Question

What is the slope of the line normal to the tangent line of `f(x) = xtanx-xsin(x-pi/3)` at `x= (15pi)/8`?

Answer 1

Slope of the normal is `-0.15`

Explanation:

`f(x)= x tan x- x sin(x-pi/3)` at `x= (15pi)/8~~ 5.89( 2 dp)`

Slope of the tangent is

`f^'(x)= x sec ^2 x + tan x - {x cos(x-pi/3)+ sin(x-pi/3)}`

`f^'(5.89)= 5.89 * sec ^2 5.89 + tan 5.89 - {5.89* cos(5.89-pi/3)+ sin(5.89-pi/3)}` or

`f^'(5.89)= 5.89* sec ^2 5.89 + tan 5.89 - {5.89* cos 4.84+ sin 4.84}` or

`f^'(5.89)~~6.71`, slope of the tangent is `m_t~~ 6.71`

Let slope of the normal be `m_n :. m_t*m_n= -1`

or `m_n= -1/m_t = - 1/6.71~~ -0.15`

Slope of the normal is `-0.15` [Ans]