Print Page - hyunah's Specialist Questions

Title: hyunah's Specialist Questions
Post by: hyunah on June 01, 2014, 11:57:05 am

how do i know that tan(pi/2 - x) = cot (x) ?
how would i prove these complementary angles. i'm struggling to remember them... any tips please?

Title: Re: questions
Post by: Rectophobia on June 04, 2014, 09:09:24 pm

You can reason it by saying that sin(pi/2 - x) = cos(x) and cos(pi/2 - x) = sin(x) as they are complimentary.

It would then follow that tan(pi/2 - x) = [sin(pi/2 - x)/cos(pi/2 - x)] = cos(x)/sin(x) = cot(x).

Other than this rough proof, the only one I can come up with involves expressing sine and cosine in terms of their exponential form. I've attached this proof, but it isn't needed for VCE.

P.S. Sorry for the shitty effects on the camera. The image wasn't perfect so I had to distort it in order for all the lettering to be visible.

P.S.S. I just realised That you could do it in the following way:

tan(pi/2 - x)

= sin(pi/2 - x)/cos(pi/2 - x)

= [sin(pi/2)cos(x) - cos(pi/2)sin(x)]/[cos(pi/2)cos(x) + sin(pi/2)sin(x)] (by compound angle formula)

= cos(x)/sin(x) (after simplification)

=cot(x)

Title: Re: questions
Post by: hyunah on July 05, 2014, 01:20:26 pm

thanks XD

Title: Re: questions
Post by: hyunah on July 05, 2014, 01:22:15 pm

how do i diff sin^-1(x)

isn't it just
u =1 u' = 0
v = sin(x) v' = cos(x)
dy/dx = (v'u - u'v)/v^2
?

Title: Re: questions
Post by: keltingmeith on July 05, 2014, 01:52:55 pm

Not even close. sin^-1 is inverse sine, not 1/sin. Consult your textbook and/or formula sheet on how to diff it.

Title: Re: questions
Post by: hyunah on July 05, 2014, 02:16:38 pm

oh cause normal when i find a function eg (1/x)^-1 isn't it x?
i guess its a different story with sin?

Title: Re: questions
Post by: keltingmeith on July 05, 2014, 02:27:29 pm

Yep - circular functions are just stupid like that. Their reciprocal are given entirely different names, which you should be aware of of you're doing spec - sec (for cos), cosec (for sin) and cot (for tan).

Title: Re: questions
Post by: hyunah on July 05, 2014, 02:30:58 pm

yes thank you

Title: Re: questions
Post by: hyunah on July 05, 2014, 02:31:26 pm

Find the particular solution. Of f’(x) = sqroot(1-y^2) if f(0) = ½
I got y = sin(x+pi/6)
Is the largest domain R
The second part of the question (whats the largest domain) is worth 3 marks… am I missing something?
Cause with the question after it
It says to find the general solution of f’(x) = 2/(7-4y)
Which I got x = 7/2y – y^2 +c
And part b worth also 3 marks in this question is to find the largest domain
I had to go y = ____________
Then have whatever in the sqroot >0 and solve for x to get the largest domain being x<(16c+49)/16
I get why that is worth 3 marks, but in the other one… just writing R because the domain of sin(x+pi/6) is R seems like im missing something?
Am I right?

Title: Re: questions
Post by: hyunah on July 05, 2014, 10:46:11 pm

Quote from: hyunah on July 05, 2014, 02:31:26 pm

Find the particular solution. Of f’(x) = sqroot(1-y^2) if f(0) = ½
I got y = sin(x+pi/6)
Is the largest domain R
The second part of the question (whats the largest domain) is worth 3 marks… am I missing something?
Cause with the question after it
It says to find the general solution of f’(x) = 2/(7-4y)
Which I got x = 7/2y – y^2 +c
And part b worth also 3 marks in this question is to find the largest domain
I had to go y = ____________
Then have whatever in the sqroot >0 and solve for x to get the largest domain being x<(16c+49)/16
I get why that is worth 3 marks, but in the other one… just writing R because the domain of sin(x+pi/6) is R seems like im missing something?
Am I right?

Title: Re: questions
Post by: hyunah on July 15, 2014, 04:10:48 pm

can someone please help with question 13c)
I thought I could just do vectorGP . vectorGQ but that doesn't equal 0?

Title: Re: questions
Post by: keltingmeith on July 16, 2014, 06:24:17 pm

Quote from: hyunah on July 15, 2014, 04:10:48 pm

can someone please help with question 13c)
I thought I could just do vectorGP . vectorGQ but that doesn't equal 0?

You're completely right there - maybe double check you haven't made an arithmetic error?

