Short Answer

Prove that the tangent vector is always orthogonal to the position vector for the vector-valued function.
$r (t) = <\sin t, \sin t \cos t, \cos^{2} t>$

Ans:

$r (t) \cdot r^{'} (t) = (i \sin t + j \sin t \cos t + k \cos^{2} t) \cdot (i \cos t + j \cos 2 t - k \sin 2 t) r (t) \cdot r^{'} (t) = 0$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given information:

$r (t) = <\sin t, \sin t \cos t, \cos^{2} t>$

Step 2. Proving that the tangent vector is always orthogonal to the position vector for the vector-valued function:

$r (t) = <\sin t, \sin t \cos t, \cos^{2} t> r (t) = i \sin t + j \sin t \cos t + k \cos^{2} t$

Differentiate with respect to t

$r^{'} (t) = i \cos t + j \cos 2 t - k \sin 2 t$

Find $r (t) \cdot r^{'} (t)$

$r (t) \cdot r^{'} (t) = (i \sin t + j \sin t \cos t + k \cos^{2} t) \cdot (i \cos t + j \cos 2 t - k \sin 2 t) r (t) \cdot r^{'} (t) = 0$

Thus, the tangent vector is always orthogonal to the position vector for the vector-valued function.

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Short Answer

Prove that the tangent vector is always orthogonal to the position vector for the vector-valued function.
$r (t) = <\sin t, \sin t \cos t, \cos^{2} t>$

Step by Step Solution

Step 1. Given information:

Step 2. Proving that the tangent vector is always orthogonal to the position vector for the vector-valued function:

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Prove that the tangent vector is always orthogonal to the position vector for the vector-valued function.
$r (t) = <\sin t, \sin t \cos t, \cos^{2} t>$