Chapter 11: Q.52 (page 872)

Use the given acceleration vectors $a (t) = v' (t) = r " (t)$ and initial conditions in Exercises to find the position function $r (t)$ .
$a (t) = <\sin 3 t, t^{2}, \cos 3 t>, v (0) = <0, 0, 0>, r (0) = <2, - 5, 3>$

Short Answer

Expert verified

$r (t) = (\frac{- \sin 3 t}{9} + \frac{t}{3} + 2) i + (\frac{t^{4}}{12} - 5) j + (\frac{- \cos 3 t}{9} + \frac{28}{9}) k r (t) = <\frac{- \sin 3 t}{9} + \frac{t}{3} + 2, \frac{t^{4}}{12} - 5, \frac{- \cos 3 t}{9} + \frac{28}{9}>$

Step by step solution

Given Information

Consider,

$a (t) = <\sin 3 t, t^{2}, \cos 3 t>, v (0) = <0, 0, 0> r (0) = <2, - 5, 3>$

The objective is to find the position function $r (t)$ .

Since $a (t) = v' (t) = r " (t)$ , we have $v (t) = \int a (t) d t$ and $r (t) = \int v (t) d t$

Now

$v (t) d t = \int a (t) d t = \int <\sin 3 t, t^{2}, \cos 3 t> d t = \int (\sin 3 t i + t^{2} j + \cos 3 t k) d t = \frac{- \cos 3 t}{3} i + \frac{t^{2}}{3} j + \frac{\sin 3 t}{3} k + c$

Where c is a constant

$v (t) = (\frac{- \cos 3 t}{3} + c_{1}) i + (\frac{t^{3}}{3} + c_{2}) j + (\frac{\sin 3 t}{3} + c_{3}) k$

Where $c_{1}, c_{2}$ and $c_{3}$ are scalars

$v (0) = (\frac{- 1}{3} + c_{1}) i + c_{2} j + c_{3} k . . . . . . . . . . . . . . (1)$

Expression

But as per problem, $v (0) = (0, 0, 0) = 0 i + 0 j + 0 k . . . . . . . . . . . . (2)$

Comparing (1) and (2) equations

$\frac{- 1}{3} + c_{1} = 0, c_{2} = 0, c_{3} = 0 c_{1} = \frac{1}{3} s o v (t) = (\frac{- \cos 3 t + 1}{3}) i + \frac{t^{2}}{3} j + \frac{\sin 3 t}{3} k$

Now

$r (t) = \int v (t) d t = \int (\frac{- \cos 3 t + 1}{3}) i + \frac{t^{2}}{3} j + \frac{\sin 3 t}{3} k d t r (0) = c_{4} i + c_{5} j + (\frac{- \cos 0}{9} + c_{6}) k r (0) = c_{4} i + c_{5} j + (\frac{- 1}{9} + c_{6}) k . . . . . . . . . . . . (3) r (0) = (2, - 5, 3) = 2 i - 5 j + 3 k . . . . . . . . . . . . . (4)$

Comparing (3) and (4) equations,

$c_{4} = 2, c_{5} = - 5 \frac{- 1}{9} + c_{6} = 3$

$C_{6} = 3 + \frac{1}{9} C_{6} = \frac{28}{9}$

$r (t) = (\frac{- \sin 3 t}{9} + \frac{1}{3} + 2 \frac{t^{3}}{12} 5, \frac{- \cos 3 t}{9} + \frac{28}{9})$

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Use the given acceleration vectors $a (t) = v' (t) = r " (t)$ and initial conditions in Exercises to find the position function $r (t)$ .
$a (t) = <\sin 3 t, t^{2}, \cos 3 t>, v (0) = <0, 0, 0>, r (0) = <2, - 5, 3>$

Short Answer

Step by step solution

Given Information

Expression

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Use the given acceleration vectors a(t)=v'(t)=r"(t)and initial conditions in Exercises to find the position function r(t).a(t)=sin3t,t2,cos3t,v(0)=0,0,0,r(0)=2,-5,3

Short Answer

Step by step solution

Given Information

Expression

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Applied Mathematics

Decision Maths

Calculus

Geometry

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

Use the given acceleration vectors $a (t) = v' (t) = r " (t)$ and initial conditions in Exercises to find the position function $r (t)$ .
$a (t) = <\sin 3 t, t^{2}, \cos 3 t>, v (0) = <0, 0, 0>, r (0) = <2, - 5, 3>$