PERTEMUAN XXII
Differensiasi Vector
Suatu vector valued function dalam dimensi dua dan tiga masing-masing
dinyatakan sebagai
r(t) = x(t)i + y(t)j
dan
r(t) = x(t)i + y(t)j + z(t)k
Untuk informasi lebih lanjut dapat click disini
Contoh
r(t) = (t - 1)i + t2 j
mempunyai grafik yang sama dengan
y = (x + 1)2
Misalkan r merupakan vektor posisi dari suatu objek yang bergerak
sepanjang curva C maka r akan merupakan fungsi dari waktu t yaitu
r = r(t). Bila P merupakan vektor posisi pada saat t dan Q vektor posisi
pada saat t+ δt maka kecepatan rata-rata adalah :
Definisi Kecepatan ( Velocity)
v(t) = r'(t) = x'(t)i + y'(t)j + z'(t)k
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Pada gambar berikut kecepatan rata-rata :
Percepatan
Contoh
Cari kecepatan dari vektor posisi
r(t) = 3ti + 2t2j - sin t k
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Jawab :
v(t) = r′(t) = 3i + 4tj + cos t k
Contoh
Andaikan
r(t) = 3i + 2t j + cos t k
Cari kecepatan saat t = /4 detik.
Jawab :
Dari vector kecepatan
v(t) = r'(t) = 2 j - sin t k
didapat
v(π/4) = 2 j -
/2 k
Speed = || v || =
Contoh
Cari kecepatan dan percepatan dari :
r(t) = 4t i + t2 j
untuk t = -1. Gambarkan
Jawab :
v(t) = r'(t) = 4 i + 2t j substitusi t = -1 didapat
v(-1) = 4 i - 2j
Percepatan adalah
a(t) = v'(t) = 2j
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Projectile Motion
Contoh You are a anti-missile operator and have spotted a missile
heading
towards you at the position
re = 1000i + 500j with velocity ve = -30i + 3j
You can fire your anti-missile at 100 meters per second. At what
angle should you fire it so that you intercept the missile. Assume that
gravity is the only force acting on the projectiles.
Jawab :
The acceleration vector of the enemy missile is
ae(t) = -9.8 j
Integrating, we get the velocity vector
ve(t) = v1 i + (v2 - 9.8t) j
Setting t = 0 and using the initial velocity of the enemy missile gives
ve(t) = -30 i + (3 - 9.8t) j
Now integrate again to find the position function
re(t) = (-30t + r1) i + (-4.9t2 + 3t + r2) j
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Again setting t = 0 and using the initial conditions gives
re(t) = (-30t + 1000) i + (-4.9t2 + 3t + 500) j
The acceleration of your anti-missile-missile is also
ay(t) = -9.8 j
Integrating, we get the velocity vector
vy(t) = v1 i + (v2 - 9.8t) j
Since the magnitude of our velocity is 100, we can say
vy(0) = 100 cos  i + 100 sin  j
So that
vy(t) = 100 cos  i + (100 sin  - 9.8t) j
Now integrate again to find the position function
ry(t) = (100t cos  + r1) i + (-4.9t2 + 100t sin  + r2) j
Our anti-missile-missile starts out at base, so the initial position is the
origin. All the constants are zero.
ry(t) = (100t cos ) i + (-4.9t2 + 100t sin ) j
Since we want to intercept the enemy missile, we set the position
vectors equal to each other.
(100t cos ) i + (-4.9t2 + 100t sin ) j
= (-30t + 1000) i + (-4.9t2 + 3t + 500) j
Equating coefficients gives
100t cos  = -30t + 1000
-4.9t2 + 100t sin  = -4.9t2 + 3t + 500
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The first equation gives
1000
t =
100cos  + 30
Simplifying the second equation and substituting gives
100000 sin 
3000
=
100cos  + 30
+ 500
100cos  + 30
Clear denominators to get
100000 sin  = 3000 + 50000 cos  + 15000
At this point we use a calculator to solve for to
 = .62535 radians
Contoh
Cari
XXII-6
RUMUS
Contoh
Cari :
Jawab :
Soal-soal
XXII-7
Jawab :
Contoh
Find the velocity vector v(t) if the position vector is
r(t) = 3ti + 2t2j - sin t k
Jawab :
v(t) = r′(t) = 3i + 4tj + cos t k
Contoh
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Andaikan
r(t) = 3i + 2t j + cos t k
Cari kecepatan saat t = /4 detik.
