ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: tcg93 on October 22, 2010, 03:30:33 pm
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This is from insight 2010 E2
How do you get the (h-25) part it in the first line of the solutions for part ii?
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The radius at any point is that horizontal line which you derive from pythagerous' theorem. You know the hypotenuse is 25, but that vertical line is something minus h.
that something if you think carefully is also the initial radius of 25.
So therefore the radius at anypoint is:
square root (625- (25-h)^2)
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This question did my head in for 20 minutes...