Chapter 11: Q. 34 (page 872)

Find the velocity and acceleration vectors for the position vectors given in Exercises 30–34
$r (t) = \cos 3 t i + \sin 4 t j + t k$

Short Answer

Expert verified

The velocity and acceleration vectors are:

$\begin{aligned} v (t) = ⟨ - 3 s i n 3 t, 4 c o s 4 t, 1 ⟩ \\ a (t) = ⟨ - 9 c o s 3 t - 16 s i n 4 t, 0 ⟩ \end{aligned}$

Step by step solution

Step 1. Given data

The given position vector is $r (t) = \cos 3 t i + \sin 4 t j + t k$

We have to find the velocity and acceleration vectors.

Step 2. Velocity vector

The velocity vector v(t) is given by

$v (t) = \frac{d}{d t} r (t) = x^{'} (t) i + y^{'} (t) j + z^{'} (t) k = ⟨ x^{'} (t), y^{'} (t), z^{'} (t) ⟩$

Therefore,

$\begin{aligned} v (t) & = \frac{d}{d t} (r (t)) \\ = \frac{d}{d t} (\cos 3 t i + \sin 4 t j + t k) \\ = \frac{d}{d t} (\cos 3 t) i + \frac{d}{d t} (\sin 4 t) j + \frac{d}{d t} (t) k \\ = - 3 \sin t i + 4 \cos 4 t j + k \\ = ⟨ - 3 \sin 3 t, 4 \cos 4 t, 1 ⟩ \end{aligned}$

Hence the velocity vector $v (t) = ⟨ - 3 s i n 3 t, 4 c o s 4 t, 1 ⟩$ .

Step 3. Acceleration vector

The acceleration vector a(t) is given by

$a (t) = \frac{d}{d t} v (t) = x^{''} (t) i + y^{''} (t) j + z " (t) k = ⟨ x^{''} (t), y^{''} (t), z^{''} (t) ⟩$

Therefore,

$\begin{aligned} a (t) & = \frac{d}{d t} (v (t)) \\ = \frac{d}{d t} ⟨ - 3 \sin 3 t, 4 \cos 4 t, 1 ⟩ \\ = ⟨ \frac{d}{d t} (- 3 \sin 3 t), \frac{d}{d t} (4 \cos 4 t), \frac{d}{d t} (1) ⟩ \\ = ⟨ - 9 \cos 3 t, - 16 \sin 4 t, 0 ⟩ \end{aligned}$

Hence the acceleration vector $a (t) = ⟨ - 9 c o s 3 t - 16 s i n 4 t, 0 ⟩$

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Find the velocity and acceleration vectors for the position vectors given in Exercises 30–34
$r (t) = \cos 3 t i + \sin 4 t j + t k$

Short Answer

Step by step solution

Step 1. Given data

Step 2. Velocity vector

Step 3. Acceleration vector

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One App. One Place for Learning.

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Find the velocity and acceleration vectors for the position vectors given in Exercises 30–34
$r (t) = \cos 3 t i + \sin 4 t j + t k$