a curve in the plane is defined parametrically by the equations x = √(1 - 3t) and y = 3/t. find the value of…

Explanation:

Step1: Differentiate x with respect to t

Using the chain - rule, if $x=\sqrt{1 - 3t}=(1 - 3t)^{\frac{1}{2}}$, then $\frac{dx}{dt}=\frac{1}{2}(1 - 3t)^{-\frac{1}{2}}\times(-3)=-\frac{3}{2\sqrt{1 - 3t}}$.

Step2: Differentiate y with respect to t

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

Step3: Use the chain - rule for parametric differentiation

The formula for $\frac{dy}{dx}$ in parametric form is $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. So, $\frac{dy}{dx}=\frac{-\frac{3}{t^{2}}}{-\frac{3}{2\sqrt{1 - 3t}}}=\frac{2\sqrt{1 - 3t}}{t^{2}}$.

Step4: Substitute t = - 1 into $\frac{dy}{dx}$

When $t=-1$, we have $\frac{dy}{dx}=\frac{2\sqrt{1-3\times(-1)}}{(-1)^{2}}=\frac{2\sqrt{4}}{1}=\frac{2\times2}{1}=4$.

Answer:

A. 4