecturer
• ATAR Notes Legend
• Posts: 4416
• Electrical Engineer by day, AN Enthusiast by Night
• Respect: +217
##### Re: 3U Maths Question Thread
« Reply #1189 on: January 12, 2017, 02:09:19 am »
0
For this question I'm not sure what I have to do

Cool question! Think of this like a production line - The composite function is really just two functions performed one after the other. The question is asking what can come out the end of the conveyor belt, given what we are putting in to the conveyor belt (that is, the domain given the range).

Consider the first as an example. Putting the numbers -2, -1, 0, 1, 2 into the first box, $f(x)=x^2$, yields the numbers 0, 1 and 4. When you put those numbers in the second box, $g(x)=2x-3$, you get:

$g(0)=-3\\g(1)=-1\\g(4)=5$

So the range of the composite function in the first case is -3, -1, 5. Note that it might be a little weird writing a discrete range like this, but since we have a finite set of numbers given in the domain, we'll get a finite set of numbers as our range
« Last Edit: January 12, 2017, 02:11:06 am by jamonwindeyer »

#### bsdfjn;lkasn

• Trendsetter
• Posts: 166
• Respect: 0
##### Re: 3U Maths Question Thread
« Reply #1190 on: January 12, 2017, 08:48:15 am »
0

$(1)^2+(2)^2\text{ gives}\\ a^2+b^2=R^2(\cos^2\alpha+\sin^2\alpha)=R^2\\ (2)\div (1) \text{ gives}\\ \frac{b}{a}=\frac{\sin \alpha}{\cos \alpha}=\tan \alpha$

Hey, thanks for the detailed response, I haven't been taught this thoroughly yet so your answer really helped.
I was just wondering how these lines of working help me get R = 5? I understand how to get the answer by equating coefficients and using the identity but is the first (quoted) line the beginning a different method for obtaining the same answer? If not, would these lines be helpful in solving other questions based off the same congruency statement?
« Last Edit: January 12, 2017, 08:50:58 am by bsdfjn;lkasn »

#### RuiAce

• Moderator
• ATAR Notes Legend
• Posts: 3595
• It will be ok
• Respect: +95
##### Re: 3U Maths Question Thread
« Reply #1191 on: January 12, 2017, 09:52:45 am »
0
Hey, thanks for the detailed response, I haven't been taught this thoroughly yet so your answer really helped.
I was just wondering how these lines of working help me get R = 5? I understand how to get the answer by equating coefficients and using the identity but is the first (quoted) line the beginning a different method for obtaining the same answer? If not, would these lines be helpful in solving other questions based off the same congruency statement?
Those lines of working seek to prove the formula I quoted at the very start.
$\text{By formula}\\ a\sin x + b \cos x = R \sin (x+\alpha)\\ \text{where }R=\sqrt{a^2+b^2}\\ \tan \alpha=\frac{b}{a}\text{ and }0 <\alpha <\frac{\pi}{2}$
In your case, since a = 4 and b = 3, clearly upon substitution we have R = √(4^2+3^2) = 5.

This is not HSC content. This belongs in the preliminary course. If you are not familiar with it then you will need to go back to your notes regarding preliminary trigonometry.

Interested in Mathematics Tutoring? Message for details!

ATAR: 98.60
Currently studying: Bachelor of Actuarial Studies/Bachelor of Science (Advanced Maths) @ UNSW

#### Rathin

• Trendsetter
• Posts: 118
• Arthur Phillip High School
• Respect: +2
##### Re: 3U Maths Question Thread
« Reply #1192 on: January 12, 2017, 09:57:43 am »
0
Hey, thanks for the detailed response, I haven't been taught this thoroughly yet so your answer really helped.
I was just wondering how these lines of working help me get R = 5? I understand how to get the answer by equating coefficients and using the identity but is the first (quoted) line the beginning a different method for obtaining the same answer? If not, would these lines be helpful in solving other questions based off the same congruency statement?

A small rule to remember when converting to the auxiliary form:
acosx+bsinx ≡ Rcos(x-a)
acosx-bsinx ≡ Rcos(x+a)
asinx+bcosx ≡ Rsin(x+a)
asinx+bcosx ≡ Rsin(x-a)

The part of Rui's answer you quoted is just the derivation. R is denoted as the amplitude in the auxiliary form and is given by square rooting the squared the two coefficients. E.g  3cosx+4sinx ≡ Rcos(x-a) is the same as to match the correct auxiliary form 4sinx+3cosx ≡ Rsin(x+a)..where R is given by sqrt(4^2 + 3^2) =5 and the auxiliary angle which is responsible for a lateral shift is given by tan^-1(3/4) and note that the values for the angle are absolute as we are trying to find the smallest angle.

« Last Edit: January 12, 2017, 10:10:21 am by Rathin »
2017 HSC
4u | 3u | Physics | Biology | Adv Eng | PDHPE

#### RuiAce

• Moderator
• ATAR Notes Legend
• Posts: 3595
• It will be ok
• Respect: +95
##### Re: 3U Maths Question Thread
« Reply #1193 on: January 12, 2017, 09:58:57 am »
0
So the question you asked 3cosx+4sinx ≡ Rsin(x+a)..the transformation is actually wrong..it should be 3cosx+4sinx ≡ Rcos(x-a)
No, to this bit only. Because you can simply rewrite the LHS as 4sinx+3cosx
Interested in Mathematics Tutoring? Message for details!

