anonymous
  • anonymous
can someone help me with qs 6 please??? http://www.ocr.org.uk/download/pp_10_jun/ocr_57857_pp_10_jun_gce_475201.pdf its finding the equation of the curve thanks :)
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
dy/dx=6x^2 +12x^(1/2)
anonymous
  • anonymous
integrate both sides
anonymous
  • anonymous
y = 2x^3 +12x^(3/2) / (3/2) +C\[y=2x^3 +8x^{3/2} +C\]

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anonymous
  • anonymous
thats as far as i got :S
anonymous
  • anonymous
when x=4,y=10 sub in to find the constant
anonymous
  • anonymous
constant?
anonymous
  • anonymous
\[x^{3/2} = \sqrt{x^3}\]
anonymous
  • anonymous
so 10 = 2(64) +8 ( sqrt(64)) +C
anonymous
  • anonymous
10 = 3(64) +C C= 10 -3(64) then use a calculator to find that value
anonymous
  • anonymous
thanks you i understand and worked out the answer whoop whoop :) its -182 right

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