2. (5 points) Find the parametric equation of the tangent line to the curve FM = (1/F 3, ln(t2 + 3), t) at (2,1n4, XXXXXXXXXXpoints) Find the velocity and acceleration at t = 2 of the object with the...

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2. (5 points) Find the parametric equation of the tangent line to the curve FM = (1/F 3, ln(t2 + 3), t) at (2,1n4,1) 3. (4 points) Find the velocity and acceleration at t = 2 of the object with the position function F(t) =< 2="" sin="" t,="" 4t2="" +="" t,="" 5et2="">. Answer: < 2="" cos="" 2,="" 17,="" 20e4="">, < —2="" sin="" 2,="" 8,="" 90e4=""> 4. (6 points) Given an acceleration 5(t) =< 1,="" t,="" 4t="">, initial velocity < 20,0,0=""> and initial position < 0,="" 0,="" 0="">, find the velocity and position vectors for t > 0. Answer: (t + 20, g, 2t2), (q+ 20t, it3) 5. (4 points) Find the arc length: (t) .< et,="" e-t,l/lt="">, for 0 < x="">< 1n3.="" answer:="" 6.="" (4="" points)="" find="" the="" length="" of="" the="" complete="" cardioid="" r="4" +="" 4="" sin="" 0.="" answer:="" 32="" 7.="" (5="" points)="" find="" the="" unit="" tangent="" vector="" t="" and="" the="" curvature="" k="" for="" f(t)="">< 4+t2,="" t,="" 0="">. Answer: 2 (4t2+1)3/2 8. (20 points) Find the limit, if exists, or show that the limit does not exist. (a) lim X4 — x2y6 — 4x. Answer: — (x,y)—,(1,-1) x5 + xy4 + 3 x3+ X2y + x2+xy+x+y 3 (b) (x,y)-,( lim . Answer: 1,-1) x2y + xy2 + 3x + 3y x2y2 (c) (x,y1)i—,(o,o) x2 + 2y3' Answer: 0 sin (x2 + y2) (d) lira Answer: 0 (x,y)-(0.o) ?s2 + y2 X4y (e) (x,yli)—,m(0,0) 5y2 Answer: does not exist (f) fim(0,0,0) 3x2 x2y + /i 4y2 — 3 x . Answer: Does not exist (x,y,z)—> _z_ 4


Answered Same DayDec 26, 2021

Answer To: 2. (5 points) Find the parametric equation of the tangent line to the curve FM = (1/F 3, ln(t2 + 3),...

David answered on Dec 26 2021
119 Votes
Ans. 6
The given centroid,
4 4sin r
Calculate the length of the cardioid,

Use the formula,
2
2 rr
d
ds
d
 
   

 

Now,
2 2
r
4cos
r 16 16sin 32sin
d
d


 

  

So,
2 216 16sin 32sin 16cos
32 32sin
4 2 2sin
ds
ds
ds
  


   
 
 

Now,
2
0
2
5/2
0
2
3/2
0
2
0
4 2 2sin
2 sin 1
2
2 cos
4
2
16sin
4
32
s d
s d
s d
s
s




 
 
 

 
 
 
 
  
 
 
  
 







Ans. 7
The given vector,
2( ) 4 ,t t t r
The unit of the vector,
24
ˆ ,
2 2
r t t
r
r
 
   
 

Calculate the above equation w. r.t. t
Differentiate the vector,
'
2
' 2
1 1
r ( ) (2 ),
2 2
1 4 1
( ) 2
2 2
t t
t
r t t
 
  
 

  

The magnitude of the curvature,
2
'
4 2 2
4 1
( ) 2
( ) 16 8
t
r t
k
r t t t t

 
   


Ans. 12
(a)
The given function,
2 2( , ) 2 4f x y x y x y   
Calculate the directional derivative,
So, the partial derivative,
2 2
2 4
f
x
x
f
y
y

 


 


So,
ˆ ˆ(1, 1) 0 2i j  
So, calculate the direction derivative in the direction of (3,4)
ˆ ˆ ˆ(3, 4) 2 .( )
(3, 4) 2
j ui vj
v
   
  

The unit vector for (3, 4),
3 4
(3, 4) ,
5 5
 
  
 

So,
The d., d in direction of...
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