Question

Consider the tangent line to the curve y = 6 sin(x) at the point (π/6, 3). (a) Find a unit vector that is parallel to the tangent line to the curve at the given point.

Answer 1

`y=(\pi )/(6)+5.196(x-3)`

Explanation:

We have given the equation y = 6 sin (x)

On differentiating both side
`(dy)/(dx)=m=6cosx`

As it passes through the point
`((\pi )/(6),3)`

So
`(dy)/(dx)=6cos(\pi )/(6)=5.196`

Now the unit vector is parallel to the tangent so m will be 5.196

This passes through the point
`((\pi )/(6),3)`

So unit vector will be
`y-(\pi )/(6)=5.196(x-3)`

`y=(\pi )/(6)+5.196(x-3)`