Tangents with Parametric Equations — Question 8

Question 8 – Solution

Solution

1. A regular vertical tangent would require $$x'(t)=t-2=0 \qquad\text{and}\qquad y'(t)=t^2-4\ne0.$$ The first condition gives the only candidate, $t=2$.

2. At $t=2$, $$(x'(2),y'(2))=(0,0).$$ Thus the particle is momentarily stopped, and the ordinary vertical-tangent test does not apply.

3. For $t\ne2$, factor $y'$ and simplify the slope: $$\frac{dy}{dx} =\frac{t^2-4}{t-2} =\frac{(t-2)(t+2)}{t-2} =t+2.$$ Consequently, $$\lim_{t\to2}\frac{dy}{dx}=4,$$ which is finite rather than infinite. The path has a tangent of slope $4$ at the stationary point, not a vertical tangent.

4. The direction also changes across $t=2$. Just before $2$, both $x'$ and $y'$ are negative, so the particle moves down and left. Just after $2$, both are positive, so it moves up and right. At $t=2$ itself, the velocity is zero and therefore has no instantaneous direction.

5. Hence $$\boxed{\text{the path has no vertical tangent}.}$$ The only candidate is a stationary point where the particle reverses direction along a tangent of slope $4$.