9. Find the tangential component aT and normal component aN for the curve given byr(t) 3ti - tj + tk.10. Let a(t) 2ti + e'j+ cos (t) k denote the acceleration of a moving particle. If the initialv(0)= i+2j- k, find the particle's velocity v(t) at any time t.V2-xIn(-1)(a) Find the domain of f (x, y)H(b) Sketch the graph of f(x, y) = 6-x-2y.Find the limit of show it does not exists.4(a)lim(x,y)(0,0) y8(b)ry ylim1)2 +y(a.y)(1,0) (cu ve, then the arc length is always increasing, so s' (t)> 0 for t > a. Last, if)= 1 for all t, thenst) Il r'(u) l du =sents the arc length as long as a = 0a1 du = t- a,

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Asked Jun 12, 2019
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9. Find the tangential component aT and normal component aN for the curve given by
r(t) 3ti - tj + tk.
10. Let a(t) 2ti + e'j+ cos (t) k denote the acceleration of a moving particle. If the initial
v(0)= i+2j- k, find the particle's velocity v(t) at any time t.
V2-x
In(-1)
(a) Find the domain of f (x, y)
H
(b) Sketch the graph of f(x, y) = 6-x-2y.
Find the limit of show it does not exists.
4
(a)
lim
(x,y)(0,0) y8
(b)
ry y
lim
1)2 +y
(a.y)(1,0) (
cu ve, then the arc length is always increasing, so s' (t)> 0 for t > a. Last, if
)= 1 for all t, then
st) Il r'(u) l du =
sents the arc length as long as a = 0
a
1 du = t- a,
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9. Find the tangential component aT and normal component aN for the curve given by r(t) 3ti - tj + tk. 10. Let a(t) 2ti + e'j+ cos (t) k denote the acceleration of a moving particle. If the initial v(0)= i+2j- k, find the particle's velocity v(t) at any time t. V2-x In(-1) (a) Find the domain of f (x, y) H (b) Sketch the graph of f(x, y) = 6-x-2y. Find the limit of show it does not exists. 4 (a) lim (x,y)(0,0) y8 (b) ry y lim 1)2 +y (a.y)(1,0) ( cu ve, then the arc length is always increasing, so s' (t)> 0 for t > a. Last, if )= 1 for all t, then st) Il r'(u) l du = sents the arc length as long as a = 0 a 1 du = t- a,

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Expert Answer

Step 1

For a given curve r(t), the tangential component aT and the normal component aN are given by expressions shown on the white board. We therefore need to find:

  • r'(t) and |r'(t)|
  • r''(t)
  • r'(t).r"(t)
  • r'(t) x r''(t)
  • |r'(t) x r''(t)|
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r'() "(t) ат |r'(t)| r'()xr "( аy

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Step 2

Please see the white board for some of the intermediate calculations.

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r(t) 3ti- k r'(t) 3i-j+2tk r"(t) 2k And |r ()32 12 (2r) = 10+4r

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Step 3

Please see white board for so...

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r'(t)r"()(i-j+2rk).2k = 4t r'(f)xr"(t)=(i-j+2tk)x 2k = 2(ix k)- 2(jx k) = -2 j-2i r)xr"()2i-2j r()xr"()-2)+(-2) =2/2_

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