a curve in the plane is defined parametrically by the equations x = 8e^3t and y = cos(4t). find dy/dx…

Explanation:

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

Using the chain - rule, if $x = 8e^{3t}$, then $\frac{dx}{dt}=8\times3e^{3t}=24e^{3t}$.

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

If $y=\cos(4t)$, then $\frac{dy}{dt}=- 4\sin(4t)$ by the chain - rule.

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

The formula for $\frac{dy}{dx}$ when $x = x(t)$ and $y = y(t)$ is $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. Substituting $\frac{dy}{dt}=-4\sin(4t)$ and $\frac{dx}{dt}=24e^{3t}$ into the formula, we get $\frac{dy}{dx}=-\frac{4\sin(4t)}{24e^{3t}}=-\frac{\sin(4t)}{6e^{3t}}$.

Answer:

C. $-\frac{\sin(4t)}{6e^{3t}}$