13.4. Curvature
Curvature
Let F(t) be a vector values function. We say it is regular if F’(t)=0
Let F(t) be a vector valued function which is arclength parametrized, which means ||F t ||  1 for all t.
Then, we can check the change of this unit vector. We call it curvature. Let T(t) = F’(t). Then
 t 
dT
ds
Example 1) F(t) = <2t, 1-2t, t>
Then, in the last class, we showed that F(s) = <2s/3, 1 – 2s/3, s/3> is arclength parametrization. Therefore,
T t 
F t
dt
= <2/3, -2/3, 1/3> and hence
= <0,0,0>, k(s) = 0.
ds
|| F t ||
Remind that we never uses the arclength parametrization directly. What we need is the definition of the
arclength function s,
Now, apply the chain rule on (1)
Therefore,
More precisely, from T(t) = F’(t)/||F’(t)||,
Example 2) Let F(t) = <-r cos t , r sin t>. Then
F’(t) = <r sin t , r cos t> and tangent vector T(t) = <sin t , cos t>. Therefore,
k(t) = ||<cos t , -sin t > || / ||<r sin t , r cos t >|| = 1/r.
Remark, we may consider then a point on the curve with curvature k looks like a circle with radius 1/k.
If the curvature at purple point is k, then the curve at that points look like a red circle with radius 1/k.
Theorem. (Curvature formula) If F(t) is regular (F’(t) is always nonzero vector), then
1) T and T’ are orthogonal. Consider derivative of dot product of T itself. Then,
T.T = ||T|| = 1. Thus, (T.T)’ = 0. On the other hand, by product rule on dot product,
(T.T)’ = T.T’ + T”.T = 2T.T’. Therefore, T.T’ = 0 and hence T and T’ are orthogonal.
2) By 1), T and T’ are orthogonal and hence they will make a right rectangle. Thus
T T
 T
T
 ||T ||   t ||F t ||
From F’(t) = ||F’(t)|| T(t), F’’(t) = ||F’(t)||’ T(t) + ||F’(t)|| T’(t)
3)
F t
F
t  ||F t || T t
4) Finally,  t 
||F t || T t  ||F t ||T t  ||F t || T t
2
F t F t
T T
.

3
||F t ||
||F t ||
Example. (curvature of a graph) If F(t) = <t, f(t), 0>, then
Proof) F’(t) = <1, f’(t), 0>, f’’(t) = <0, f’’(t), 0> and hence F’(t) x F’’(t) = <0, 0, f’’(t)>
Example) F t  < t,t 2 , t 3 > . Let’s find the curvature.
F t  <1, 2t, 3t 2 >, F
t  <0, 2, 6t> . Thus, F
F  <6t 2 ,
||F t ||  1  4t 2  9t 4 . Therefore,
 t 
36t 4  36 t 2  4
1  16t 4  81 t 8
3
6t, 2> and
T t
Black – trace of F(t). Orange – Tangent vector. Yellow – Normal vector
Red – x,y,z axis. Blue – xy – projection(y = x^2), Purple – xy- projection (z = x^3)
Normal vector
We checked the curvature on previous subsection. Now, we define the normal vector, which is the
direction of acceleration, or the direction of the center of “circle” which fits at given point.
Definition. Let F(t) be a parametrization of a curve. Then, we define a normal vector at F(t) to be
N t  T t \ ||T t ||
Example. (Helix) Let F t  < sint , cos t , t> . Then
F t  < cos t ,
So, T
t 
sin t , 1>, ||F t ||  2 , T t  1\ 2 < cos t ,
1
2<
\
sin t ,
cos t , 0>, N t  <
sin t ,
cos t , 0>
sin t , 1>
13.5 Motion in three-space
Let’s recall that we will consider that a vector-valued function F t represents the position of a particle
at time t. Then, as we did, V t  F t , sp t  ||F t ||, A t  F t (Check that any lower case
function means a scalar-valued function and Upper case function means a vector valued function.)
Example. Let F t  t i
t 3 j  lnt k Find the acceleration.
Solution. It requires to find the second derivative of given function Thus,
t  <1,
F
1
3t 2 , >, F
t
t  <0,
6t,
1
>A t
t2
Example. The term unit circular motion : F t  R < cos t , sin t> motion on a circle with radius R
and angular speed w. Let’s calculate the acceleration.
V t F
t  R <
sin  t , cos t>, A t  F
t  R 2 <
cos  t ,
sin t>
2
||F t ||  R, ||V t ||  R | |, ||A t ||  R |  2 |
||A t || 
V t
F t
Example. From fundamental Theorem of Calculus, we can find the velocity and position function from
acceleration function. Let A t  <2,12t,0>, V 0  <7,0,0>, F 0  <2,0,9> . Then,
t
t
A u du  V t
0
V 0 , V u du  F t
F 0
0
Thus, we can find
V t  <2t  7, 6t 2 , 0>, F t  <t 2  7t  2, 2t 3 , 9>
Exanmple. Let’s start from how to determine the functions. A bullet fired to the sky with the angle 60
degree and initial speed v 0 . Then, what the vector-valued function representing the bullet’s motion?
We know the initial speed and direction, which represent the initial velocity. Also, we know the starting
point and hence initial position. Also, there is only force acting on the bullet, which is gravity. So, we
have information
A t  <0,
g>, V 0  <v 0 cos