Jawab :
Dari vector kecepatan
v(t) = r'(t) = 2 j - sin t k
didapat
v(p/4) = 2 j -
/2 k
Speed = || v || =
Contoh
Cari kecepatan dan percepatan dari fungsi posisi :
r(t) = 4t i + t2 j
untuk t = -1. Gambarkan sketchnya.
Jawab :
The velocity vector is
v(t) = r'(t) = 4 i + 2t j
Plugging in -1 for t gives
v(-1) = 4 i - 2j
Take another derivative to find the acceleration.
a(t) = v'(t) = 2j
Below is a picture of the vectors.
XXII-9
Projectile Motion
Contoh
You are a anti-missile operator and have spotted a missile heading
towards you at the position
re = 1000i + 500j
with velocity
ve = -30i + 3j
You can fire your anti-missile at 100 meters per second. At what angle
should you fire it so that you intercept the missile. Assume that gravity
is the only force acting on the projectiles.
Jawab :
The acceleration vector of the enemy missile is
ae(t) = -9.8 j
Integrating, we get the velocity vector
ve(t) = v1 i + (v2 - 9.8t) j
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Setting t = 0 and using the initial velocity of the enemy missile gives
ve(t) = -30 i + (3 - 9.8t) j
Now integrate again to find the position function
re(t) = (-30t + r1) i + (-4.9t2 + 3t + r2) j
Again setting t = 0 and using the initial conditions gives
re(t) = (-30t + 1000) i + (-4.9t2 + 3t + 500) j
The acceleration of your anti-missile-missile is also
ay(t) = -9.8 j Integrating, we get the velocity vector
vy(t) = v1 i + (v2 - 9.8t) j
Since the magnitude of our velocity is 100, we can say
vy(0) = 100 cos θ i + 100 sin θ j
So that
vy(t) = 100 cos θ i + (100 sin θ - 9.8t) j
Now integrate again to find the position function
ry(t) = (100t cos θ + r1) i + (-4.9t2 + 100t sin θ + r2) j
Our anti-missile-missile starts out at base, so the initial position is the
origin. All the constants are zero.
ry(t) = (100t cos θ) i + (-4.9t2 + 100t sin θ) j
Since we want to intercept the enemy missile, we set the position
vectors equal to each other.
(100t cos θ) i + (-4.9t2 + 100t sin θ) j
= (-30t + 1000) i + (-4.9t2 + 3t + 500) j
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Equating coefficients gives
100t cos θ = -30t + 1000
-4.9t2 + 100t sin θ = -4.9t2 + 3t + 500
The first equation gives
1000
t =
100cos  + 30
Simplifying the second equation and substituting gives
100000 sin 
3000
=
100cos  + 30
+ 500
100cos  + 30
Clear denominators to get
100000 sin θ = 3000 + 50000 cos θ + 15000
At this point we use a calculator to solve for to
θ = .62535 radians
Integral Vektor
Contoh.
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Hitung
Jawab :
Soal-soal
Selesaikan :
 ( (sin t)i + 2t j - 8t k) dt
3
Jawab :
(
(sin t)dt i) + (
2t dt j) - (
8t3 dt k)
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= (-cost + c1)i + (t2 + c2)j + (2t4 + c3)k
(3)
giving a surprising connecting between the area of a region and the line
integral around its boundary. For a plane curve specified parametrically as
x(t) , y(t) , equation (3) becomes
Contoh Dengan menggunakan rumus di atas hitung luas ellips berikut :
x2
y2
+
4
= 1
9
Jawab :
Ubah x dan y dalam bentuk parameter sebagai berikut :
r(t) = 2 cos t i + 3 sin t j
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r'(t) = -2 sin t i + 3 cos t j
maka diperoleh :
= 12π/2
=
6π
Definisi Panjang Busur
Contoh
Suppose that
r(t) = 3t i + 2j + t2k
Set up the integral that defines the arc length of the curve from 2 to 3.
Then use a calculator or computer to approximate the arc length.
Jawab :
We use the arc length formula
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Notice that we could do this integral by hand by letting t = 9/2 tan ,
however the question only asked us to use a machine to approximate
the integral. A TI 89 calculator gives
s = 5.8386
For the student . . . .
