#help - dy/dx
27 messages · Page 1 of 1 (latest)
root x can be written as 1/2
dydx 3root x + 1 = dy/dx 3x^1/2 + 1 = 3/2 x^-1/2
so our gradient
is 3/2 x^-1/2
5x - 10y + 7 = 0
10 y = 5x + 7
y = 1/2x + 7/10
perpendicular gradient to this is -2
= 3/(2 root x)
oh and to find the normal
we do 1/3/(2rootx)
which simplifies to
.- (2 root x / 3)
now - (2root x / 3) = -2x
when u solve it
u should get
x = 9
now to find the y coordinate
slot it into the curve equation
3 root 9 + 1 = 3x3 + 1 = 10
(9, 10)
so the gradient is -2 x
and the point is (9, 10)
y = -2x + c
10 = -18 + c
y = -2x + 28