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Mathematics 14 Online
OpenStudy (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 :)

OpenStudy (anonymous):

dy/dx=6x^2 +12x^(1/2)

OpenStudy (anonymous):

integrate both sides

OpenStudy (anonymous):

y = 2x^3 +12x^(3/2) / (3/2) +C\[y=2x^3 +8x^{3/2} +C\]

OpenStudy (anonymous):

thats as far as i got :S

OpenStudy (anonymous):

when x=4,y=10 sub in to find the constant

OpenStudy (anonymous):

constant?

OpenStudy (anonymous):

\[x^{3/2} = \sqrt{x^3}\]

OpenStudy (anonymous):

so 10 = 2(64) +8 ( sqrt(64)) +C

OpenStudy (anonymous):

10 = 3(64) +C C= 10 -3(64) then use a calculator to find that value

OpenStudy (anonymous):

thanks you i understand and worked out the answer whoop whoop :) its -182 right

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