Math 2433 Notes – Week 2
12.6 Lines
Vector form: r (t ) = ( x0 i + y0 j + z0k ) + t (d1i + d 2 j + d3k )
x(t ) = x0 + td1
Scalar form: y (t ) = y0 + td 2
z (t ) = z0 + td3
x − x0 y − y0 z − z0
=
Symmetric form: =
d1
d2
d3
Examples:
Parallel lines have direction vectors that are scalar multiples of each other.
6) Give a vector parameterization for the line that passes through P(1, 2, -2) and is parallel to the line:
3( x − 1) = 2( y − 2) = 6( z + 2)
If we have two lines, we can determine if they are parallel, coincident, skew, or intersecting.
Parallel –
Coincident –
Skew –
Intersect –
7) Determine whether the lines l1 and l2 are parallel, coincident, skew, or intersecting. If they
intersect, find the point of intersection:
Once again, two lines (l1: r(t) = r0 + t d and l2: R(u) = R0 + u D) intersect if there exists numbers t
and u such that r(t) = R(u)
8) Determine whether the lines l1 and l2 are parallel, coincident, skew, or intersecting. If they
intersect, find the point of intersection:
l1 : x1 (t ) =1 + t , y1 (t ) =−1 − t , z1 (t ) =−4 + 2t
l2 : x2 (u ) =
1 − u , y2 (u ) =
1 + 3u , z2 (u ) =
2u
If two lines are not parallel and do not intersect, then they are skew.
9) Determine whether the lines l1 and l2 are parallel, coincident, skew, or intersecting. If they
intersect, find the point of intersection:
l1 : x1 (t ) =3 + 2t , y1 (t ) =−1 + 4t , z1 (t ) =2 − t
l2 : x2 (u ) =3 + 2u , y2 (u ) =2 + u , z2 (u ) =−2 + 2u
To find the angle between two lines:
cos θ = u d ⋅ u D
Distance from a point to a line:
d ( P, l ) =
P0 P1 × d
d
10) Find the distance from P(2, 0, 2) to the line through P0(3, -1, 1) parallel to i - 2j - 2k.
12.7 Planes
Let N = Ai + Bj + Ck be the nonzero vector perpendicular to a plane:
The equation of the plane containing P( x0 , y0 , z0 ) with normal N = Ai + Bj + Ck is
A( x − x0 ) + B ( y − y0 ) + C ( z − z0 ) =
0
Examples:
1) Find an equation for the plane which passes through the point P(2, 4, 0) and is
perpendicular to i - 3j + 2k.
2) Find an equation for the plane which passes through the point P(3, -2, 3) and is parallel
to the plane:
4x + 3y + 4z + 6 =
0
3) Find an equation for the plane which passes through the point P(1, 3, 4) and contains
the line:
l : {x(t )= 3t , y (t )= 4t , z (t )= 2 + 2t}
4) Find the unit normal vectors to the plane:
3x − 4 y + 2 z − 2 =
0
5) Write the equation of the plane in intercept form and find the points where it intersects
the coordinate axes.
4 x − 2 y + 5z =
20
Angle between two planes:
cosθ = u N1 ⋅ u N2
6) Find the angle between the planes.
2x − 2 y + 2z + 5 =
0 and 2 x − y + z − 3 =
0
| Ax + By1 + Cz1 + D |
Distance from a point to a plane d ( P1 ,℘) = 1
A2 + B 2 + C 2
7)Find the distance from the point P(1, -1, 2) to the plane
2x − y + 2z + 3 =
0
8) Find an equation in x, y, z for the plane that passes through the given points.
P(1, -1, 2), Q(3, -1, 3), R(2, 0, 2)
Two planes will be parallel if their normal vectors are scalar multiples of each other. If two planes
are not parallel, then they will intersect in a line. The line of intersection will have a direction vector
equal to the cross product of their norms.
9) Find a set of scalar parametric equations for the line formed by the two intersecting
planes.
x + y + z +1=
0 and 3 x − 3 y + 3 z + 6 =
0
13.1 Vector Functions
Vector functions in  3 are in the form: f (t ) = f1 (t )i + f 2 (t ) j + f3 (t )k where f1, f2, and f3 are the
component functions.
The scalar functions in the case of a line would take the form:
f1 (t ) =
x0 + d1t f 2 (t ) =
y0 + d 2t f3 (t ) =
z 0 + d 3t
We can think of the vector f1 (t )i + f 2 (t ) j + f3 (t )k or ( f1 (t ), f 2 (t ), f3 (t ) ) as the position of a moving
particle at time t. In order to sketch the graph of a vector function, it is easier to look at it in terms of
the scalar vectors:
=
x(t ) f=
y (t ) f=
z (t ) f3 (t )
1 (t )
2 (t )
The domain of a vector function is the set of all t values for which all the component functions are
defined. It is the largest possible interval for which all three components are defined.
