Question

Show that the equation of the tangent to the curve y = xtanx at x=x/4 (2+pi)x-2y=pi^2/4 ? Thank you

Answer 1

Tangent is `(2+pi)x-2y=pi^2/4`

Explanation:

At `x=pi/4`, we have `y=pi/4tan(pi/4)=pi/4*1=pi/4`

Hence we are seeking tangent to curve `y=xtanx` at `(pi/4,pi/4)`

As slope of tangent is given value of first derivative at that point,

let us find derivative of `y=xtanx`

`(dy)/(dx)=1*tanx+x*sec^2x`

and its value at `(pi/4,pi/4)` is `1*tan(pi/4)+pi/4*sec^2(pi/4)`

= `1*1+pi/4^(sqrt2)^2=1+pi/4*2=1+pi/2`

Now equation of a line passing through `(pi/4,pi/4)` and having a slope `1+pi/2` is

`y-pi/4=(1+pi/2)(x-pi/4)` and multiplying each term by `2`

`2ycolor(red)(-pi/2)=(2+pi)xcolor(red)(-pi/2)-pi^2/4`

or `(2+pi)x=2y+pi^2/4`

or `(2+pi)x-2y=pi^2/4`