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

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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, XXXXXXXXXXpoints) 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, it XXXXXXXXXXpoints) 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: XXXXXXXXXXpoints) Find the unit tangent vector T and the curvature K for f(t) =< 4+t2, t, 0 >. Answer: 2 (4t2+1)3/ XXXXXXXXXXpoints) 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

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David · Accepted Answer

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.,

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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