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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: sxcalexc on November 06, 2008, 08:40:55 pm

Title: Last Minute Q
Post by: sxcalexc on November 06, 2008, 08:40:55 pm: In regards to last question on the 2007 exam 1. I understand how to use the distance between points formula, but not how they worked it out from the square root. For the first solution what is d(l^2)/dx ? How can you differentiate like that? For their second solution, how did they arrive at the minimum so easily (without chain rule differentiation). Thanks guys
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Title: Re: Last Minute Q
Post by: /0 on November 06, 2008, 08:47:15 pm: Not so sure about the first one, but the second one uses the fact that the turning point of a parabola is halfway between its x-intercepts.

$l = \sqrt{5x^2-40x+100}$

Since the $x^2$ coefficient is positive, the turning point will give the minimum value of $l$ . By taking the average of x-intercepts $-2$ and $10$ , the x-coordinate of this turning point is found.
Title: Re: Last Minute Q
Post by: sxcalexc on November 06, 2008, 08:49:36 pm: Quote from: DivideBy0 on November 06, 2008, 08:47:15 pm
Not so sure about the first one, but the second one uses the fact that the turning point of a parabola is halfway between its x-intercepts.

$l = \sqrt{5x^2-40x+100}$

Since the $x^2$ coefficient is positive, the turning point will give the minimum value of $l$ . By taking the average of x-intercepts $-2$ and $10$ , the x-coordinate of this turning point is found.

Ahh nice one, cheers /0!
Title: Re: Last Minute Q
Post by: shinny on November 06, 2008, 08:50:03 pm: For the first one, since you're trying to find the maximum of $l$ , it's easier to just square it and then differentiate to find the maximum of $l^{2}$ as the maximum's of both functions are equal for their x-values, as squaring the largest number in a function will lead to it being the largest number in the next as well (as long as this number is greater than 1 of course). And yeh, only reason you do this is just to make it easier to differentiate, since having a polynomial/no chain is clearly better :]
Title: Re: Last Minute Q
Post by: sxcalexc on November 06, 2008, 08:52:53 pm: Quote from: shinjitsuzx on November 06, 2008, 08:50:03 pm
For the first one, since you're trying to find the maximum of $l$ , it's easier to just square it and then differentiate to find the maximum of $l^{2}$ as the maximum's of both functions are equal for their x-values, as squaring the largest number in a function will lead to it being the largest number in the next as well (as long as this number is greater than 1 of course). And yeh, only reason you do this is just to make it easier to differentiate, since having a polynomial/no chain is clearly better :]
Ah.. I think I get it.. , so the x value remains the same but not the y?
Title: Re: Last Minute Q
Post by: vce01 on November 06, 2008, 08:53:40 pm: ughh i did not get that question at all, can someone explain it? :(
Title: Re: Last Minute Q
Post by: shinny on November 06, 2008, 08:54:20 pm: Yep pretty much, but you're only interested in the x value to know when $l$ is at a maximum anyway so it's all good.
Title: Re: Last Minute Q
Post by: onlyfknhuman on November 07, 2008, 08:05:38 am: lol that is so confusing i rather do it the old way.