=
lim f (t ) L iff =
lim f1 (t ) L=
L=
L3
1 , lim f 2 (t )
2 , lim f 3 (t )
t −>t0
t −>t0
t −>t0
t −>t0
So, if we can find the limit, we can find the derivative:
A vector function f is differentiable at t if
f (t + h) − f (t )
exists.
h −> 0
h
lim
And if the limit exists then f '(t ) = lim
h −> 0
f (t + h) − f (t )
h
Or rather, given f (t ) = f1 (t )i + f 2 (t ) j + f3 (t )k , f ′(t ) = f1 ′(t )i + f 2′(t ) j + f3 ′(t )k
Note, if any component of f (t ) is not differentiable, then f '(t ) DNE.
Example: Find the derivative of
Integration:
b
 b
 b

∫a f (t )dt =  ∫a f1 (t )dt  i +  ∫a f 2 (t )dt  j +  ∫a f3 (t )dt  k
b
1
Example: Find ∫ f (t )dt
0
13.2 Differentiation formulas
Properties of the derivatives:
Let r and u be differentiable vector-valued functions of t, and let f be a differentiable real-valued
function of t, and let c be a scalar.
d
[cr(t )] = cr′(t )
dt
d
2. [r (t ) ± u(t ) ] =r′(t ) ± u′(t )
dt
d
3. [ f=
(t )r (t ) ] f (t )r′(t ) + f ′(t )r (t )
dt
d
4. [r (t ) ⋅ u(t ) ] =r (t ) ⋅ u′(t ) + r′(t ) ⋅ u(t )
dt
d
5. [r (t ) × u(t ) ] =r (t ) × u′(t ) + r′(t ) × u(t )
dt
d
6. [r ( f (t )) ] = r′( f (t )) f ′(t )
dt
7. If r (t )=
⋅ r (t ) c, then r (t ) ⋅ r′(t ) = 0
1.
d
1
Example: If r (t ) = i − j + ln t k and u(t ) = t 2 i − 2tj + k , find [r (t ) ⋅ u(t ) ]
dt
t
Example:
Find r(t) given that r '' (t) = a + t b for all real t, r ' (0) = c and r(0) = d
13.3 Curves
The parameterization of the curve represented by the vector-valued function
r (t ) = x(t )i + y (t ) j + z (t )k
is smooth on an open interval I if 𝑥𝑥′(𝑡𝑡), 𝑦𝑦′(𝑡𝑡), and 𝑧𝑧′(𝑡𝑡) are continuous on I and r′(t ) ≠ 0 for any
value of t on the interval I.
Let C be a smooth curve represented by r on an open interval I. The unit tangent vector T(t) at
t is defined to be
r '(t )
=
T(t )
, r '(t ) ≠ 0
r '(t )
A tangent line to a curve at a point is the line passing through the given point that is parallel to
the tangent vector.
Examples:
Find the tangent vector r′(t ) and the unit tangent vector T(t ) at the point where t =
parameterize the tangent line at the indicated point.
r (t ) = 2 cos(t )i + 2sin(t ) j + t k
π
4
and
Now if T '(t ) = 0 then the unit tangent vector does not change direction.
If T '(t ) ≠ 0 , then we can find the principal normal vector:
T '(t )
N(t ) =
T '(t )
Example: Find the principle normal for r (t=
) 2 cos(t )i + 2sin(t ) j + t k , =
t
π
4
The plane determined by the unit tangent vector and the principal normal vector is called the
osculating plane.
Continuing the problem above:
Intersecting Curves:
cosθ =
r1′ (t1 ) ⋅ r2′ (u2 )
r1′ (t1 ) r2′ (u2 )
Example:
Find the point at which the curves intersect and find the angle of intersection.
