Author Topic: VCE Specialist 3/4 Question Thread! (Read 2612673 times) Tweet Share

DSubShell · « **Reply #4470 on:** January 24, 2015, 06:19:46 pm »

Quote from: Spxtcs on January 24, 2015, 05:16:47 pm

maybe more experienced posters could pitch in on this but personally I find that a bit of a confusing method since you get the answer by deriving anyway.

I'm pretty sure they use their method to make the derivate easier to do by hand. There is less potential for errors if you don't have to do the chain rule that the square root poses.

Personally I don't think it's necessary, but for some people it might be easier

Chang Feng · « **Reply #4471 on:** January 24, 2015, 08:00:21 pm »

Ahh okay thanks for the clear up.
And also how would you do question 9 that is attached too?
And question 10.b) too?
Thank you a lot.

Chang Feng · « **Reply #4472 on:** January 24, 2015, 08:12:03 pm »

And here is the attachment for the above questions.
Thanks again.

Eiffel · « **Reply #4473 on:** January 24, 2015, 09:51:55 pm »

Hi Guys, bit of a problem here

$| (z-w)( \bar{z} - \bar{w} ) = r^{2} |$

show that, the above, defines a circle on the argand diagram with centre at w and a radius of r. I got to a centre of a, b (where w = a + bi ) but i had a minus and not a plus as required in a circle.

brightsky · « **Reply #4474 on:** January 24, 2015, 09:58:28 pm »

Quote from: Eiffel on January 24, 2015, 09:51:55 pm

Hi Guys, bit of a problem here

$| (z-w)( \bar{z} - \bar{w} ) = r^{2} |$

show that, the above, defines a circle on the argand diagram with centre at w and a radius of r. I got to a centre of a, b (where w = a + bi ) but i had a minus and not a plus as required in a circle.

$(z-w)(\bar{z} - \bar{w}) = r^{2}$

$(z-w)(\overline{z-w}) = r^2$

$|z-w|^2 = r^2$

$|z-w| = r$

Read the final equation out loud: "The distance from z to the complex number w is r". It should be clear now that the graph that we want is a circle centred at the complex number w, with a radius of r.

Eiffel · « **Reply #4475 on:** January 24, 2015, 10:45:37 pm »

Quote from: brightsky on January 24, 2015, 09:58:28 pm

$(z-w)(\bar{z} - \bar{w}) = r^{2}$

$(z-w)(\overline{z-w}) = r^2$

$|z-w|^2 = r^2$

$|z-w| = r$

Read the final equation out loud: "The distance from z to the complex number w is r". It should be clear now that the graph that we want is a circle centred at the complex number w, with a radius of r.

Alright, cheers brightsky

Though, how did you manage to

$(z-w)(\overline{z-w}) = r^2$
$|z-w|^2 = r^2$

get from that - to that.

keltingmeith · « **Reply #4476 on:** January 24, 2015, 11:03:09 pm »

Quote from: Eiffel on January 24, 2015, 10:45:37 pm

Alright, cheers brightsky

Though, how did you manage to

$(z-w)(\overline{z-w}) = r^2$
$|z-w|^2 = r^2$

get from that - to that.

$z\bar{z}=|z|^2$ , where z is a complex number. If you let z-w=u, then you get:

$(z-w)(\overline{z-w})<br />\\ =u\bar{u}<br />\\ =|u|^2<br />\\ =|z-w|^2$

kinslayer · « **Reply #4477 on:** January 25, 2015, 01:41:43 am »

Quote from: brightsky on January 24, 2015, 09:58:28 pm

$|z-w| = r$

Read the final equation out loud: "The distance from z to the complex number w is r". It should be clear now that the graph that we want is a circle centred at the complex number w, with a radius of r.

This interpretation of the modulus function is extremely useful. Get used to it! It will serve you well.

Eiffel · « **Reply #4478 on:** January 25, 2015, 06:48:00 pm »

how does
$4cos(2x)sin(2x) <br />= 2sin(4x)$

brightsky · « **Reply #4479 on:** January 25, 2015, 07:13:35 pm »

sin (2*BLAH) = 2 cos (BLAH) sin (BLAH)

Eiffel · « **Reply #4480 on:** January 25, 2015, 07:33:03 pm »

Haha sorry brightsky, i don't understand your mathematical jargon

double angle formula? since this is a methods question originally, is this needed for methods?

e.g. would the first part suffice as the final answer?

pi · « **Reply #4481 on:** January 25, 2015, 07:59:15 pm »

To avoid confusion in the future, shall we post "methods questions" in the Methods board perhaps?

Eiffel · « **Reply #4482 on:** January 25, 2015, 10:10:50 pm »

How could you double differentiate this?

$\frac {\cos(2x)}{x}$

Spoiler

\answer is

$\frac{-4x^{2}\cos(2x) + 4x\sin(2x) + 2\cos(2x)}{x^{3}}$

keltingmeith · « **Reply #4483 on:** January 25, 2015, 10:17:15 pm »

Quote from: Eiffel on January 25, 2015, 10:10:50 pm

How could you double differentiate this?

$\frac {\cos(2x)}{x}$

Spoiler
\answer is

$\frac{-4x^{2}\cos(2x) + 4x\sin(2x) + 2\cos(2x)}{x^{3}}$

You know this has steps, yeah?

pi · « **Reply #4484 on:** January 25, 2015, 10:22:49 pm »

Eiffel in your textbook, after product rule, they'll have "quotient rule". Have a look into that, show us your attempt and someone will guide you if you have troubles!

(But personally I'd product rule that for the first time at least)

Search the forums now!

We have moved!

Author Topic: VCE Specialist 3/4 Question Thread! (Read 2612673 times) Tweet Share

DSubShell

Re: Specialist 3/4 Question Thread!

Chang Feng

Re: Specialist 3/4 Question Thread!

Chang Feng

Re: Specialist 3/4 Question Thread!

Eiffel

Re: Specialist 3/4 Question Thread!

brightsky

Re: Specialist 3/4 Question Thread!

Eiffel

Re: Specialist 3/4 Question Thread!

keltingmeith

Re: Specialist 3/4 Question Thread!

kinslayer

Re: Specialist 3/4 Question Thread!

Eiffel

Re: Specialist 3/4 Question Thread!

brightsky

Re: Specialist 3/4 Question Thread!

Eiffel

Re: Specialist 3/4 Question Thread!

pi

Re: Specialist 3/4 Question Thread!

Eiffel

Re: Specialist 3/4 Question Thread!

keltingmeith

Re: Specialist 3/4 Question Thread!

pi

Re: Specialist 3/4 Question Thread!

Recent Posts

User:
Password: