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Help with this proving.

(1/1)

nickglyn:
Hey so I'm having trouble with this whole question. Can you show me proving? Thanks heaps.

kb123:

--- Quote from: nickglyn on October 20, 2016, 05:19:57 pm ---Hey so I'm having trouble with this whole question. Can you show me proving? Thanks heaps.

--- End quote ---

Hi!

Here is (i) and (ii), will be back with (iii) soon :)

jamonwindeyer:

--- Quote from: nickglyn on October 20, 2016, 05:19:57 pm ---Hey so I'm having trouble with this whole question. Can you show me proving? Thanks heaps.

--- End quote ---

Ergh, this one is nasty ;)

Okay, so let's consider the distance \(L\) in two parts: The length of rope PM and the distance PX.

Let's start with PX. That's just pythagoras on the triangle on the left:



Now, PM is the length of the rope in total, minus the length of rope AP:



Now we can find AP using pythagoras theorem, noting that \(AX=1-x\):



So, \(PM=2-\sqrt{1-2x+r^2}\).

Putting all that together:



And that is the proof. From here it is just normal (albeit tedious) differentiation work.



Then you just use the factorisation given to solve for the numerator equal to zero:



So \(x=1\) is one obvious candidate; but it's actually incorrect, the rope would just be hanging that way, it violates our conditions!

Note that here I deviate from the answer above, because the factorisation above was incorrect. For the right bracket, use the quadratic formula:



Note that we don't care about the negative answer here, since x>0 ;D

Now this is the value we want, and we do need to prove that it is a maximum. You could do this by differentiating again (10/10 disgusting) or more nicely, just substitution into the first derivative. Tedious algebra, but the usual technique! ;D

(Ps - Great answer kb123! Only uploaded my Part 1 and 2 because I had it ready ;))

PS - This is far more difficult than anything you will get in 2U these days, this is from a 2003 HSC Paper ;D

jamonwindeyer:

--- Quote from: kb123 on October 20, 2016, 05:54:06 pm ---Here's my result for (iii), not sure if it would get both marks though... But i did it aha

Can't wait to finally finish the mathematics HSC.... :)

--- End quote ---

Almost! Check your factorisation in Part (iii), not quite correct, super close! We just need the quadratic, and yes, we do need to prove it is a maximum for full marks :)

kb123:

--- Quote from: jamonwindeyer on October 20, 2016, 06:05:52 pm ---Almost! Check your factorisation in Part (iii), not quite correct, super close! We just need the quadratic, and yes, we do need to prove it is a maximum for full marks :)

--- End quote ---

oh! thanks for picking up on that... ooops

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