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1. The vector function r ( ) t
G
is shown in Figure 1. The slope of this function at the point ( −4,1 )is
−. Suppose that the point ( −4,1 )is generated at t = t 0.
a. What are T ˆ ( t 0 ), N ˆ ( t 0 ), and B ˆ ( t 0 )?
Following the direction arrow, the motion is left to right at t = t 0 Using the given slope, a
vector that points parallel to the tangent line in the direction of motion is −4,3. Hence
T t
. This is the blue vector in Figure 1; its length has been
exaggerated for visibility purposes.
The slope of the normal line at t = t 0 is
. From the concavity of the curve, N t ˆ( 0 )clearly
points down and left. Hence, ( 0 )
N t
. This is the pink vector in
Figure 1; its length has been exaggerated for visibility purposes.
From the right hand rule and the fact that B ˆ ( t 0 )is a
unit vector we have B t ˆ^ ( 0 )= 0,0,1.
b. r ′′( t 0 )
G
has been added in Figure 2. Estimate aT ( t 0 )
and aN ( t 0 ).
From Figure 2, aT ( t 0 )< 0 because the projection of
the acceleration vector onto the unit tangent vector
(the projection is shown in green) points opposite the
unit tangent vector (shown in pink). aT ( t 0 ) is the
magnitude of the green vector. Hence
2 2 aT t 0 (^) ≈ − 7 + 5 ≈ − 8.6.
From Figure 2, aN ( t 0 )is the length of the
projection of the acceleration vector onto the unit
normal vector (the projection is shown in orange).
Hence, ( )
2 2 aN t 0 (^) ≈ 3 + 4 = 5
Figure 1
Figure 2
x
y
x
y
c. Is the speed increasing or decreasing at t = t 0? What is the magnitude of the rate of
change in speed?
The speed is clearly decreasing at t = t 0 (as indicated by the negative tangential component
of acceleration). The magnitude of decrease is aT ( t 0 ) ≈ 8.6.
d. The curvature of r ( ) t
G
at ( −4,1 ) is ( 0 )
κ t =. At what point does the center of the
osculating circle through ( −4,1 )lie?
0 0
t t
= =. The center lies at the “point”
r t + κ t N t = − + − − = − −
G
.
e. What are parametric equations that model the osculating circle?
One set of parametric equations that model the osculating circle are:
cos , sin 8 24 6 24
x = − + t y = − + t
The circle is (partially) shown in Figure 3.
f. Graphically determine the equation of the rectifying
plane to r ( ) t
G
at ( −4,1 ).
The rectifying plane is perpendicular to the unit
normal vector. Consequently it contains the tangent
line and is perpendicular to the osculating plane (which in this case is the xy -plane). Hence
(in this case) the rectifying plane has the same equation as the tangent line (which is the
decreasing dotted line in figure 3); i.e. it is
y = − x −.
2. The equation of the curve in Figure 1 is r ( ) t = 4 2 sin ( ) t , 2 cos 3( t )
G
and the point ( −4,1)
is generated at
t
Figure 3
x
y
3. The vector function r ( ) t
G
is shown in Figure 3. The slope of this function at the point ( −4,3)
is 6. Suppose that the point ( −4,3 )is generated at t = t 0.
a. What are T ˆ ( t 0 ), N ˆ ( t 0 ), and B ˆ ( t 0 )?
T t = , ( 0 )
N t
= , B t ˆ^ ( 0 )= 0,0, − 1
b. r ′′( t 0 )
G
has been added in Figure 2. Estimate aT ( t 0 )and aN ( t 0 ).
The pink vector in Figure 4 shows the
direction of T t ˆ( 0 ) while the blue vector
shows a t ( 0 )
G
. Since these vectors point in
opposite directions (from a 90
O perspective),
aT ( t 0 )must be negative.
aT ( t 0 ) is represented by the length of the
green line segment in Figure 4 and a N ( t 0 ) is
represented by the length of the red line segment.
