The parametric equations of a curve are x=t-e^-t and y=t-e^t.

Question

find dy/dx in terms of t and hence find the crdinates of the stationary point?

Answer 1

The curves are  x = t - e^ -t  and y = t -e^t

dx /dt = 1 - (- e^ - t) = 1 + e^ ( -t )

dy /dt = 1 - e^(t)

Now dy/dx = (dy /dt) (dt /dx)

dt / dx = 1/ (1 + e^(-t))

dy / dx =  (1 - e ^ t)(1/1 + e^(-t))

dy /dx = (1 - e^ t)/(1 +e^(-t))

Now  make dy / dx= 0

(1 - e^ t)/(1 +e^(-t)) = 0

(1 - e ^ t) = 0

e ^ t =1

apply log on both sides 

t log e = log 1

t = 0.

Hence the value of t = 0.

Now solving for the stationary points:

x = t - e^ -t  = 0 - e^(-0) =  - 1

y = t -e^t = 0- e^(0)= 0 - 1 =  - 1.

Hence the stationary point  is  (- 1, 1) .