Question

The parametric equation of a curve are `x=t+e^-t`, `y=1-e^-t`, where t takes all real values. How do you find dy/dx in terms of t, and hence find the value of t, for which the gradient of the curve is 1, giving your answer in logarithmic form?

Answer 1

`t = ln 2`

Explanation:

`dy/dx = (dy/dt) / (dx/dt) = (e^-t)/(1-e^-t)`

`dy/dx = 1 implies 1-e^-t = e^-t`

`e^t = 2`

`t = ln 2`