, v 0 sin >, F 0  <0,0>
3
3
By integrating twice, we get
V t  <2t  7, 6t 2 , 0>, F t  <t 2  7t  2, 2t 3 , 9>
Understanding acceleration vector
We have some relation between tangent vector, normal vector and acceleration vector. Let’s do some
differentiation.
V t  F t  ||F t || T t
A t  ||F t || T t  ||F t || T t
And
N t 
T t
T
t
, t 
||T t ||
||F t ||
Combining these two lines to get
A t  ||F t || T t   t ||F t || N t
2
Recall that the tangent vector and the normal vector are orthogonal. Thus, the acceleration vector is in the
plane whose orthogonal system consists of the tangent and the normal vectors.
So, we decompose the acceleration vector into two pieces,
A t  aT T t  a N N t
and call the coefficient of the tangent vector as tangential components and the coefficient of the normal
vector as normal component of acceleration vector. Using previous works, we can find those component
using the dot product
aT  A t T t 
A t V t
||A t V t ||
, aN  A t N t 
V t
||V t ||
More precisely for the second equality,
A V  ||V|| A T  ||V|| AT T  A N N
T  ||V|| A N N T
||A V||  ||V|| aN
From the fact that T,N are a unit vector, we can find
||A t ||  |aT |  |a N |
2
Example. Let F t  <t 2 , 2t, lnt> . Then, F
2
2
1
t  <2t , 2, >, F
t
t  <2,0,
1
>
t2
By previous result, we can decompose
F
t  <2,0,
1
>
t2
4t
1
t3
4t  4 
2
1
t2
T t 
4
t4
36

t2
16
4t  4 
1
t2

2
N t
*Osculating circle.
Let C be a curve described by a vector valued function F(t) and P = F(a) be a point on it. We consider
that it looks like a circle with radius 1/k near point. Then, which circle is most similar? We call such
circle “osculating circle”. Then it is a circle whose radius is 1/k and the center lies on the line defined by
the point P and the direction vector N.
Thus, the center of osculating circle is F(a) + 1/k N(a).
14.1 Functions of Two variables
Usually a function is decided by lot of factor. Now, we will consider a function (Not vector valued) and
do differentiations with respect to different variables
Example. Let f x,y  x  y  1 .Then, it is a function of two variable, is defined for all x,y values. For
this function, we can consider the graph, which is a set of points (x,y,z) satisfying
z  f x,y  x  y  1 . In other words, the graph is the set of points
x,y,z |z  x  y  1 
x,y,x  y  1
Definition. Let F(x,y) be a function of two variables. Then, the graph of z=F(x,y) is a set of points
(x,y,f(x,y)) in 3D space. We say the Horizontal trace at height c is the intersection with the graph and
horizontal plane z=c.
HT(F,c) = {(x,y,c) | f(x,y) = c}
We also define the vertical trace in the plane x=a (plane y=b, respectively) is the intersection with the
graph and vertical plane x=a (plane y=b, respectively).
VT(F,x=a) = {(a,y,z) | f(a,y) = z}
VT(F,y=b) = {(x,b,z) | f(x,b) = z}
Example : Let f x,y  x 2 y2 . Then, the graph is a set of points (x,y,x^2 – y^2). The horizontal trace
at 1 is the set of points (x,y,1) satisfying x^2 – y^2 = 1. Thus, it is a hyperbola in the plane z=1. The
vertical trace in x=1 is the set of points (1,y,x) satisfying 1-v^2 = z, which is a parabola in the plan x=1.
Example. Let f(x,y) = x+y +4. Then, the graph of this function is a line. Consider any trace. Then, it will
fix any of x,y or z values, which generates a line. More ideally, the trace is a intersection with the graph
and plan. Thus, trace of a plane is the intersection of two plan, which is a line, whole plane or empty set.
Now, give another name on vertical trace. We call the level curve of f(x,y) is the projection of the
horizontal trace to xy-plane. The contour map is the set of level curve with contour interval m. For
example, the contour map of f(x,y) with contour inverval 2 is the union of level curves f(x,y)=0,2,4,6 …,
which looks like real map of US.
14.2 Limit and continuity in several variables
Limit of function : lim
X
P
f X  Q means that we expect that f(X) goes closer to Q if X approaches to
P with any pass possible. It’s like a putting. If you aim the hole correctly, then the ball should dive into
the hole.
Mathematically, we define the limit as follows: define an open disk and a punctured disk.
D P,r  Q | |P
Q|<r , D P,r  D P,r \ P
Then, we say
lim
X
P
f X Q
if for any > 0 , there is a >0 such that
|f R
Q| < for all R D P,
Example : Let f x,y  x 2  y 2 . Then,
|R
O| <
|f R
O| <
f x,y  0 : for > 0 , let    . Then,
lim
x,y
0,0
x 2  y2 <
x 2  y2 <
2
And hence
Example : Let g x,y 
y2
. Then,
x 2  y2
x 2  y2
lim
x,y
2
0  x 2  y 2 <  
2
g x,y does not exist. If we follow (t,t) curve, then g
0,0
approaches to ½ . If we follow other curve (2t,t), then g goes to 1/3.
Prop. Limit Laws
1.
2.
3.
lim f x,y  g x,y  lim f x,y  lim g x,y
x,y
P
lim f x,y
x,y
P
x,y
g x,y  lim f x,y
lim k f x,y  k
x,y
P
P
x,y
P
lim f x,y
x,y
P
x,y
P
lim g x,y
x,y
P
4.
lim f x,y
f x,y
x,y
P
lim

