Question

How do I find the equation of tangent to the curve at the point corresponding the given value of the parameter. `x=e^sqrtt, y=t-lnt, t=3`?

Answer 1

The given parametric equation of the curve `x=e^sqrtt, y=t-lnt`

So `(dx)/(dt)=e^sqrttxx1/(2sqrtt)`

and

`(dy)/(dt)=1-1/t`

`(dy)/(dx)=(1-1/t)/(e^sqrttxx1/(2sqrtt))`

Slope of the tangent at parametric point `t=3`

`m=[(dy)/(dx)]_(t=3)=(1-1/3)/(e^sqrt3xx1/(2sqrt3))`

`=>m=4/(sqrt3e^sqrt3)`

The coordinates of the point of contact at `t=3` in Cartesian system
are `(e^sqrt3,3-ln3)`

Hence the equation of the tangent at desired point will be

`y-(3-ln3)=m(x-e^sqrt3)`

`=>y-(3-ln3)=4/(sqrt3e^sqrt3)(x-e^sqrt3)`

`=>y=(4x)/(sqrt3e^sqrt3)-4/sqrt3+(3-ln3)`