A moving particle starts ✓ (0) = (0,0,07 It's Find 2 Given acceleration its velocity f(tu) is with at initial 2 (t) = 4 t↑ + 6 + ₁ + k and 10 an _initial Position Velocity (0) 7(0) = T-J+F. position at time t.
A moving particle starts ✓ (0) = (0,0,07 It's Find 2 Given acceleration its velocity f(tu) is with at initial 2 (t) = 4 t↑ + 6 + ₁ + k and 10 an _initial Position Velocity (0) 7(0) = T-J+F. position at time t.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
![E
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Activity #3
A moving particle starts
1
It's
Find
2) Given f(x,y)
R
and
4tm <0'0'0) = (0) A
что
4+ $79+ $7h - 12
acceleration
its.
mulas
3) Find lim
4 Find lim
(0/0) & (x²x)
D
5) Find lim
(x, y) => (0, 0)
(0'0) = (x^²x)
1) act) = 54t, 6t₁ | 7
x (1) = ((26²3)
{
velocity and
%
166
5
T
=(7) 1
Y
is
91
2h+₂x+1
horizontal traces then
)
X
htzł
7
U
2
ty
at
initial
t
17.7
<7 278 2727=1711
0'0'0 = 6021
3+ (-1 = (0) 1
2+ $7+ £279. + √₂72
(7) u = $7+ √√√ 75+ √₂ 724
PULL
* CO
r(t) = ² t²³ ↑ + £²³² Ĵ + t² k + D
8
Position at
h+2x
+-+8+48 +1 -71 [2h+₂2[h=2]] (0] (11-27)
+ + (1+²)j + (1) I
Andres Mora
unit soorit
if it exists
V(0)
initial
velocity
If it exists
)
6
domain, vertical traces
graph the surface.
0 = V(0) = (> (=0
X (1+2₂7) + 5 (1-₁7) + (1+
O
(
O
7 Just t
2
Position
P
224+6-1 = (0) 1
1²-√²+R = V/O=D
]
+ 11
}
[
delete
2](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5b55a0fb-5dff-4587-97c0-c3ec381e9c7e%2F0c40df0c-5ec5-4e46-87e4-2849dbe53abb%2Fcjzqcii_processed.jpeg&w=3840&q=75)
Transcribed Image Text:E
$
Activity #3
A moving particle starts
1
It's
Find
2) Given f(x,y)
R
and
4tm <0'0'0) = (0) A
что
4+ $79+ $7h - 12
acceleration
its.
mulas
3) Find lim
4 Find lim
(0/0) & (x²x)
D
5) Find lim
(x, y) => (0, 0)
(0'0) = (x^²x)
1) act) = 54t, 6t₁ | 7
x (1) = ((26²3)
{
velocity and
%
166
5
T
=(7) 1
Y
is
91
2h+₂x+1
horizontal traces then
)
X
htzł
7
U
2
ty
at
initial
t
17.7
<7 278 2727=1711
0'0'0 = 6021
3+ (-1 = (0) 1
2+ $7+ £279. + √₂72
(7) u = $7+ √√√ 75+ √₂ 724
PULL
* CO
r(t) = ² t²³ ↑ + £²³² Ĵ + t² k + D
8
Position at
h+2x
+-+8+48 +1 -71 [2h+₂2[h=2]] (0] (11-27)
+ + (1+²)j + (1) I
Andres Mora
unit soorit
if it exists
V(0)
initial
velocity
If it exists
)
6
domain, vertical traces
graph the surface.
0 = V(0) = (> (=0
X (1+2₂7) + 5 (1-₁7) + (1+
O
(
O
7 Just t
2
Position
P
224+6-1 = (0) 1
1²-√²+R = V/O=D
]
+ 11
}
[
delete
2
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