Math 208 Worksheet on 10.8
1) Consider
r (t )  5t 2 , t  A) Find the point(s) where the curve has the largest
curvature.
i)
first find the curvature as a function of t
ii) Take the derivative
d
dt
iii) find the critical point by computing its derivative = 0)
iv) Apply the first derivative test to make sure it is indeed the
maximum
2) Consider r (t )  2 cos t ,2 sin t ,6 
a. Express the arclength from (2,0,6) as a function of t and
reparametrize the as a function of s.
Steps
i) Replace t by another variable, say u to write r(u)
ii) Apply the formula s 

t
t0
2
2
2
 dx   dy   dz 
        du
 du   du   du 
iii) Solve for t using ii)
iv) Substitute iii) into r(t)
b. Find the arclength from (2,0,6) to (0,2,6)
3) Let
r(t )  t 2 , t , 3t  1  . Find B (the binormal vector) at t = 0.
4) Consider
r(t )  et cos 2ti  2 j  e2t sin 2tk Reparametrize r(t) as a function of
arclength s starting at t=0.
5) Find the maximum curvature of y  ln x What happens as x  
?
6) Find the maximum curvature of
y  e x . What happens as x   ?
Answers
1) First compute the curvature. Then find the maximum by computing its derivative
and find the critical points. Perform the first derivative test to show it indeed gives
the maximum value.
r ' ' (t )  10, 0,0,  .

r 'r"
r
3

r(t )  5t 2 , t , 0  .
Then
r ' (t )  10t , 1,0  and
We obtain
 10t , 1, 0    10,0,0 
 10t ,1,0 

10
( 100t 2  1) 3
 10(100t 2  1)

3
2
, and
 3000t
'
(100t  1)
2
5
2
. To find the critical points, set the numerator = 0 and
denominator = 0. t=0 is the only critical point and by the first derivative test, t = 0
produces the maximum curvature. Max curvature is 10.
r (u )  2 cos u,2 sin u,6 
2) First observe that the position (2,0,6) is attained at t = 0.
t
s(t )   r' (u ) du  
t
0
(2 sin u ) 2  (2 cos u ) 2  0 du 2t  t 
0
into r(t) gives
t

2
s
s
r ( s )  2 cos ,2 sin ,6 
2
2
. In part a) we obtained
s
2
. Substitute this
b) Observe that (0,2,6) is attained when
s (t )  2t .
Therefore

s( )  
2
3) First find the tangent vector and the normal vector.
r' (t )  2t , 1, 3 
1


 2t ,1,3 
r' (t )  2t , 1, 3 
4t 2  10
T(t ) 
20

T' (t )
N(t ) 

T' (t )
(4t 2  10)
3
2
,
4t
4t  10
2
,
12t
4t  10
2


0
T' (t )
20
10


3
2
20
10
3
2
,0,0 
 1,0,0  .
,0,0 
(It is too messy to compute the magnitude of T’(t). But at t = 0, the y-comp and the zcomp of N(t)=0 and the x-comp is positive. Since N(t) is a unit vector, it must be
<1,0,0> at t = 0. ) The tangent vector at t = 0 is
Therefore, at t = 0,
4) I)
B(t )  0,
1
10
,
r (u )  eu cos u,2, eu sin u 
3
10
 0,
1
10
,
   1,0,0  0,
3
10
3
10

,
1
10
.
dr
 eu cos u  eu sin u,0, eu sin u  eu cos u  .
du
t
t
0
0
s(t )   (eu cos u  eu sin u)2  (eu sin u  eu cos u)2 du  
2e2u du
s 2
) Thus
2
s 2
s 2
s 2
s 2
r( s)  ln(
) cos(ln(
),2, ln(
) sin(ln(
)
2
2
2
2
t
 2  (eu )du  2 ( et  1) . s  2et  2  t  ln(
0
5) To find the maximum curvature, first compute the curvature function. Then find the
maximum by finding its critical points (compute its first derivative) . To compute
the curvature function, it is best to use the formula

f " ( x)
equation y = f(x) is given. Since
3
2
since the
(1  f ' ( x) )
1
1
1
x2
f ' ( x)  , f " ( x)   2 ,  
3
x
x
1 2
(1  2 )
x
2
. You
should simplify the expression before computing the derivative: Multiply each term
1
x3
x
3
3
2
x2
by x . Notice that since x  (x ) , i  
. Use the

3
3
3
1 2 x
2
(1  2 )
( x  1) 2
x
 2x 2  1
quotient rule to differentiate the function:  ' 
. Set this equal zero, we
5
3
2
( x 2  1) 2
get
x
1
. We take the positive only since the domain of the function y = ln x is
2
x>0. By conducting the first derivative test (make a sign graph), we can see that
x
the maximum curvature is attained at
1
2
1
x3
x
x2


 lim   0
3
3
3
x 
1 2 x
2
2
(1  2 )
( x  1)
x
(which means the curve looks straight when x is large)
6) Since
f ' ( x)  e x , f " ( x)  e x  
ex
3
2x 2
. Use the quotient rule to differentiate the
(1  e )
3
2x 2
1
3
e (1  e )  (1  e 2 x ) 2 2e 2 x e x
[e x (1  e 2 x )  3e3 x ] e x (1  2e 2 x )
2


function:  ' 
5
5 .
(1  e 2 x )3
2x 2
2x 2
(1  e )
(1  e )
1
Set this equal zero, we get x   ln 2 . Making a sign graph the max is attained at
2
1
x   ln 2
3
lim   0
x
x 
(this means the curve looks straight when x is large)