Write the equations of the tangent and normal

Write the equations of the tangent and normal to the curve $x=2t-t^{2},\ y=3t-t^{3}$ at the points $t=0$ and $t=1$.

Step 1: Parametric derivatives

\[ \dot x = 2-2t, \qquad \dot y = 3-3t^{2}, \qquad \frac{dy}{dx} = \frac{3(1-t)(1+t)}{2(1-t)}. \]

For $t\neq1$ this simplifies to

\[ \frac{dy}{dx} = \frac{3(1+t)}{2}. \tag{1} \]

Step 2: At $t=0$

Point $M_{0}=(0,0)$; slope from (1): $k_{t}=\dfrac{3}{2}$, $k_{n}=-\dfrac23$.

\[ \text{Tangent: } y=\frac32x; \qquad \text{Normal: } y=-\frac23x. \tag{2} \]

Step 3: At $t=1$

Point $M_{1}=(1,2)$. Here $\dot x(1)=\dot y(1)=0$, so (1) was obtained by cancelling the common factor $(1-t)$; the true slope is the limit

\[ \lim_{t\to1}\frac{dy}{dx} = \lim_{t\to1}\frac{3(1+t)}{2} = 3, \]

so $k_{t}=3$, $k_{n}=-\dfrac13$.

\[ \text{Tangent: } y-2=3(x-1)\Rightarrow y=3x-1; \qquad \text{Normal: } y-2=-\frac13(x-1)\Rightarrow y=-\frac13x+\frac73. \tag{3} \]

Step 4: Verification

\[\text{Tangent}\cdot\text{normal slopes: } \tfrac32\cdot\!\left(-\tfrac23\right)=-1\ \checkmark; \qquad 3\cdot\!\left(-\tfrac13\right)=-1\ \checkmark\]

\[\text{Both lines pass through their points: direct substitution}\ \checkmark\]