Multivariable Calculus
Math LCU
LAB
Curvature
b
1.
Find the arc length L =

r (t) dt for the curves
text 13.3
t=a


(a) r (t )  t 2i  2t j  ln t k , 1  t  e
*(b) r (t ) t cos t i tsin t j  t k , 0  t  1.
(c) the cycloid r ( )  a   sin  , 1  cos , 0    2
Hint: Use the identity cos  1  2sin  2 
2
2.
3.
Reparametrize the curves with respect to arc length measured from t = 0 in the direction of

increasing t, that is, find r t (s) .


(a) r (t )  3sin t , 4t , 3 cos t
*(b) r (t )  cos3 t , sin 3 t , cos 2t
Find the unit tangent vector T  t  , the unit normal vector N  t  and use the formula
κ t  =
T t 
to find the curvature for the following curves.
r (t)
(a) r (t )  sin  4t  , 3t, cos  4t 
4.
Use the formula κ  t  =
*(b) r (t )  t 2 , 2t, ln t 
r (t) × r (t)
r (t)
3

(a) the helix r (t )  cos t , sin t , t
(c) r (t )  sin t, cos t, sin t 
5.
Use the formula  (x) =

(c) r (t )  6t , 3 2 t 2 , 2t 3
to find the curvature for

(b) r (t )  1  t , 1  t , 3t 2
(d) the cycloid r ( )  a   sin  , 1  cos  and  ( ) .
y
to find the curvature of the plane curves
3
2
 1 + y 2 
(a) y  sin x and obtain  (0),  ( 2 ) and  ( ) .
(b) y  e x and find the maximum curvature and the point P on the curve where it occurs.
6.
7.
8.
Find the velocity and the position vectors of a particle, as functions of time, given the acceleration



vector and initial conditions a(t )  i + 2 j + 2t k , v (0) = 0 , r (0)  i  k .

If the position function of a particle is r (t )  t 2 , 5t , t 2  16t , when is the speed a minimum?
Find the tangential and normal components of the acceleration vector for the following curves.


(a) r (t )  ( t 2 + 4)i  ( 2t  3) j
(b) the cycloid r (t )  a t  sin t , 1  cos t

(c) r (t )  t 3i  t 2 j  t k





r (t )  r (t )
r (t )  r (t )
r (t )
aT 
aN 


r (t )
r (t )
A

r( t )
9.
Prove Kepler's Second Law :
"A planet sweeps out equal areas in equal times"
 


dA

Hint: Let h  r (t )  r (t ) , use differentiation to show that the vector h is constant in time and that
dt
1
2

h .
_________________________________________________________________________________________________________________________
answers:
(b) L  
1(a) L  e2
1
2  t 2 dt 
0
3
2
 3 1
  1.5245 via the trigonometric substitution t  2 tan
 2 
 ln 

s 4
 s 
2(a) s  5t and r  t ( s )   3sin   , s , 3cos   
5 5
 5 

 2  32  2  32
4 
(b) s  25 sin2 t and r  t ( s )   1  s  ,  s  , 1  s 
5
5
5 





3(a) T  t   15  4cos  4t  , 3,  4sin  4t  , N  t     sin  4t  , 0,cos 4t  , k 
(b) T  t  
 2t
1
1  2t 2 2t 
2t

2t , 2,  , N  t    2 , 2 , 2  , k 
2

2
2t  1 
t
2
t

1
2
t

1
2
t

1


 2t  1
(c) T  t  
1
1
  2t ,
 1, 2 t , t 2  , N  t  

1  t2 
2 1  t 2  
(c) k 
16
25
t
y
2
1
2
4(a) k 
(b) k 
1  cos t 
2
3
2 1  t 2  , 2t  , k 
4(d)
2
6 1  t

4a
1  18t 
3
2
(d) k 
2
radius of curvature is 4a
1
1
, k   
4a
2 2a 1  cos t
k 0  0
sin x
1  cos x 
2
e
(b) k 
2
x
1  e 
2x
y
3
3
2
y
, k  2   1
k    0
kmax 
2
27
y  ex
, maximum curvature when k   x   0  x   12 ln2 ,
and kmax 
1
2
cycloid curve
2 2
3
2
2
5(a) k 
(c) L= 8a
2
27
x
5(b)
6. r  t    12 t 2  1 , t 2 , 13 t 3  1
7. v  t   r  t   8t 2  64t  281 , minimum when
8(a) aT 
2t
t 1
2
, aN 
9. Using product rule
dA

dt
1
2
2
t 1
2
(b) aT 
d
8t 2  64t  281  0  t  4
dt
a sin t
a
1  cos t
, aN 
2 1  cos t
2
(c) aT 
18t 3  4t
9t  4t  1
4
2
, aN 
dh
 r   t   r   t   r  t   r   t   0  h  c
constant in time
dt
 zero since r  t  is parallel to r  t  (opposite direction)
area of parallelogram determined by the vectors r  t  and r  t   12 r  t   r  t  
1
2
h  constant
2 9t 4  9t 2  1
9t 4  4t 2  1
x