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Arte-miy333 [17]
2 years ago
11

You are given the parametric equations x=2cos(θ),y=sin(2θ). (a) List all of the points (x,y) where the tangent line is horizonta

l. In entering your answer, list the points starting with the smallest value of x. If two or more points share the same value of x, list those points starting with the smallest value of y. If any blanks are unused, type an upper-case "N" in them. Point 1: (x,y)= ( , )
Mathematics
1 answer:
vladimir1956 [14]2 years ago
6 0

Answer:

The solutions listed from the smallest to the greatest are:

x:  -\sqrt{2}   -\sqrt{2}  \sqrt{2}  \sqrt{2}

y:      -1         1     -1     1

Step-by-step explanation:

The slope of the tangent line at a point of the curve is:

m = \frac{\frac{dy}{dt} }{\frac{dx}{dt} }

m = -\frac{\cos 2\theta}{\sin \theta}

The tangent line is horizontal when m = 0. Then:

\cos 2\theta = 0

2\theta = \cos^{-1}0

\theta = \frac{1}{2}\cdot \cos^{-1} 0

\theta = \frac{1}{2}\cdot \left(\frac{\pi}{2}+i\cdot \pi \right), for all i \in \mathbb{N}_{O}

\theta = \frac{\pi}{4} + i\cdot \frac{\pi}{2}, for all i \in \mathbb{N}_{O}

The first four solutions are:

x:   \sqrt{2}   -\sqrt{2}  -\sqrt{2}  \sqrt{2}

y:     1        -1        1     -1

The solutions listed from the smallest to the greatest are:

x:  -\sqrt{2}   -\sqrt{2}  \sqrt{2}  \sqrt{2}

y:      -1         1     -1     1

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I think it is D:{5,-infinity}
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