HSC 2012 MX1 Marathon #1 (archive) (1 Viewer)

Sidekickk

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Re: 2012 HSC MX1 Marathon

Differentiate y=3sqrroot2 using the First Principles method.
Hey, there's something i can do in this thread lol.

Answer: 1!!

i am a beast

Edit: oops. i got it wrong. its 0 :/
 

Examine

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Re: 2012 HSC MX1 Marathon

Differentiate with respect to x to the following

(x-3) sqrroot(x-1)
 

deswa1

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Re: 2012 HSC MX1 Marathon

Differentiate with respect to x to the following

(x-3) sqrroot(x-1)
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{3x-5}{2\sqrt{x-1}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{3x-5}{2\sqrt{x-1}}" title="\frac{3x-5}{2\sqrt{x-1}}" /></a>

A question from our 3U test today: Prove using calculus that the shortest diagonal of a rectangle occurs when the rectangle is a square.
 

seanieg89

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Re: 2012 HSC MX1 Marathon

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{3x-5}{2\sqrt{x-1}}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{3x-5}{2\sqrt{x-1}}" title="\frac{3x-5}{2\sqrt{x-1}}" /></a>

A question from our 3U test today: Prove using calculus that the shortest diagonal of a rectangle occurs when the rectangle is a square.
For this question to make sense you also need to stipulate that the collection of rectangles you are looking at has fixed area or fixed perimeter (either will do).

As stated, there is no rectangle with shortest diagonal length as one can simply draw a smaller rectangle...
 

deswa1

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Re: 2012 HSC MX1 Marathon

For this question to make sense you also need to stipulate that the collection of rectangles you are looking at has fixed area or fixed perimeter (either will do).

As stated, there is no rectangle with shortest diagonal length as one can simply draw a smaller rectangle...
Whoops my bad. There is a fixed perimeter.
 

Shadowless

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Re: 2012 HSC MX1 Marathon

MATHS.png

BTW: How do you get integral symbols and fractions INTO YOUR POST?
 

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