Hence, aT ( t 0 )≈ − 1.2and aN ( t 0 )≈ 3.
c. Is the speed increasing or decreasing at t = t 0? What is the magnitude of the rate of
change in speed?
Since a t ( 0 )< 0
G
, the speed is decreasing at t = t 0. The magnitude of the rate of change in
speed is aT ( t 0 ) ≈ 1.25.
d. Graphically determine the equations of the normal and osculating planes to r t ( )
G
at ( −4,3 ).
Since the motion is confined to the xy -plane, that plane is the osculating plane; so the
equation of the osculating plane is z = 0. Since the normal plane is perpendicular to the
osculating ( xy ) plane and the normal plane contains the normal line, the normal plane and
normal line must, in this situation , have the same equation. So the equation of the normal
plane is ( )
1 1 7 4 3 6 6 3
y = − x + + ⇒ y = − x +
Figure 4
4. The equation of the curve in Figure 3 is ( ) ( )
2 4,3 2 sin 16
t r t t
π
π
G
and the point ( −4,3)
is generated at 4
t
a. Symbolically verify your answers to parts a, b, and c of question 3.
Store the formula for r t ( )
G
as y t ( ). Store the formula for T t ˆ( ) as z t ( ).
Store r ′( t 0 )
G
as x and r ′′( t 0 )
G
as xx.
Verify ( )
0 0 0 0
T
r t r t a t r t
G G
G.
Verify ( )
0 0 0 0
N
r t r t a t r t
′ × ′′
G G
G.
Verify speed ′^ ( t 0 ) = aT ( t 0 ).
Verify ( )
0 0 0
T t N t T t
. Verify ( ) ( ) ( )
0 0 0 B ˆ^ t = T t ˆ^ × N t ˆ.
Verify T t ˆ( 0 ).
b. What are the curvature and unit normal vector at 2
t
c. At what point does the center of the osculating circle to r ( t )
G
at 2
t
= lie?
The radius of the osculating circle is 1 1 (^2 2 )
2
π ρ π κ
⎛ ⎞ ⎜ ⎟ =^ = ⎝ ⎠ ⎛^ ⎞ ⎜ ⎟ ⎝ ⎠
. The center lies at the “point”
r N
⎜ ⎟ +^ ⎜ ⎟ ⋅^ ⎜ ⎟=^ −^ +^ ⋅^ −^ −
G
d. On what sphere does this osculating circle form a great circle? (The equator is an example
of a great circle.)
The osculating circle lies on the sphere
2 2 2 9 15 1
4 4 8
x y z
.
Store the formula for T t ˆ( ) as z t ( ).
Find ( )
0 0 0
T t N t T t
.
Store the formula for r t ( )
G
as y ( ) t.
Store r ′( t 0 )
G
as x and r ′′( t 0 )
G
as xx.
Find
0 0 (^0 ) 0
r t r t t r t
′ × ′′
G G
G
.
e. Find parametric equations for this osculating circle. (Hint: Find equations for the
intersection between the sphere from question d and the osculating plane.)
From the plane equation, z = y + 6. Substituting for z in the sphere equation we get:
2 2 2 2 2 2
2 2
2 2
x y y x y y
x y
x y
We’ll maintain this relationship if we define 8 x = cos( t ) and ( )
4 sin 4
y t
⎜ +^ ⎟=
. This
gives us
cos( )
t x = and
sin ( ) 9
t y = −. Using z = y + 6 we have
sin ( ) 15
t z = +.
Therefore, parametric equations that model the osculating circle are:
cos ( )
t x = ,
sin ( ) 9
t y = − , and
sin ( ) 15
t z = +.
In Figure K2, I’ve zoomed in around the top of the ellipse (which is the point that occurs at
t
=. The function is shown in red, the osculating plane in green, the sphere we found in
blue, and the osculating circle in black.
Figure K
