1. **State the problem:** Given functions $f(x(t)) = t^2 - 3t + 2$ and $y(t) = t^3 - 4t^2 + 1$, find the derivatives $\dot{x}$ and $\dot{y}$ at $t=2$, and determine the nature (local max, min, or inflection) of the point on the curve at $t=2$. 2. **Interpretation:** Since $f$ is given as a function of $x(t)$ but explicitly depends on $t$, assume $x(t) = f(t) = t^2 - 3t + 2$ to find $\dot{x}$ (rate of change of $x$ with respect to $t$). 3. **Find $\dot{x}$:** Differentiate $x(t)$: $$\dot{x} = \frac{d}{dt}(t^2 - 3t + 2) = 2t - 3.$$ Evaluate at $t=2$: $$\dot{x}(2) = 2(2) - 3 = 4 - 3 = 1.$$ 4. **Find $\dot{y}$:** Differentiate $y(t)$: $$\dot{y} = \frac{d}{dt}(t^3 - 4t^2 + 1) = 3t^2 - 8t.$$ Evaluate at $t=2$: $$\dot{y}(2) = 3(2)^2 - 8(2) = 3(4) - 16 = 12 - 16 = -4.$$ 5. **Summarize derivatives:** At $t=2$, $\dot{x} = 1$ and $\dot{y} = -4$. 6. **Determine the curve nature at $t=2$: ** To classify the point, compute second derivatives: $$\ddot{x} = \frac{d}{dt}(\dot{x}) = \frac{d}{dt}(2t - 3) = 2,$$ $$\ddot{y} = \frac{d}{dt}(\dot{y}) = \frac{d}{dt}(3t^2 - 8t) = 6t - 8.$$ At $t=2$, $$\ddot{y}(2) = 6(2) - 8 = 12 - 8 = 4.$$ 7. Compute curvature information to classify the point. The determinant of acceleration vector for parametric curves helps detect curvature: $$D = \dot{x} \ddot{y} - \dot{y} \ddot{x} = (1)(4) - (-4)(2) = 4 + 8 = 12.$$ Since $D > 0$, the curve is concave upward at $t=2$, which indicates a local minimum. **Final answers:** $$\dot{x} = 1, \quad \dot{y} = -4,$$ At $t=2$, the point on the curve is a **Local Minimum**.