r1 (t ) = e3t i + 4sin(t + π / 2) j + ( t 2 − 1) k
(
)
r2 (u ) = ui + 4 j + u 2 − 2 k
13.4 Arc Length
Recall from calc II: (9.8) Arc length and speed:
b
=
L (c )
∫
 x ' ( t )  +  y ' ( t )  dt
∫
1 +  f ' ( x )  dx
a
b
L=
(c )
2
2
2
a
β
=
L (c )
∫
α
 ρ (θ )  +  ρ ' (θ )  dθ
2
2
Speed along a curve:
0 to t: s
Path ( x ( t ) , y ( t ) ) takes from
=
t
∫
 x ' ( u )  +  y ' ( u )  du
2
2
0
So, v=
'(t )
( t ) s=
 x ' ( t )  +  y ' ( t ) 
2
2
Now apply this to a path C traced out by x = x(t ) , y = y (t ) and z = z (t ) for t ∈ [ a, b ] and we have:
L(C=
)
∫ [ x′(t )] + [ y′(t )] + [ z′(t )] dt
b
2
2
a
Or
b
L(C ) = ∫ r ′(t ) dt
a
Example:
Find the length of the curve given by
r=
(t ) 2t i+ 5 ln(sec t ) j + t k
from=
t 0 to
π
4
2
Speed:
Let r(t) = x(t) i + y(t) j + z(t) k, t ∈ [a, b] be a continuously differentiable curve. If s is the
length of the curve from the tip of r(a) to the tip of r(t), then the speed is:
ds
=
dt
2
2
 dx   dy   dz 
 dt  +  dt  +  dt 
2
Note: a curve is said to be a unit speed curve if r′ ( t ) is a unit vector for all t
13.5 Curvilinear Motion; Curvature
We can describe the position of a moving object at time t by a radius vector r(t). As t ranges over a
time interval I , the object traces out some path C : r(t) = x(t) i + y(t) j + z(t) k, t ∈ I
If r is twice differentiable, then we can find
′′ ( t ) v=
′ ( t ) a ( t ) - acceleration
r′ ( t ) = v ( t ) - velocity and r=
Also, v ( t ) = r′ ( t ) is the speed at time t. The magnitude of the velocity vector is thus the rate of
change of arc distance with respect to time. This is why we call it the speed of the object.
Example:
Sketch the curve. Then compute and sketch the acceleration vector at the indicated points.
r(t) = t3 i + t j, t real; at t = −1/2 , 1/2, 1
The Curvature of a Plane Curve
In this figure we have a curve through point P with
tangent line l that intersects the x-axis at angle φ
dφ
(the magnitude of the rate of change
ds
of the angle per unit of arc length) is called the
curvature.
Then κ =
If we have the tangent vector T, then κ =
If a curve is given by y = y ( x) , then
dT
ds
κ=
y′′
1 + ( y′ )2 


3/2
r ( t ) x ( t ) i + y ( t ) j and
Now, if we have a curve given parametrically, =
Note:
• Along a straight line, the curvature is constantly zero.
• Along a circle of radius r the curvature is constantly l/r.
Also, the reciprocal of the curvature is called the radius of curvature: ρ =
1
κ
And the point at the distance ρ from the curve in the direction of the principal normal is called the
center of curvature.
Examples:
1. Find the curvature of the given curve: y = ln(sec( x))
2. Express the curvature in terms of t.
r (=
t ) 2t i + 3t 3 j
The Curvature of a Space Curve
A space curve bends in two ways. It bends in the osculating plane (the plane of the unit tangent T and
the principal normal N) and it bends away from that plane. The first form of bending is measured by
the rate at which the unit tangent T changes direction. The second form of bending is measured by the
rate at which the vector T × N changes direction. We will concentrate here on the first form of
bending, the bending in the osculating plane. The measure of this is called curvature.
If the space curve is given in terms of a parameter t, C : r(t) = x(t) i + y(t) j + z(t) k, t ∈ [ a, b ]
Then the curvature
is: κ
=
dT / dt
=
ds / dt
Example: Calculate the curvature of r (=
t ) cos ti + tj + sin tk
Components of Acceleration
dr / dt
v
=
, so: v =
dr / dt ds / dt
Given r(t) = x(t) i + y(t) j + z(t) k=
=> T
This means that a ==
v′(t ) T′(t )
So, a(t ) N(t ) T′(t )
=
ds
d  ds 
+ T(t )   and since N(t ) =
dt
dt  dt 
T′(t )
⇒ T′(t ) = N(t ) T′(t )
T′(t )
ds
d 2s
+ T(t ) 2
dt
dt
The acceleration vector lies in the osculating plane, the plane of T and N. The tangential component
of acceleration is given by:
aT =
And this depends only on the change of speed of the object; if the speed is constant, the tangential
component of acceleration is zero and the acceleration is directed entirely toward the center of
curvature of the path.
The normal component of acceleration is
aN =
And depends both on the speed of the object and the curvature of the path. At a point where the
curvature is zero, the normal component of acceleration is zero and the acceleration is directed
entirely along the path of motion. If the curvature is not zero, then the normal component of
acceleration is a multiple of the square of the speed.
If we take the dot product of T with a, we get
So, aT =
v •a
ds / dt
and aN =
v×a
ds / dt
Which means we can write: κ =
v×a
( ds / dt )
3
Examples:
1. Interpret r(t) as the position of a moving object at time t. Determine the tangential and normal
components of acceleration .
r=
(t ) cos(t )i + tj + sin(t )k
2. The vector function r(t) = (cos t + t sin t)i + (sin t − t cos t)j +
3 2
t k determines a curve C
2
in space.
(a) Find the unit tangent vector T(t) and the principal normal vector N.
(b) Find the curvature, κ of C.
(c) Determine the tangential and normal components of acceleration; express the acceleration
vector, a(t) in terms of T and N.