Chapter 11: Q. 33 (page 872)

Find the velocity and acceleration vectors for the position vectors given in Exercises 30–34
$33. r (t) = ⟨ \sin t, \cos t, 2 \sin 2 t ⟩$

Short Answer

Expert verified

The velocity and acceleration vectors are;

$v (t) = ⟨ \cos t, - \sin t, 4 \cos 2 t ⟩ a (t) = ⟨ - \sin t, - \cos t, - 8 \sin 2 t ⟩$

Step by step solution

Step 1. Given data

The given position vector is $r (t) = ⟨ \sin t, \cos t, 2 \sin 2 t ⟩$

We have to find the velocity and acceleration vectors.

Step 2. Velocity vector

The velocity vector v(t) is given by

$v (t) = \frac{d}{d t} r (t) = ⟨ \frac{d}{d t} x (t), \frac{d}{d t} y (t), \frac{d}{d t} z (t) ⟩$

Therefore,

$\begin{aligned} v (t) & = \frac{d}{d t} r (t) \\ = \frac{d}{d t} ⟨ \sin t, \cos t, 2 \sin 2 t ⟩ \\ = ⟨ \frac{d}{d t} \sin t, \frac{d}{d t} \cos t, \frac{d}{d t} 2 \sin 2 t ⟩ \\ = ⟨ \cos t, - \sin t, 4 \cos 2 t ⟩ \end{aligned}$

Hence the velocity vector $v (t) = ⟨ \cos t, - \sin t, 4 \cos 2 t ⟩$

Step 3. Acceleration vector

The acceleration vector a(t) is given by

$a (t) = \frac{d}{d t} v (t)$

Therefore,

$\begin{aligned} a (t) & = \frac{d}{d t} v (t) \\ = \frac{d}{d t} ⟨ \cos t, - \sin t, 4 \cos 2 t ⟩ \\ = ⟨ \frac{d}{d t} (\cos t), \frac{d}{d t} - \sin t, \frac{d}{d t} 4 \cos 2 t ⟩ \\ = ⟨ - \sin t, - \cos t, - 8 \sin 2 t ⟩ \end{aligned}$

Hence the acceleration vector $a (t) = ⟨ - \sin t, - \cos t, - 8 \sin 2 t ⟩$

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Find the velocity and acceleration vectors for the position vectors given in Exercises 30–34
$33. r (t) = ⟨ \sin t, \cos t, 2 \sin 2 t ⟩$

Short Answer

Step by step solution

Step 1. Given data

Step 2. Velocity vector

Step 3. Acceleration vector

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