Question

What is the equation of the line that is normal to `f(x)= -xsin^2x` at `x=(4pi)/3`?

Answer 1

`y+pi=(4sqrt3)/(3sqrt3+16pi)(x-(4pi)/3)`

Explanation:

The line intersects the function at:

`f((4pi)/3)=-(4pi)/3sin^2((4pi)/3)=-(4pi)/3(-sqrt3/2)^2=-pi`

To find the slope of the normal line, first find the slope of the tangent line by differentiating the function. We'll need first the product rule, then the chain rule.

`f'(x)=-(d/dxx)sin^2x-x(d/dxsin^2x)`

`color(white)(f'(x))=-sin^2x-x(2sinx)(d/dxsinx)`

`color(white)(f'(x))=-sin^2x-2xsinxcosx`

So the slope of the tangent line is:

`f'((4pi)/3)=-sin^2((4pi)/3)-2((4pi)/3)sin((4pi)/3)cos((4pi)/3)`

`color(white)(f'((4pi)/3))=-(-sqrt3/2)^2-(8pi)/3(-sqrt3/2)(-1/2)`

`color(white)(f'((4pi)/3))=-3/4-(4pi)/sqrt3=-(3sqrt3+16pi)/(4sqrt3)`

The normal line is perpendicular to the tangent line, so their slopes will be opposite reciprocals. Thus, the slope of the normal line is:

`(4sqrt3)/(3sqrt3+16pi)`

Since we know the slope of the normal line and a point that it passes through, `((4pi)/3,-pi)`, we can write the line's equation:

`y-y_0=m(x-x_0)`

`color(blue)(y+pi=(4sqrt3)/(3sqrt3+16pi)(x-(4pi)/3)`