Background . . . .
Briefly, Coriolis acceleration is the
apparent acceleration one sees when
observing particle motion from a
rotating, rather an an inertial, reference
frame. For instance, the needle of a
phonograph moves (nearly) in a straight
line, yet traces a spiral on the rotating
record.
Mathematically, this is equivalent to
considering the opposite situation, such
as an object moving along a radial line
from the center of a rotating disk. You
may visualize this phenomenon by
placing yourself at the center of a merrygo-round and then walking toward the
edge. [ Ignore the horses or any other
obstruction. ] This situation is analyzed
in the box below.
For surface geometry, study the globe
on the left. The black line gives the great
circle path a frictionless hockey puck
would follow if the Earth were not
rotating. The blue line gives the great
circle it actually follows.
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The red line shows the apparent path as seen from the Earth which rotates
underneath the blue line - much as a phonograph rotates underneath the needle.
Now click on the globe to the left to select a variety of animations.
For the mathematical analysis we must use vector calculus.
Definition of the Arc Length Function
If r(t) is a differentiable vector valued function, then the arc length function
is defined by
Contoh
Parameterizing by arc length
Find the arc length parameterization of the helix defined by
r(t) = cos t i + sin t j + t k
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Jawab :
First find the arc length function
Solving for t gives
t = s/
Now substitute back into the position equation to get
r(s) = cos(s /
Gradient
) i + sin(s /
)j+s/
k
grad w (  w )
Contoh
Bila T(x,y,z)=2x3eysin(z),
Cari  T
=6x2eysin(z)i +2x3eysin(z)j +2x3eycos(z)k
Contoh
Sebuah bukit mempunyai ketinggian
w(x,y) = x2 - y
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Find the direction that is the steepest uphill and the steepest downhill
at the point (2,3).
Jawab :
We find
grad w = ( 2x, -y ) = ( 4, -3 )
Hence the steepest uphill is in the direction
( 4, -3 )
while the steepest downhill is in the direction
- ( 4, -3 ) = ( -4, 3 )
PROBLEM 1 Please answer the following true or false. If false, explain
why
or provide a counter example. If true explain why.
A. If A, B, and C are points, v is the vector from A to B, w is the vector from B to C,
and v x w = 0 , then A, B and C are collinear.
Jawab :
True, since
v x w = ||v|| ||w|| sin
Hence
sin  = 0
So the angle is 0, thus they are collinear.
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If x = x(t), y = y(t) are parametric equations of a line then dx/dt is
a constant.
Jawab :
A. False, consider the line
x = t3,
y = t3
For this line
dx/dt = 3t2
which is not a constant.
PROBLEM 3
Use vectors to find the equation of the line that passes through the point
(2,3,4) and is perpendicular to the plane 5x - 4y + 2z = 7.
Jawab :
This line is parallel to the normal vector to the plane. We need to find the
equation of the line that passes through the point (2,3,4) and is parallel to the
vector <5,-4,2>. The vector equation is
<x,y,z> = <2,3,4> + <5,-4,2>t
The parametric equations are
x = 2 + 5t
y = 3 - 4t
z = 4 + 2t
PROBLEM 7
Use vectors to determine if the triangle with vertices (1,0,1), (2,1,0), (0,0,4)
is a right triangle.
Solution
PROBLEM 8
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Find parametric equations for the a particle moves along the line through
(1,4,2) and (3,5,7) such that it is at the point (1,4,2) when t = 0 is at the point
(3,5,7) when t = 2 and is speeding up as time progresses
Solution
If we just wanted a parametric equation of a line we would just use the
formula
v + wt where v is the vector from the origin to (1,4,2) and w is the vector
from (1,4,2) to (3,5,7). However, we want to end when t = 2, so we use
v + wt/2
This almost works, but does not speed up as time progresses. We instead
use a t2 so that it is speeding up. Since 22 = 4, we use
v + wt2/4 or
<1,4,2> + <2,1,5>t2/4
x = 1 + t2/2 y = 4 + t2/4 z = 2 + 5t2/4
(6) The scalar or dot product
http://www.netcomuk.co.uk/~jenolive/vect6.html
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http://www.falstad.com/vector3d/
TERIMA KASIH
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