Question

What is the equation of the line that is normal to `f(x)= xsqrt( 3x+2)` at `x=1 4`?

Answer 1

`y-28sqrt11=-(2sqrt11)/65(x-14)`

Explanation:

First, find the point of tangency.

`f(14)=14sqrt(42+2)=14sqrt44=28sqrt11`

The function and normal line will pass through the point `(14,28sqrt11)`.

First, find the slope of the tangent line by differentiating the function.

`f(x)=x(3x+2)^(1/2)`

To differentiate, we will have to use the product rule.

`f'(x)=(3x+2)^(1/2)d/dx(x)+xd/dx(3x+2)^(1/2)`

The two derivatives present are:

`d/dx(x)=1`

Via the chain rule:

`d/dx(3x+2)^(1/2)=1/2(3x+2)^(-1/2)d/dx(3x+2)`

`=3/2(3x+2)^(-1/2)`

Plugging these both back in to find `f'(x)`:

`f'(x)=(3x+2)^(1/2)*1+x*3/2(3x+2)^(-1/2)`

`f'(x)=sqrt(3x+2)+(3x)/(2sqrt(3x+2))`

Getting a common denominator:

`f'(x)=(2(3x+2))/(2sqrt(3x+2))+(3x)/(2sqrt(3x+2))`

`f'(x)=(9x+4)/(2sqrt(3x+2))`

The slope of the tangent line is

`f'(14)=(126+4)/(2sqrt(42+2))=130/(2sqrt44)=65/(2sqrt11)`

The slope of the normal line is the opposite reciprocal of the slope of the tangent line, since the two lines are perpendicular.

The opposite reciprocal of `65/(2sqrt11)` is `-(2sqrt11)/65`.

The equation of a line with slope `-(2sqrt11)/65` passing through `(14,28sqrt11)` is:

`y-28sqrt11=-(2sqrt11)/65(x-14)`

Graphed are the function and the normal line:

graph{(y-28sqrt11+(2sqrt11)/65(x-14))(y-xsqrt(3x+2))=0 [-72.4, 227.8, -13.4, 136.7]}