Question

find the first derivative of the parametric equations (x = 4t^{2}) and (y = 8t) is (\frac{dy}{dx}=\frac{1}{t}). find the second derivative (\frac{d^{2}y}{dx^{2}}). sl = 2 sl = 3

Explanation:

Step1: Recall the formula for the first - derivative of parametric equations

The formula for $\frac{dy}{dx}$ in parametric equations $x = x(t)$ and $y = y(t)$ is $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. Given $x = 4t^{2}$ and $y = 8t$, we have $\frac{dx}{dt}=8t$ and $\frac{dy}{dt}=8$. So, $\frac{dy}{dx}=\frac{8}{8t}=\frac{1}{t}$.

Step2: Recall the formula for the second - derivative of parametric equations

The formula for $\frac{d^{2}y}{dx^{2}}$ is $\frac{d}{dx}(\frac{dy}{dx})=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$. We know that $\frac{dy}{dx}=\frac{1}{t}$, so $\frac{d}{dt}(\frac{dy}{dx})=-\frac{1}{t^{2}}$. Since $\frac{dx}{dt}=8t$.

Step3: Calculate the second - derivative

$\frac{d^{2}y}{dx^{2}}=\frac{-\frac{1}{t^{2}}}{8t}=-\frac{1}{8t^{3}}$.

Answer:

$-\frac{1}{8t^{3}}$