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

Answer:

Step-by-step 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.