Parametric Tangents: Horizontal vs Vertical Tangents in a Trig-Based Parametric Curve

Horizontal and Vertical Tangents in a Parametric Curve

• Topic focus: horizontal tangency, with a brief discussion of vertical tangency, in the context of a parametric curve where the position is given in terms of a parameter (theta or t).
• Core definitions:
  • For a parametric curve x = x(t), y = y(t):
  • Horizontal tangent occurs when \frac{dy}{dx} = 0, with \frac{dx}{dt} \neq 0. Since \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}, this means dy/dt = 0 while dx/dt ≠ 0.
  • Vertical tangent occurs when \frac{dx}{dx} = \infty, i.e., \frac{dx}{dt} = 0 with \frac{dy}{dt} \neq 0. (Transcript notes: there is a question about the possibility of a vertical tangent; the class concludes no vertical tangent is possible in this example.)
• Transcript point: there is no t for which dx/dt = 0 while dy/dt ≠ 0 in the discussed example, so vertical tangents do not occur here.
• Context on the parameter domain:
  • When trig functions are involved, the natural domain to consider is one full trig cycle, typically from 0 to 2π (a single rotation).
  • There could be more than one horizontal tangent in that interval due to periodicity of trig functions.

Question 38: Can I have a horizontal tangency?

• The instructor frames the problem around Question 38, emphasizing trig behavior (0 to 2π) and the possibility of multiple horizontal tangents within one cycle.
• Two candidate forms for dy/dt (as discussed by students):
  • Option A: \frac{dy}{dt} = 4 \cos(2\theta)
  • Option B: \frac{dy}{dt} = 4 \sin^{2}\theta
• For dx/dt, a common factor mentioned is \frac{dx}{dt} = -\sin\theta (or a form with a common factor involving sine). This is the derivative with respect to theta (t in the discussion).
• Teachers’ approach: They highlighted the two expressions for dy/dt and used dx/dt = - sin(\theta) as the common thread to test horizontal tangency.
• Decision about which form to use: The group discussed both representations (one in terms of 2θ, the other in terms of θ); choosing dy/dt as either form is a matter of representation, and the chosen form will influence where dy/dt = 0 occurs.
• Specific plan mentioned: After a break, the plan was to continue with the sine-squared form as a tool for the analysis, though not claiming it is better or worse overall.
• Practical note about the class discussion:
  • Private chat responses were used to gather inputs from students (e.g., two answers for dy/dt were shared via chat):
  • One answer: \frac{dy}{dt} = 4 \cos(2\theta)
  • Another answer (written in a single theta form): presumably \frac{dy}{dt} = 4 \sin^{2}\theta (transcript suggests a variant like this; students offered different representations).
  • The instructor highlighted these options and asked for agreement or disagreement on whether changes were necessary.
• Key takeaway from the dx/dt discussion:
  • The recurring theme for dx/dt is the sine function with a negative sign: \frac{dx}{dt} = -\sin\theta. This is essential when checking the condition dx/dt ≠ 0 for horizontal tangents.
• Interim conclusion before break:
  • The instructor chose to pursue the sine-squared option as a computational tool after the break, with the plan to revisit the horizontal tangency calculation.
• Break and return: The instructor indicated breaks (start at roughly 02:02, return at 02:07) and resumed after the break; students were asked to come back with results from their chat discussions.

Analyzing horizontal tangency with the two candidate dy/dt forms

• General check for horizontal tangency: Solve dy/dt = 0 while ensuring dx/dt ≠ 0.

• Case 1: If \frac{dy}{dt} = 4 \cos(2\theta)

  • Set dy/dt = 0: \cos(2\theta) = 0. Solutions: 2\theta = \frac{\pi}{2} + k\pi \quad\Rightarrow\quad \theta = \frac{\pi}{4} + \frac{k\pi}{2}.
  • Within one full cycle 0 ≤ θ < 2π, this yields: \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}.
  • Check dx/dt: \frac{dx}{dt} = -\sin\theta. At these θ values, sin(θ) ≠ 0, so dx/dt ≠ 0. Hence horizontal tangents occur at these four θ values within one 0 to 2π cycle.
  • Implication: There can be four horizontal tangents in one trig cycle for this form of dy/dt.

• Case 2: If \frac{dy}{dt} = 4 \sin^{2}\theta

  • Set dy/dt = 0: \sin^{2}\theta = 0 \quad\Rightarrow\quad \sin\theta = 0.
  • Solutions: \theta = k\pi \quad\Rightarrow\quad \theta = 0, \pi, (2\pi).
  • Check dx/dt: \frac{dx}{dt} = -\sin\theta. At θ = 0 or π, sin(θ) = 0, so dx/dt = 0.
  • Since dx/dt must be nonzero for a horizontal tangent, these points do not yield horizontal tangents.
  • Note: Endpoints like θ = 2π may coincide with θ = 0 (same point on the curve), so in a 0 ≤ θ < 2π interval you typically count distinct θ values that yield valid tangents.
  • Conclusion for this case: No horizontal tangents arise from this form within the interval due to dx/dt = 0 at the dy/dt zeros.

Summary of implications for Question 38

• Depending on which expression you use for dy/dt, you get different results for horizontal tangency:
  • If dy/dt = 4 cos(2θ): there are four horizontal tangents in 0 ≤ θ < 2π.
  • If dy/dt = 4 sin^{2}θ: no horizontal tangents in 0 ≤ θ < 2π (because dx/dt = 0 at the zeros of dy/dt).
• The discussion emphasizes that trig problems can yield multiple tangents within one period, due to periodicity.
• The approach used in class:
  • Start with dy/dx = (dy/dt) / (dx/dt).
  • Identify candidate t-values by setting dy/dt = 0, then verify dx/dt ≠ 0.
  • Consider the domain 0 ≤ t ≤ 2π to account for one full cycle of trig functions and count all distinct tangents in that period.
• Practical considerations for exams or problem sets:
  • When faced with dy/dt expressions that can be written in multiple trigonometric forms (e.g., using 2θ or θ), you may choose the form that simplifies finding zeros of dy/dt, but remember to verify dx/dt ≠ 0 for the resulting t-values.
  • Be mindful of endpoints that may correspond to the same point on the curve (e.g., θ = 0 and θ = 2π).

Connections to the broader approach and next steps

• The core method is grounded in the chain rule: \frac{dy}{dx} = \frac{dy/dt}{dx/dt}. This is the standard way to analyze tangents to a parametric curve.
• The discussion connects to prior lectures on horizontal vs vertical tangents and the interpretation of when tangents exist (dx/dt and dy/dt conditions).
• Real-world relevance: Understanding where a parametric curve has horizontal or vertical tangents helps in sketching curves, optimizing paths, and understanding geometry of parametric shapes in physics and engineering.

Closing notes from the session

• The instructor acknowledged the class's effort in contributing responses via chat and used those to guide the discussion before the break.
• The plan was to continue after the break with the sine-squared approach as a tool, then revisit the results and consolidate understanding of horizontal tangency for the problem.
• Time cue from the lecture: the instructor noted a break around 02:02 and resumed at 02:07, indicating an extended interactive session with student input before continuing.