Find an equation of the normal line to the curve y = √x that is parallel to the line 6x + y = 1.?

Answer 1

hint:

gradient of the normal to the curve = -6

--->gradient of the tangent = 1/6

Answer 2

y = sqrt(x)

y' = -0.5/sqrt(x)

6x + y = 1

Gradient -6

Gradient of normal 1/6

1/6 = -0.5/sqrt(x)

x = (6*-0.5)^2

 = 9

Answer 3

The slope of the given line is -6, so you need a point on y = sqrt(x) where the derivative is +1/6.

The derivative is dy/dx = 1/[2*sqrt(x)], and if you set this equal to 1/6, the value of x must be 9.

So the line you're looking for goes through the point (9,3), and its equation is 6x + y = 57.