Table of Contents

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Recall that if we have a parametric curve in the form $x = f(t)$ and $y = g(t)$ and we want find a derivative for this curve, dy/dx, then it follows that:
(1)Horizontal Tangents
Recall that with a function f(x), horizontal tangents arise when f'(x) = 0, that is, the numerator of f'(x) is 0. This is similar for parametric curves.
For a parametric curve with a derivative defined by:
(2)If $\frac{dy}{dt} = 0 \quad \mathbf{and} \quad \frac{dx}{dt} ≠ 0$, then a horizontal tangent exists for the specific value of t that satisfies this condition.
Example 1
Find any horizontal tangents for the parametric curve defined by $x = 2t$ and $y = t^2$.
This is a rather simple example, but let's first differentiate to obtain dx/dt, and dy/dt:
(3)Now let's set dy/dt = 0. We thus get $2t = 0$, so t = 0. When t = 0, dx/dt ≠ 0 since dx/dt is just a constant. Hence there exists a horizontal tangent at t = 0. We can plug this into our parametric equations to get the coordinates (0, 0) where this occurs.
Vertical Tangents
When we have a function f(x), there were no such things as vertical tangents (although there was vertical asymptotes). Nevertheless, parametric curves are not necessarily functions, and vertical tangents can exist. If $\frac{dy}{dx} ≠ 0 \quad \mathbf{and} \quad \frac{dx}{dt} = 0$, then a vertical tangent exists for the specific value of t that satisfies this condition.
Example 2
Find any vertical tangents for the parametric curve defined by $x = \cos \theta$ and $y = \sin \theta$.
Finding dx/dθ and dy/dθ, we obtain:
(4)Hence dx/dθ = 0 when θ = 0 and π. Also note that dy/dθ ≠ 0 when θ = 0 and when θ = π, hence there exists two vertical tangents when θ = 0 and when θ = π. We can plug these into our parametric equations to get the xycoordinates (1, 0) and (1, 0)
Exceptions
So what are we supposed to do if when dy/dt = 0, dx/dt = 0, or if when dx/dt = 0, dy/dt = 0? Well we have to use limits to determine what happens at these points.
Example 3
Find any vertical tangents for the parametric curve defined by $x = r(\theta  \sin \theta)$ and $y = r(1  \cos \theta)$.
First let's differentiate dx/dθ and dy/dθ to obtain dy/dx.
(5)Therefore:
(6)So dx/dθ = 0 when θ = 0, π, 2π, … But notice that when θ = 0, π, 2π, …, then dy/dθ = 0, since sin(0) = 0, sin(π) = 0, …. So we need to use our knowledge of limits in order to figure out what is happening here. Let's just use 2π as an example,
(7)We can use L'Hospital's Rule to evaluate this limit further, first from the right side:
(8)Now let's look at the limit from the left:
(9)Since 2π exists as a point on our curve, it thus follows that when theta is 2π, a vertical tangent exists.