Q. 32 Find the velocity and accelerati... [FREE SOLUTION]

Chapter 11: Q. 32 (page 872)

Find the velocity and acceleration vectors for the position vectors given in Exercises 30鈥�34
$32 . r (t) = 鉄� \sec t, \frac{1}{t}, e^{t} \ln t 鉄�$

Short Answer

Expert verified

The velocity and acceleration vectors are;

$\begin{aligned} v (t) = 鉄� \sec t \tan t, \frac{鈭� 1}{t^{2}}, e^{t} (\frac{1}{t} + \ln t) 鉄� \\ a (t) = 鉄� \sec^{3} + \sec t \tan^{2} t, \frac{2}{t^{3}}, e^{t} (\frac{鈭� 1}{t^{2}} + \frac{2}{t} + \ln t) 鉄� \end{aligned}$

Step by step solution

Step 1. Given data

The given position vector is $r (t) = 鉄� \sec t, \frac{1}{t}, e^{t} \ln t 鉄�$

We have to find the velocity and acceleration vectors.

Step 2. Velocity vector

The velocity vector v(t) is given by

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

Therefore,

$\begin{aligned} v (t) & = \frac{d}{d t} r (t) \\ = \frac{d}{d t} 鉄� \sec t, \frac{1}{t}, e^{t} \ln t 鉄� \\ = 鉄� \frac{d}{d t} (\sec t), \frac{d}{d t} \frac{1}{t}, \frac{d}{d t} e^{t} \ln t 鉄� \\ = 鉄� \sec t \tan t, \frac{鈭� 1}{t^{2}}, \frac{e^{t}}{t} (\ln t) (e^{t}) 鉄� \\ = 鉄� \sec t \tan t, \frac{鈭� 1}{t^{2}}, e^{t} (\frac{1}{t} + \ln t) 鉄� \end{aligned}$

Hence the velocity vector $v (t) = 鉄� \sec t \tan t, \frac{鈭� 1}{t^{2}}, e^{t} (\frac{1}{t} + \ln t) 鉄�$

Step 3. Acceleration vector

The acceleration vector a(t) is given by

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

Therefore,

$\begin{aligned} a (t) & = \frac{d}{d t} v (t) \\ = \frac{d}{d t} 鉄� \sec t \tan t, \frac{鈭� 1}{t^{2}}, e^{t} (\frac{1}{t} + \ln t) 鉄� \\ = 鉄� \sec^{3} t + \sec t \tan^{2} t, \frac{2}{t^{3}}, e^{t} (\frac{鈭� 1}{t^{2}} + \frac{1}{t}) + (\frac{1}{t} + \ln t) e^{t} 鉄� \\ = 鉄� \sec^{3} t + \sec t \tan^{2} t, \frac{2}{t^{3}}, e^{t} (\frac{鈭� 1}{t^{2}} + \frac{2}{t} + \ln t) 鉄� \end{aligned}$

Hence the acceleration vector $a (t) = 鉄� \sec^{3} t + \sec t \tan^{2} t, \frac{2}{t^{3}}, e^{t} (\frac{鈭� 1}{t^{2}} + \frac{2}{t} + \ln t) 鉄�$

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91影视

Find the velocity and acceleration vectors for the position vectors given in Exercises 30鈥�34
$32 . r (t) = 鉄� \sec t, \frac{1}{t}, e^{t} \ln t 鉄�$

Short Answer

Step by step solution

Step 1. Given data

Step 2. Velocity vector

Step 3. Acceleration vector

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91影视

Find the velocity and acceleration vectors for the position vectors given in Exercises 30鈥�3432.r(t)=鉄�sect,1t,etlnt鉄�

Short Answer

Step by step solution

Step 1. Given data

Step 2. Velocity vector

Step 3. Acceleration vector

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Geometry

Calculus

Pure Maths

Statistics

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

Find the velocity and acceleration vectors for the position vectors given in Exercises 30鈥�34
$32 . r (t) = 鉄� \sec t, \frac{1}{t}, e^{t} \ln t 鉄�$