a curve in the plane is defined parametrically by the equations x = 7ln(t) and y = √(1 - 4t). find dy/dx…

Answer:

A. $-\frac{2t}{7\sqrt{1 - 4t}}$

Explanation:

Step1: Find $\frac{dx}{dt}$

Differentiate $x = 7\ln(t)$ with respect to $t$. Using the formula $\frac{d}{dt}\ln(t)=\frac{1}{t}$, we get $\frac{dx}{dt}=\frac{7}{t}$.

Step2: Find $\frac{dy}{dt}$

Differentiate $y=\sqrt{1 - 4t}=(1 - 4t)^{\frac{1}{2}}$ with respect to $t$. Using the chain - rule $\frac{d}{dt}u^n=nu^{n - 1}\frac{du}{dt}$, where $u = 1-4t$ and $n=\frac{1}{2}$. So $\frac{dy}{dt}=\frac{1}{2}(1 - 4t)^{-\frac{1}{2}}\times(-4)=-\frac{2}{\sqrt{1 - 4t}}$.

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

Substitute $\frac{dy}{dt}$ and $\frac{dx}{dt}$ into the formula: $\frac{dy}{dx}=\frac{-\frac{2}{\sqrt{1 - 4t}}}{\frac{7}{t}}=-\frac{2t}{7\sqrt{1 - 4t}}$.