m2 integration

To Hermen Tam: Normal is perpendicular to the tangent but you have found the equation of the tangent- -

i) dy/dx = 6 / √(4x+3)
For P(3,3)
dy/dx = 6 / √(4*3+3) = 6 / √15
∵Normal is perpendicular to the tangent at P(3,3)
Let m be the slope of the normal
m * 6 / √15 = -1
m = -√15 / 6
∴The equation of the normal is
(y-3)/(x-3) = -√15 / 6
6y-18= -√15 x + 3√15
√15 x + 6y = 3√15 + 18
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ii) dy/dx = 6 / √(4x+3)
y=∫ 6/ √(4x+3) dx
=6 ∫ (4x+3)^(-1/2) dx
=(6/4) ∫ (4x+3)^(-1/2) d(4x+3)
=(3/2) [(4x+3)^(1/2) / (1/2)] + C
=3(4x+3)^(1/2) + C
=3√ (4x+3) +C
Put (3,3) into y=3√(4x+3) + C
3=3√(4*3+3) + C
C=3-3√15
∴The equation of the curve is
y= 3√ (4x+3) + 3-3√15

a curve is such that dy/dx = 6/ (4x+3)^1/2 and P(3,3) is a point on the curve.

i) find the equation of the normal to the curve at P , in the form of ax+by=c
dy/dx at (3,3) = 6/ (4*3+3)^1/2
=6/ sqrt(15)
the equation of the normal
(y-3) / (x-3) = 6/ sqrt(15)
y-3 = 6/sqrt(15) *x - 18/sqrt(15)
y-6/sqrt(15) *x=3-18/sqrt(15)
sqrt(15)*y-6x = 3*sqrt(15)-18

ii) find the equation of the curve

dy/dx = 6/ (4x+3)^1/2
dy = 6/ (4x+3)^1/2 dx
integrate both sides:∫dy = ∫6/ (4x+3)^1/2 dx
y= 6∫dx/ (4x+3)^1/2
let k=4x+3
dk/dx=4
dx=dk/4
y=6∫dk/4/ (k)^1/2
y=6/4 ∫k^(-1/2)dk
=3/2 k^(1/2)/(1/2) +C
=3/2 *2 k^(1/2) +C
=3k^(1/2)+C
=3(4x+3)^(1/2)+C
Put (3,3) into the equation to find C
3=3sqrt(15)+C
C=3-3sqrt(15)
The equation of the curve is:
y=3(4x+3)^(1/2)+3-3sqrt(15)