Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.


e. There are two points on the curve x=−4 cos t, y=sin t, for 0≤t≤2π, at which there is a vertical tangent line.

Verified step by step guidance

1. First, recall that a vertical tangent line occurs where the derivative \( \frac{dy}{dx} \) is undefined or infinite. This happens when \( \frac{dx}{dt} = 0 \) but \( \frac{dy}{dt} \neq 0 \).
2. Given the parametric equations \( x = -4 \cos t \) and \( y = \sin t \), compute the derivatives with respect to \( t \):
3. \[ \frac{dx}{dt} = -4 (-\sin t) = 4 \sin t \]
4. \[ \frac{dy}{dt} = \cos t \]
5. Set \( \frac{dx}{dt} = 0 \) to find potential points of vertical tangents:
6. \[ 4 \sin t = 0 \implies \sin t = 0 \]
7. Solve for \( t \) in the interval \( 0 \leq t \leq 2\pi \): \( t = 0, \pi, 2\pi \). For each \( t \), check if \( \frac{dy}{dt} \neq 0 \) to confirm vertical tangents:
8. \[ \frac{dy}{dt} = \cos t \]
9. Evaluate \( \cos t \) at these points:
10. - At \( t=0 \), \( \cos 0 = 1 \neq 0 \)
11. - At \( t=\pi \), \( \cos \pi = -1 \neq 0 \)
12. - At \( t=2\pi \), \( \cos 2\pi = 1 \neq 0 \)
13. Since \( \frac{dy}{dt} \neq 0 \) at all these points, vertical tangents occur at \( t=0, \pi, 2\pi \). However, note that \( t=0 \) and \( t=2\pi \) correspond to the same point on the curve because the parametric equations are periodic with period \( 2\pi \).
14. Therefore, there are exactly two distinct points on the curve where vertical tangent lines occur: one at \( t=0 \) (or \( 2\pi \)) and one at \( t=\pi \).

Key Concepts

Parametric Curves and Their Derivatives

Parametric curves are defined by functions x(t) and y(t). To analyze their behavior, especially tangents, we compute derivatives dx/dt and dy/dt. These derivatives help determine the slope of the tangent line at any point on the curve.

Vertical Tangent Lines in Parametric Form

A vertical tangent line occurs where the slope of the tangent is undefined, which happens when dx/dt = 0 but dy/dt ≠ 0. Identifying such points requires solving dx/dt = 0 and checking the corresponding dy/dt values.

Analyzing the Interval for Parameter t

Since the curve is defined for t in [0, 2π], it is important to find all values of t within this interval that satisfy the conditions for vertical tangents. This ensures the correct number of points with vertical tangents is identified.