Respuesta :
Answer:
The solutions listed from the smallest to the greatest are:
x: [tex]-\sqrt{2}[/tex] [tex]-\sqrt{2}[/tex] [tex]\sqrt{2}[/tex] [tex]\sqrt{2}[/tex]
y: -1 1 -1 1
Step-by-step explanation:
The slope of the tangent line at a point of the curve is:
[tex]m = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }[/tex]
[tex]m = -\frac{\cos 2\theta}{\sin \theta}[/tex]
The tangent line is horizontal when [tex]m = 0[/tex]. Then:
[tex]\cos 2\theta = 0[/tex]
[tex]2\theta = \cos^{-1}0[/tex]
[tex]\theta = \frac{1}{2}\cdot \cos^{-1} 0[/tex]
[tex]\theta = \frac{1}{2}\cdot \left(\frac{\pi}{2}+i\cdot \pi \right)[/tex], for all [tex]i \in \mathbb{N}_{O}[/tex]
[tex]\theta = \frac{\pi}{4} + i\cdot \frac{\pi}{2}[/tex], for all [tex]i \in \mathbb{N}_{O}[/tex]
The first four solutions are:
x: [tex]\sqrt{2}[/tex] [tex]-\sqrt{2}[/tex] [tex]-\sqrt{2}[/tex] [tex]\sqrt{2}[/tex]
y: 1 -1 1 -1
The solutions listed from the smallest to the greatest are:
x: [tex]-\sqrt{2}[/tex] [tex]-\sqrt{2}[/tex] [tex]\sqrt{2}[/tex] [tex]\sqrt{2}[/tex]
y: -1 1 -1 1
Using derivatives of the parametric curves and the solving a trigonometric equation, it is found that the points (x,y) are:
[tex]x = 2\cos{\left(\frac{k\pi}{4}\right)}, k = 1, 3, 5, \cdots[/tex]
[tex]y = \sin{\left(\frac{k\pi}{2}\right)}, k = 1, 3, 5, \cdots[/tex]
Given two parametric curves x(t) and y(t), the slope of the tangent line is given by:
[tex]m = \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]
The tangent line is horizontal for:
[tex]\frac{dy}{dt} = 0, \frac{dx}{dt} \neq 0[/tex]
In this problem, since we are working with trigonometric functions sine and cosine, they will never be simultaneously 0, hence, the equation of interest is:
[tex]\frac{dy}{dt} = 0[/tex]
Then:
[tex]y(t) = \sin{2\theta}[/tex]
[tex]\frac{dy}{d\theta} = 2cos{2\theta}[/tex]
Hence:
[tex]2\cos{2\theta} = 0[/tex]
[tex]\cos{2\theta} = 0[/tex]
[tex]\cos{2\theta} = \cos{\left(\frac{k\pi}{2}\right)}, k = 1, 3, 5, \cdots[/tex]
[tex]2\theta = \frac{k\pi}{2}[/tex]
[tex]\theta = \frac{k\pi}{4}, k = 1, 3, 5, \cdots[/tex]
Hence, the points (x,y) are:
[tex]x = 2\cos{\left(\frac{k\pi}{4}\right)}, k = 1, 3, 5, \cdots[/tex]
[tex]y = \sin{\left(\frac{k\pi}{2}\right)}, k = 1, 3, 5, \cdots[/tex]
A similar problem is given at https://brainly.com/question/13442736
