Question

How do you find equation of tangent line and normal line of curve y=6cosx at (π/3,3)?

Answer 1

Equation of tangent line is `(y-3)=-sqrt3/2(x-pi/3)`

and equation of normal line is `(y-3)=2/sqrt3(x-pi/3)`

Explanation:

As at `x=pi/3`, `y=6cos(pi/3)=6xx1/2=3`, this confirms that the point `(pi/3,3)` lies on curve `y=6cosx`.

Now we should first find the slope of the tangent, as given slope and point, we can get the equation of the tangent from point slope form of the equation. As normal is at a right angle to tangent, its slope will be `-1` divided by slope of tangent.

Slope of tangent at `(pi/3,3)` is given by value of `(dy)/(dx)` at `x=pi/3`.

As `(dy)/(dx)=-sinx`, the slope will be `-sin(pi/3)=-sqrt3/2`. Hence slope of tangent is `-sqrt3/2` and slope of normal is `2/sqrt3`.

Hence, equation of tangent line is `(y-3)=-sqrt3/2(x-pi/3)`

and equation of normal line is `(y-3)=2/sqrt3(x-pi/3)`