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 :)

Answer 1

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

Answer 2

integrate both sides

Answer 3

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

Answer 4

thats as far as i got :S

Answer 5

when x=4,y=10

sub in to find the constant

Answer 6

constant?

Answer 7

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

Answer 8

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

Answer 9

10 = 3(64) +C
C= 10 -3(64)

then use a calculator to find that value

Answer 10

thanks you i understand and worked out the answer
whoop whoop :)

its -182 right