Answer to Question #271914 in Calculus for Angel Nodado

Question

Find the slope of the curve and the equation of tangent line of the parametric equation to the given point.

x=ln t , y = t ^ - 1 , when t = 2

Answer 1

Given that $x=ln(t),y=t^{-1}$

$\therefore\frac{dx}{dt}=\frac{1}{t}, \frac{dy}{dt}=-\frac{1}{t^2}$

So, $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\$

$\Rightarrow \frac{dy}{dx}=-\frac{1}{t}$

At $t=2, \frac{dy}{dx}=-\frac{1}{2}$

The slope of curve at $t=2$ is $-\frac{1}{2}$.

At $t=2, x= ln(2),y=\frac{1}{2}$

We can write:

$t=e^x,t=\frac{1}{y}$

Therefore, equation of curve is: $y=e^{-x}$

Equation of tangent at $t=2$ is:

$y-\frac{1}{2}=\frac{1}{2}(x-ln(2))$

$\therefore y=\frac{x}{2}+\frac{1-ln(2)}{2}$