Parametric Equations and Curves — Question 7

Question 7 – Solution

Solution

1. Factor the coordinate equations: $$x=t^2-1=(t-1)(t+1), \qquad y=t^3-t=t(t-1)(t+1).$$

2. Both coordinates equal zero when $t=1$ and when $t=-1$: $$\mathbf r(1)=(0,0), \qquad \mathbf r(-1)=(0,0).$$ Because two distinct parameter values produce the same point, the self-intersection is $(0,0)$.

3. Differentiate both coordinates: $$\frac{dx}{dt}=2t, \qquad \frac{dy}{dt}=3t^2-1.$$ Since $dx/dt\ne0$ at $t=\pm1$, the tangent slope is $$\frac{dy}{dx}=\frac{3t^2-1}{2t}.$$

4. At $t=1$, $$\frac{dy}{dx}=\frac{3(1)^2-1}{2(1)}=1,$$ so one tangent line is $y=x$.

5. At $t=-1$, $$\frac{dy}{dx}=\frac{3(-1)^2-1}{2(-1)}=-1,$$ so the other tangent line is $y=-x$.

6. Therefore, the curve crosses itself at the origin with tangent directions of slopes $$\boxed{1\text{ and }-1}.$$