a curve in the plane is defined parametrically by the equations x = ln(1 - 2t) and y = 1 / 3t. find dy/dx…

Explanation:

Step1: Differentiate $x$ with respect to $t$

Using the chain - rule, if $x=\ln(1 - 2t)$, then $\frac{dx}{dt}=\frac{-2}{1 - 2t}$.

Step2: Differentiate $y$ with respect to $t$

If $y = \frac{1}{3t}=\frac{1}{3}t^{-1}$, then $\frac{dy}{dt}=-\frac{1}{3t^{2}}$.

Step3: Use the formula $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$\frac{dy}{dx}=\frac{-\frac{1}{3t^{2}}}{\frac{-2}{1 - 2t}}=\frac{1 - 2t}{6t^{2}}$.

Answer:

D. $\frac{1 - 2t}{6t^{2}}$