Question

Find an equation of the tangent to the curve at the given point by two methods:

(a) without eliminating the parameter and
(b) by first eliminating the parameter.

x = 1 + ln T, y = t² + 2; (1,3)

Answer 1

Explanation:

a) with parameter

dx/dt = 1/t and dy/dt = 2t

When x =1, we get t = 1

So slope of tangent
`= 2t^2 at t=1 = 2`

Equation of tangent is

`y-3 = 2(x-1)\\y=2x+1`

b) Eliminate parameter

`x=1+ln t\\lnt = x-1\\t = e^(x-1)`

Substitute in y

`y = e^(2x-2) +2`

`y' = e^(2x-2) (2)`

When x=1, we have

y' =2

So same equation as above.