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 :)

Question

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 :)

anonymous · Answer

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

anonymous · Answer

integrate both sides

anonymous · Answer

y = 2x^3 +12x^(3/2) / (3/2) +C$$y=2x^3 +8x^{3/2} +C$$

anonymous · Answer

thats as far as i got :S

anonymous · Answer

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

anonymous · Answer

constant?

anonymous · Answer

$$x^{3/2} = \sqrt{x^3}$$

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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