ATAR: 98.60
Currently studying: Bachelor of Actuarial Studies/Bachelor of Science (Advanced Maths) @ UNSW

#### Rathin

• Trendsetter
• Posts: 118
• Arthur Phillip High School
• Respect: +2
##### Re: 3U Maths Question Thread
« Reply #1194 on: January 12, 2017, 10:00:39 am »
0
No, to this bit only. Because you can simply rewrite the LHS as 4sinx+3cosx

EDIT: Original Post has been modified.
« Last Edit: January 12, 2017, 10:04:58 am by Rathin »
2017 HSC
4u | 3u | Physics | Biology | Adv Eng | PDHPE

#### shreya_ajoshi

• Forum Regular
• Posts: 69
• Respect: 0
##### Re: 3U Maths Question Thread
« Reply #1195 on: January 12, 2017, 03:17:28 pm »
+1
How many numbers greater than 4000 can be formed from the figures 3,5,7,8,9? (repetitions not allowed)

In how many ways can the letters of the word permute be arranged if (i) the first and last places are occupied by consonants, (ii) the vowels and consonants occupy alternate places?

#### RuiAce

• Moderator
• ATAR Notes Legend
• Posts: 3595
• It will be ok
• Respect: +95
##### Re: 3U Maths Question Thread
« Reply #1196 on: January 12, 2017, 03:32:45 pm »
0
How many numbers greater than 4000 can be formed from the figures 3,5,7,8,9? (repetitions not allowed)

In how many ways can the letters of the word permute be arranged if (i) the first and last places are occupied by consonants, (ii) the vowels and consonants occupy alternate places?
Question 1 is too arbitrary. Is there a restriction on how many digits that can be used? (Since you gave 5 digits for a 4-digit number.)
_______________________
$\text{7 letters: }\_\,\_\,\_\, \_\, \_\, \_\, \_\\ \text{Note that }E\text{ is repeated once.}$
$\text{In i), we note that out of the 7 letters, 4 are consonants}\\ \text{So we pick two consonants to occupy those spots: }4\times 3\\ \text{Then we have 5 letters (with a repeated }E\text{) which we arrange however we like: }\frac{5!}{2!}\\ \therefore 4\times 3\times \frac{5!}{2!}$
$\text{In ii), this is FORCED:}\\ \mathcal{C}\mathcal{V} \mathcal{C}\mathcal{V} \mathcal{C}\mathcal{V} \mathcal{C}\\ \text{where }\mathcal{C}\text{ is a consonant}\\ \text{and }\mathcal{V}\text{ is a vowel.}\\ \text{It now falls to just split the consonants and vowels up among only themselves.}$
$\therefore \underbrace{4!}_{\text{consonants}} \times \underbrace{\frac{3!}{2!}}_{\text{vowels}}$
Interested in Mathematics Tutoring? Message for details!

ATAR: 98.60
Currently studying: Bachelor of Actuarial Studies/Bachelor of Science (Advanced Maths) @ UNSW

#### anotherworld2b

• Forum Obsessive
• Posts: 483
• Respect: 0
##### Re: 3U Maths Question Thread
« Reply #1197 on: January 12, 2017, 05:27:39 pm »
0
Thank you for your help jamon. Can i please have help with these two questions?

#### bsdfjn;lkasn

• Trendsetter
• Posts: 166
• Respect: 0
##### Re: 3U Maths Question Thread
« Reply #1198 on: January 12, 2017, 06:18:47 pm »
0
(I know this is a really simple question but I don't know what the t-formulae is and what it's used for)

Substituting the t-formulae for cos x and sin x into the equation 3sinx+cosx = 4 produces what quadratic equation?

#### RuiAce

• Moderator
• ATAR Notes Legend
• Posts: 3595
• It will be ok
• Respect: +95
##### Re: 3U Maths Question Thread
« Reply #1199 on: January 12, 2017, 06:23:29 pm »
0
(I know this is a really simple question but I don't know what the t-formulae is and what it's used for)

Substituting the t-formulae for cos x and sin x into the equation 3sinx+cosx = 4 produces what quadratic equation?
$\text{Assuming }0\le x \le 2\pi\text{ for demonstration's sake.}$
$\text{Let }t=\tan \frac{x}{2}\text{ and note we have to check }x=\pi \text{ separately:}\\ 3\sin \pi + \cos \pi = -1 \neq 4\\ \text{So }x=\pi\text{ is not important}\\ \text{(The check was necessary as }\tan \frac{\pi}2\text{ is undefined)}$
\text{Then, }\sin x = \frac{2t}{1+t^2},\, \cos x=\frac{1-t^2}{1+t^2}\\ \text{Hence our equation becomes}\\ \begin{align*}\frac{6t}{1+t^2}+\frac{1-t^2}{1+t^2}&=4\\ 6t + 1-t^2 &= 4+4t^2\\ 0&=5t^2 -6t+3 \end{align*}
$\text{This can be chucked into the quadratic formula.}$
This method is kinda inferior to the auxiliary angle transformation for trigonometric equations if you ask me. It's used in Extension 2 for some bizarre integrals.
Interested in Mathematics Tutoring? Message for details!

ATAR: 98.60
Currently studying: Bachelor of Actuarial Studies/Bachelor of Science (Advanced Maths) @ UNSW