Represent the plane curve by a vector-valued function. x216−y24=1

Step 1 x216−y24=1 let x=t ∴t216−y24=1 y24=t216−1 y24=t2−1616 y2=4×t2−1616 y2=t2−164 y=t2−164 y=12t2−16 Step 2 let r(t) be a vector feild function ∴r(t)=(t)i+(12t2−16)j

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High School
Calculus and Analysis
Calculus 2
Parametric equations

Ashley Searcy

Answered question

2021-11-29

Represent the plane curve by a vector-valued function.

\frac{x^{2}}{16} - \frac{y^{2}}{4} = 1

Answer & Explanation

giskacu

Beginner2021-11-30Added 22 answers

Step 1

\frac{x^{2}}{16} - \frac{y^{2}}{4} = 1

let

x = t

∴ \frac{t^{2}}{16} - \frac{y^{2}}{4} = 1

\frac{y^{2}}{4} = \frac{t^{2}}{16} - 1

\frac{y^{2}}{4} = \frac{t^{2} - 16}{16}

y^{2} = 4 \times \frac{t^{2} - 16}{16}

y^{2} = \frac{t^{2} - 16}{4}

y = \sqrt{\frac{t^{2} - 16}{4}}

y = \frac{1}{2} \sqrt{t^{2} - 16}

Step 2
let r(t) be a vector feild function

∴ r (t) = (t) i + (\frac{1}{2} \sqrt{t^{2} - 16}) j

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