x,y
P g x,y
lim g x,y
x,y
P
Continuity : we say that a function is continuous at P if the expected value at P by limit is exactly the
function value.
lim f X  f P
X
Example.
example.
P
f x,y  x 2  y 2 is continuous at (0,0) since f(0,0) = 0 and the limit is also 0 be previous
14.3 Partial derivatives
We might define the derivative of given two-variable-function as follows;
f
x,y 
f x  h1 , y  h2
lim
h1 h2
f x,y
h1  h2
2
0,0
2
However, the problem is that it is quite hard to find the above limit. Thus, we cannot define the derivative
of given function.
Example. Define f(x,y) = x+ 2y. Then, if we choose a path to origin (t,0), the above limit goes to 1. But, if
we choose another path to origin (0,t), then the above limit goes to 2 so that the limit does not exist.
We can observe that the above limit varies up to the path approaching to the origin. More precisely, it
highly depends on the tangent direction of given path at the origin. Then, how to define “differentiable”
function? The basic idea is that chose two canonical directions in 2D planes; vertical and horizontal
directions, find these two directional derivatives and check whether these two derivatives can
accommodates all possible directional derivatives.
Definition. (Partial derivatives) Let f(x,y) be a function of two variables. We define the partial derivative
of f with respect to x (x, respectively) to be
f x  h,y
f
 f x  lim
h
0
x
h
f x,y
,
f x,y  h
f
 f y  lim
h
0
y
h
f x,y
Remark. The partial derivative of f with respect to x is gained when we consider the other variable y as
constant and differentiate f with x variable.
Example. Let f(x,y) = x + 2y. Then f x  1, f y  2
If g x,y  3x 2 y 3 , g x  3 y 3 2x, g y  3x 2 3 y 2
Tangent plane of the graph of f(x,y)
Think about graph of f(x,y). Let P = (a,b,f(a,b)) be a point in the graph. Now, consider two curves
containing P, (a+t, b, f(a+t,b)) and (a,b+s, f(a,b+s)).
Now, try to find the tangent line of these two curves at P. Then, we expect to have a plane containing
these two tangent lines. Later, we will define that f(x,y) is differentiable at P if any curve in the graph and
containing P has the tangent line in the tangent plane decided by two special tangent lines.
Relation with contour map.
We can estimate the partial derivative values from contour map since it might work as a table of map.
The vertical red line means that we will fix the x values to find partial derivative w.r.t y variable. And
horizontal red line means specific function values at some y values. So, it looks like that we have a table
of one variable function, f(a,y).
Higher order partial derivative.
We might define the higher order partial derivatives;
f xx 
f
f
f
f
, f xy  f x 
, f yx  f y 
, f yy 
xx
yx
xy
yy
y
x
Theorem. If f xy , f yx are continuous on an open disk, then they are same.
14.4. Differentiability, Linear Approximation and Tangent Plane
Previously, we check the brief idea of building the tangent plane of a graph of function at a point.
Definition (Linearization of the function) Let f(x,y) be a function of two variables. We define a linear
function
L x,y  f a,b  f x a,b x
a  f y a,b  y
b
And call it the linearization of f(x,y) near (a,b). It means, we will consider the tangent plane is “the
graph“ of function.
Definition.( Local linearity) We say f(x,y) is locally linear if
f x,y  L x,y   x,y
x
a  y
2
2
b
, e x,y
0 as x,y
a,b
Definition (Differentiability) We say that f(x,y) is differentiable if
1)
f x , f y exist (don’t need to be continuous)
2) f is locally linear
Theorem. f(x,y) is differentiable at P if f x , f y are continuous near P.