Question

If $y = f(x)$ defined parametrically by $x = 2t - |t - 1|$ and $y = 2t^2 + t |t|$, then

Answer 1

$$\frac{dy}{dx} = \begin{cases} \frac{2(x+1)}{9} & \text{if } x < -1 \\ \frac{2(x+1)}{3} & \text{if } -1 \le x < 2 \\ 6(x-1) & \text{if } x \ge 2 \end{cases}$$

Answer 2

To find $\frac{dy}{dx}$, we analyze the parametric equations by considering cases for $t$ due to the absolute value functions.

Case 1: $t \ge 1$ $x = 2t - (t - 1) = t + 1 \implies t = x - 1$ $y = 2t^2 + t(t) = 3t^2$ $\frac{dx}{dt} = 1$, $\frac{dy}{dt} = 6t$ $\frac{dy}{dx} = \frac{6t}{1} = 6t = 6(x - 1)$ for $x \ge 2$.

Case 2: $0 \le t < 1$ $x = 2t - (1 - t) = 3t - 1 \implies t = \frac{x + 1}{3}$ $y = 2t^2 + t(t) = 3t^2$ $\frac{dx}{dt} = 3$, $\frac{dy}{dt} = 6t$ $\frac{dy}{dx} = \frac{6t}{3} = 2t = 2\left(\frac{x + 1}{3}\right) = \frac{2(x + 1)}{3}$ for $-1 \le x < 2$.

Case 3: $t < 0$ $x = 2t - (1 - t) = 3t - 1 \implies t = \frac{x + 1}{3}$ $y = 2t^2 + t(-t) = t^2$ $\frac{dx}{dt} = 3$, $\frac{dy}{dt} = 2t$ $\frac{dy}{dx} = \frac{2t}{3} = \frac{2}{3}\left(\frac{x + 1}{3}\right) = \frac{2(x + 1)}{9}$ for $x < -1$.

Combining these cases gives the piecewise derivative.