Chapter 11: Q. 31 (page 872)

Find the velocity and acceleration vectors for the position vectors given in Exercises 30–34
$31. r (t) = t e^{t} i + t \ln t j$

Short Answer

Expert verified

The velocity and acceleration vectors are;

$v (t) = ⟨ (t + 1) e^{t}, 1 + \ln t ⟩ a (t) = ⟨ (t + 2) e^{t} \cdot \frac{1}{t} ⟩$

Step by step solution

Step 1. Given data

The given position vector is $r (t) = t e^{t} i + t \ln t j$

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,

$\begin{aligned} v (t) & = \frac{d}{d t} (r (t)) \\ = \frac{d}{d t} (t e^{t} i + t \ln t j) \\ = \frac{d}{d t} (t e^{t}) i + \frac{d}{d t} (t \ln t) j \\ = (t e^{t} + e^{t}) i + (1 + \ln t) j \\ = (t + 1) e^{'} i + (1 + \ln t) j \\ = ⟨ (t + 1) e^{t}, 1 + \ln t ⟩ \end{aligned}$

Hence the velocity vector $v (t) = ⟨ (t + 1) e^{t}, 1 + \ln t ⟩$

Step 3. Acceleration vector

The acceleration vector a(t) is given by

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

Therefore,

$\begin{aligned} a (t) & = \frac{d}{d t} (v (t)) \\ = \frac{d}{d t} ⟨ (t + 1) e^{t}, 1 + \ln t ⟩ \\ = ⟨ \frac{d}{d t} (t + 1) e^{t}, \frac{d}{d t} (1 + \ln t) ⟩ \\ = ⟨ (t + 1) e^{t} + e^{t}, \frac{1}{t} ⟩ \\ = ⟨ (t + 1 + 1) e^{t}, \frac{1}{t} ⟩ \\ = ⟨ (t + 2) e^{t} \cdot \frac{1}{t} ⟩ \end{aligned}$

Hence the acceleration vectorlocalid="1649664589547" $a (t) = ⟨ (t + 2) e^{t} \cdot \frac{1}{t} ⟩$

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Find the velocity and acceleration vectors for the position vectors given in Exercises 30–34
$31. r (t) = t e^{t} i + t \ln t j$

Short Answer

Step by step solution

Step 1. Given data

Step 2. Velocity vector

Step 3. Acceleration vector

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Find the velocity and acceleration vectors for the position vectors given in Exercises 30–34
$31. r (t) = t e^{t} i + t \ln t j$