Tangents with Parametric Equations — Question 1

Question 1 – Solution

Solution

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

2. A horizontal tangent requires $dy/dt=0$ while $dx/dt\ne0$. The candidates are $$3(t-1)(t+1)=0 \quad\Longrightarrow\quad t=-1,1.$$ At $t=-1$, $dx/dt=-4\ne0$, so this value gives a horizontal tangent. The point is $$x(-1)=3, \qquad y(-1)=2.$$ Thus the horizontal tangent line is $$\boxed{y=2}.$$

3. At $t=1$, both derivatives are zero, so the usual horizontal- or vertical-tangent tests are inconclusive. For $t\ne1$, cancel the common factor: $$\frac{dy}{dx} =\frac{3(t-1)(t+1)}{2(t-1)} =\frac32(t+1).$$ Therefore, $$\lim_{t\to1}\frac{dy}{dx}=3.$$ The point at $t=1$ is $(-1,-2)$, and its tangent line is $y+2=3(x+1)$; it is neither horizontal nor vertical.

4. A vertical tangent would require $dx/dt=0$ with $dy/dt\ne0$. The only zero of $dx/dt$ is $t=1$, where $dy/dt$ is also zero and the limiting slope is finite. Hence there are no vertical tangents.

5. The complete answer is $$\boxed{\text{horizontal tangent }y=2\text{ at }(3,2); \qquad\text{no vertical tangents}.}$$