Represent the plane curve by a vector-valued function. $\frac{{x}^{2}}{16}-\frac{{y}^{2}}{4}=1$

Ashley Searcy
2021-11-29
Answered

Represent the plane curve by a vector-valued function. $\frac{{x}^{2}}{16}-\frac{{y}^{2}}{4}=1$

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giskacu

Answered 2021-11-30
Author has **22** answers

Step 1

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

let$x=t$

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

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

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

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

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

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

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

Step 2

let r(t) be a vector feild function

$\therefore r\left(t\right)=\left(t\right)i+\left(\frac{1}{2}\sqrt{{t}^{2}-16}\right)j$

let

Step 2

let r(t) be a vector feild function

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