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I go to Fort Street high school.

A hard exam is your best friend.

I also tend to take it as it comes. I hit 4U math pretty hard on a couple days in the holidays and got about a term ahead of our topic schedule which has meant I could really ease off on doing a lot of 4U work the past couple of weeks. On average I'd do about 30-50 mins of math a night but it's...

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Selective schools ----> extension courses are much more popular as well as being much more viable, with the latter point being somewhat debatable.

Thanks for that, it's quite interesting.

Is that so? Last time I heard modesty was still modesty. Regardless of your talent you shouldn't parade here boasting about your ability. It's not cool and, by definition, it's arrogant. Like slide said, "actions speak louder than words" so wait and see your HSC results and then speak. In the...

Guys, please don't bring that arrogant shit in here. Also I don't see why you suggest at slide rule being arrogant, if anything it's a picture you've built yourself from the way people talk him up.

Yeah, our school is a bit retarded when it comes to the HSC. We don't do anything longer than an hour long exam until our trials at which point we are thrown into it. So that's all we had: 3 questions, one hour.

Here's a link to my half yearly exam (sorry about the crappy photoshack hosting): http://img43.echo.cx/img43/8459/save00105op.png It was a 1 hour exam. I've been told that the average was in the vicinity of 68% and that the top mark was 94% out of 50 students. Enjoy :)

For that one I think you should try the two circle approach again. &delta;V = &pi;(r₁² - r₂²) where r₁ = 1 and r₂ = (y-1)² --> &delta;V = &pi;(1 - [cosx -1]²) yielding: &pi;&int; 2cosx -...

For 6. you're integrating about the y-axis with perp. strips so you have to talk in terms of y again. Treat it like a (bi-quadratic?) equation. So y= 6x² - x⁴, let u=x² ---> y = 6u - u² (*note) the area of each strip will be...

You'll hate what went wrong :p. I think you assumed that when you substituted in '0' that the second bracket would simply disappear. Check out what happens when you subtract 8/3(4-y)^3/2 from 24 where y=0 ;).

for 2. (c) You're going to be integrating between 0 and 4 along the y-axis hence you need to express the volume/area of each disc in terms of y. In this case the radius of each circular slice will be 2 - &radic;(4-y) hence the area of each slice is given by &pi;(2 -...

for Ex6.1: 2. (a) you're rotating the aread between y=4-x², x=2 and y=4 about the x-axis using perpendicular slices. Each slice = &pi;r₁² - &pi;r₂² = &pi;(r₁² -r₂²) where r₁ =...

Haha, well done. That sure makes things a lot simpler.

Would you be able to hint me in the right direction? (and I'm talking a fairly direct, not so subtle hint :p)

Hey guys, this is a question I cam up against in the extension section of the cambridge 3U book that I borrowed recently. I'll try and work it up to where I got to and let you guys have at it from there. 1) First you have to prove that: 1 - t + t² - t³ + ... +...

integral of 1/(sqrt(4-x^2)) let x = 2sin&theta; ---> dx = 2cos&theta;d&theta; &int; dx/&radic;(4-x²) = &int; 2cos&theta;/&radic;(4[1-sin²&theta;]) d&theta; = &int; 2cos&theta;/&radic;[4cos²&theta;] d&theta; = &int; 1 d&theta; = &theta; since x =...

Weeks 2-3 term 3. (other :p)