10. Let a(t) = 2ti + e'j+ cos(t) k denote the acceleration of a moving particle. If the initial velocity isv(0) i+2j-k, find the particle's velocity v(t) at any time t.2-x(a) Find the domain of f(x, y) = In(a-1)(b) Sketch the graph of f(r, y) 6-x2y.12. Find the limit of show it does not exists.4(a)lim(ay)(0,0) 2 + ysxy y(b)lim(ay)(1,0) ( 1)2 +y2ve, then the arc length is always increasing, so s' (t)> 0 for t> a. Last, if= 1 for all t, thens()Il r'(u) du =h means that t represents the arc length as long as a =0.-da1 du = t - a,

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Asked Jun 12, 2019

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10. Let a(t) = 2ti + e'j+ cos(t) k denote the acceleration of a moving particle. If the initial velocity is
v(0) i+2j-k, find the particle's velocity v(t) at any time t.
2-x
(a) Find the domain of f(x, y) = In(a-1)
(b) Sketch the graph of f(r, y) 6-x2y.
12. Find the limit of show it does not exists.
4
(a)
lim
(ay)(0,0) 2 + ys
xy y
(b)
lim
(ay)(1,0) ( 1)2 +y2
ve, then the arc length is always increasing, so s' (t)> 0 for t> a. Last, if
= 1 for all t, then
s()Il r'(u) du =
h means that t represents the arc length as long as a =0.
-d
a
1 du = t - a,
help_outline

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10. Let a(t) = 2ti + e'j+ cos(t) k denote the acceleration of a moving particle. If the initial velocity is v(0) i+2j-k, find the particle's velocity v(t) at any time t. 2-x (a) Find the domain of f(x, y) = In(a-1) (b) Sketch the graph of f(r, y) 6-x2y. 12. Find the limit of show it does not exists. 4 (a) lim (ay)(0,0) 2 + ys xy y (b) lim (ay)(1,0) ( 1)2 +y2 ve, then the arc length is always increasing, so s' (t)> 0 for t> a. Last, if = 1 for all t, then s()Il r'(u) du = h means that t represents the arc length as long as a =0. -d a 1 du = t - a,

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check_circleExpert Solution
Step 1

The acceleration of the moving particle is given by,

a(t)=2ti+etj+cos(t)k      

To find the velocity at time instant t, integrate a(t) with respect to t.

a(t) 2fiecos(t)k
Ja(t)dt=(2i+ej+cos ()k)dt
v()(2H)dt [(ei)dt [(cos(t)k)d|
2i+e'j+sin (t)k +C
2
=fite'j+sin (tk+C .
...)
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a(t) 2fiecos(t)k Ja(t)dt=(2i+ej+cos ()k)dt v()(2H)dt [(ei)dt [(cos(t)k)d| 2i+e'j+sin (t)k +C 2 =fite'j+sin (tk+C . ...)

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Step 2

Where C is the integration constant.

since initial velocity is v(0)=i+2j-k.

substitute i+2j-k for v(t) and 0 for t in equation 1.

i+2j-k (0ie'j+sin(0)k+C
i +2j-k 0j+0+C
C i k
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i+2j-k (0ie'j+sin(0)k+C i +2j-k 0j+0+C C i k

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Step 3

Substitute i+j-k for C...

(t)iejsin (t)k +i +j-k
v()-(P1i(e +1)j+(sin (t)+cos())k
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(t)iejsin (t)k +i +j-k v()-(P1i(e +1)j+(sin (t)+cos())k

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