Tangents with Parametric Equations — Question 3

Question 3 – Solution

Solution

1. Differentiate both coordinates: $$\frac{dx}{dt}=2t, \qquad \frac{dy}{dt}=3t^2-3=3(t-1)(t+1).$$

2. A tangent has slope $0$ when $dy/dt=0$ and $dx/dt\ne0$. Solve $$3(t-1)(t+1)=0,$$ which gives $t=-1$ and $t=1$.

3. Check the horizontal velocity: $$x'(-1)=-2\ne0, \qquad x'(1)=2\ne0.$$ Thus both parameter values give genuine horizontal tangents.

4. Find the corresponding points: $$\mathbf r(-1)=(1,2), \qquad \mathbf r(1)=(1,-2).$$ The horizontal tangent lines are therefore $y=2$ and $y=-2$.

5. A particle is momentarily stopped only when both velocity components are zero. At $t=\pm1$, $dy/dt=0$ but $dx/dt=\pm2$ is nonzero, so the particle is moving horizontally rather than stopping.

6. Hence the requested points are $$\boxed{(1,2)\text{ and }(1,-2)},$$ and the particle is not momentarily stopped at either point.