# 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?

Jul 20, 2016

$t = \ln 2$

#### Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}} = \frac{{e}^{-} t}{1 - {e}^{-} t}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1 \implies 1 - {e}^{-} t = {e}^{-} t$

${e}^{t} = 2$

$t = \ln 2$