a curve in the plane is defined parametrically by the equations x = -2 ln(t) and y = 2t^2 + t. find the…

Explanation:

Step1: Differentiate x with respect to t

Using the derivative formula for $\ln(t)$ ($\frac{d}{dt}\ln(t)=\frac{1}{t}$), we have $\frac{dx}{dt}=-\frac{2}{t}$.

Step2: Differentiate y with respect to t

Using the power - rule $\frac{d}{dt}(at^{n})=nat^{n - 1}$, $\frac{dy}{dt}=4t + 1$.

Step3: Use the chain - rule for parametric differentiation

The chain - rule for parametric differentiation is $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. So $\frac{dy}{dx}=\frac{4t + 1}{-\frac{2}{t}}=-\frac{t(4t + 1)}{2}$.

Step4: Substitute t = 2 into the derivative

When $t = 2$, $\frac{dy}{dx}=-\frac{2\times(4\times2 + 1)}{2}=-(8 + 1)=-9$.

Answer:

C. $-9$