Title: Re: questions
Post by: Conic on July 16, 2014, 08:14:20 pm

As EulerFan said, it is most likely an arithmetic error that is causing your problem. For the vectors a and b you should have:

Spoiler

$\underset{\sim}{a}=\frac{1}{3}\left(4\underset{\sim}{i}+2\sqrt{2}\underset{\sim}{j}\right) \quad \mathrm{and} \quad \underset{\sim}{b}=\frac{1}{3}\left(2\underset{\sim}{i}+4\sqrt{2}\underset{\sim}{j}\right).$

You can find the vectors GP and GQ directly using the previous part and then find the dot product:

Spoiler

$\overrightarrow{GP}=\frac{1}{3} \left( 2\underset{\sim}{a}-\underset{\sim}{b} \right)=\frac{2}{3}\underset{\sim}{i}.$

$\overrightarrow{GQ}=\frac{1}{3} \left( 2\underset{\sim}{b}-\underset{\sim}{a} \right)=\frac{2\sqrt{2}}{3}\underset{\sim}{j}.$

$\therefore \overrightarrow{GP} \cdot \overrightarrow{GQ}= \left( \frac{2}{3}\underset{\sim}{i} \right) \cdot \left( \frac{2\sqrt{2}}{3}\underset{\sim}{j} \right)=0.$

You can also expand the expression with GP and GQ in terms of the vectors a and b and then substitute in the values:

Spoiler

$\overrightarrow{GP} \cdot \overrightarrow{GQ}=\frac{1}{3}\left( 2\underset{\sim}{a}-\underset{\sim}{b} \right)\cdot \frac{1}{3}\left( 2\underset{\sim}{b}-\underset{\sim}{a} \right)$

$= \frac{1}{9} \left( \left( 2\underset{\sim}{a}-\underset{\sim}{b} \right)\cdot \left( 2\underset{\sim}{b}-\underset{\sim}{a} \right) \right)$

$=\frac{1}{9} \left( 4 \underset{\sim}{a} \cdot \underset{\sim}{b} - 2 \underset{\sim}{a} \cdot \underset{\sim}{a} - 2 \underset{\sim}{b} \cdot \underset{\sim}{b} + \underset{\sim}{a} \cdot \underset{\sim}{b} \right)$

$=\frac{1}{9} \left( 5\underset{\sim}{a} \cdot \underset{\sim}{b}-2\underset{\sim}{a} \cdot \underset{\sim}{b}-2\underset{\sim}{b} \cdot \underset{\sim}{b}\right).$

$\mathrm{We \: have}\quad\underset{\sim}{a} \cdot \underset{\sim}{b}=\frac{8}{3},\quad \underset{\sim}{a} \cdot \underset{\sim}{a}=\frac{8}{3},\quad \mathrm{and}\quad \underset{\sim}{b} \cdot \underset{\sim}{b}=4.$

$\therefore \overrightarrow{GP} \cdot \overrightarrow{GQ}=\frac{1}{9} \left( 5\underset{\sim}{a} \cdot \underset{\sim}{b}-2\underset{\sim}{a} \cdot \underset{\sim}{b}-2\underset{\sim}{b} \cdot \underset{\sim}{b}\right)$

$= \frac{1}{9} \left( 5\cdot\frac{8}{3}-2\cdot4-2\cdot\frac{8}{3} \right)$

$=0.$

Title: Re: questions
Post by: hyunah on July 17, 2014, 11:17:51 am

thank you conic :)

is that based on the assumption that p is equal to OP which is equal to a and that q is equal to OQ which is equal to b?

cause the solution had something complex like this which i ddin't get....

Title: Re: questions
Post by: Conic on July 17, 2014, 05:57:05 pm

Quote from: hyunah on July 17, 2014, 11:17:51 am

is that based on the assumption that p is equal to OP which is equal to a and that q is equal to OQ which is equal to b?

cause the solution had something complex like this which i ddin't get....

Yes, they said "Denote vectors OP and OQ by a and b respectively", so as $p=\frac{1}{3}(4+2\sqrt{2}\:\mathrm{i})$ and $q=\frac{1}{3}(2+4\sqrt{2}\:\mathrm{i})$ , you can get the vectors a and b.

The solutions are using assumed geometry knowledge from the study design.

(http://i.imgur.com/Gue8ZtL.png)

They are basically showing that G, P and Q lie on a circle around M, and that PQ is the diameter of a circle. As in the image above, this means that PGQ is a right angle. This approach is correct, but it is a lot easier to use dot products.

Title: Re: hyunah's Specialist Questions
Post by: hyunah on August 24, 2014, 05:13:04 pm

how do i differentiate:
tan^-1(2sqrt(x/15)

the ans says: sqrt(15)/[sqrt(x)(4x+15)

Title: Re: hyunah's Specialist Questions
Post by: polar1 on August 24, 2014, 05:44:21 pm

Quote from: hyunah on August 24, 2014, 05:13:04 pm

how do i differentiate:
tan^-1(2sqrt(x/15)

the ans says: sqrt(15)/[sqrt(x)(4x+15)

Let $y=\arctan\left(2\sqrt\frac{x}{15} \right) = \arctan\left(\frac{2}{\sqrt{15}}\sqrt{x} \right)$ . Then, let $u=\frac{2\sqrt{x}}{\sqrt{15}}$ so that $y=\arctan(u)$ and $\frac{\mathrm{d}y}{\mathrm{d}u} = \frac{1}{1+u^2}$ while $\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{2}{\sqrt{15}}\times\frac{1}{2\sqrt{x}} = \frac{1}{\sqrt{15}\sqrt{x}}$ . Then, using the chain rule, we have $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{1+u^2}\times\frac{1}{\sqrt{15}\sqrt{x}} = \frac{1}{1+(4x/15)}\times\frac{1}{\sqrt{15}\sqrt{x}}$ .

Multiply the numerator and denominator of the first fraction by 15 to get $\frac{15}{15 + 4x}\times\frac{1}{\sqrt{15}\sqrt{x}} = \frac{\sqrt{15}}{\sqrt{x}(15+4x)}$

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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: hyunah on June 01, 2014, 11:57:05 am