Multivariable Calculus
Summary 4 - Curves and Motion in Space.
I. Parametric Curve
As a point moves along a curve in space, we can describe this changing of position by the
parametric equations: x=f(t), y=g(t), z=h(t)
Example: x=2+3t, y=5-4t, z=-2-5t describes points on a line in space.
Example: x=cost, y=sint, z=t, describes an helix.
Vector-valued functions: a parametric curve can be described by giving its position vector r(t) =
x(t)i +y(t)j +z(t)k or r(t) = <x(t), y(t), z(t)>
Derivative of a vector-valued function:
The derivative of a vector-valued function is defined by



r (t  t) - r (t)
, provided the limit exists.
r (t )  lim
0
t


dr

Notation: r (t ) 
 Dt r (t ) and gives the vector tangent to the curve at the point
dt
(x, y, z) determined by t.


dx dy
dz
dr
r (t )  f (t )i  g (t ) j  h(t )k  i 
j k 
dt
dt
dt
dt
II.
Differentiation formulas:




1. Dt u (t )  v (t )  u (t )  v (t )


2. Dt cu (t )  cu (t )



3. Dt h(t )u (t )  h(t )u (t )  h(t )u (t )






4. Dt u (t )  v (t )  u (t )  v (t )  u (t )  v (t )






5. Dt u (t )  v (t )  u (t )  v (t )  u (t )  v (t )
III. Velocity vector and acceleration vector: speed and scalar acceleration.

dx dy
dz


d 2x
d2y
d 2z


v (t )  r (t )  i 
j  k and a (t )  v (t )  r (t )  2 i  2 j  2 k
dt
dt
dt
dt
dt
dt
2
2

 dy 
 dx 
 dz 
speed v  v (t )         
 dt 
 dt 
 dt 
2
2
2
2
 d 2x   d 2 y   d 2z 
dv

 
 

scalar acceleration a = a (t )  
 dt 2    dt 2    dt 2   dt

 
 

-1-
Summary: Curves in space .

Position vector: r (t )  x(t )i  y(t ) j  z (t )k

Velocity vector. v (t )  x(t )i  y(t ) j  z (t )k

Acceleration vector: a (t )  x(t )i  y(t ) j  z (t )k
2
2
 dx   dy   dz 
Speed: v  
    
 dt   dt   dt 
2
2
2
 d 2x   d 2 y   d 2z 
Scalar acceleration: a   2    2    2 
 dt   dt   dt 
2
IV. Arc length. Distance s is the anti-derivative of velocity v.
The distance traveled from one point at t1 to another point at t2 is given by
s=
t2
t2
t1
t1
 v(t )dt  
2
2
2
 dx 
 dy 
 dz 
        dt  arc length
 dt 
 dt 
 dt 
V. Arc length in R 2
If y=f(x) then the function can be expressed parametrically by the equations y=g(t)
and y=h(t) where t is called a parameter.
x1  g (t1 ), y1  h(t1)  ( x1 , y1 ) is a point on the curve
x2  g (t 2 ), y 2  h(t 2)  ( x2 , y 2 ) is another point on the curve
The length of the arc from  x1 , y1  to ( x 2 , y 2 ) can be expressed in the following forms:
Parametrically by
In rec tan gular coordinates by
s
t2

t1
2
2
 dx 
 dy 
     dt
 dt 
 dt 
s
x2

x1
2
 dy 
1    dx
 dx 
or s 
y2

y1
2
 dx 
1    dy
 dy 
VI. Integration of Vector-values functions.
b
b
b
b

 r (t )dt  i  f (t )dt  j  g (t )dt  k  h(t )dt 
a
a
a
a
i[ F (b)  F (a)]  j[G (b)  G (a)]  k[ H (b)  H (a)]
[ F (b)i  G (b) j  H (b)k ]  [ F (a)i  G (a) j  H (a)k ]




R(b)  R(a) wher R (t )  r (t ) This is the fundamental theorem of calculus
for vector  valued functions
-2-




Note: r (t )   v (t )dt and v (t )   a (t )dt
Fundamental Theorem of Calculus for vector-valued functions.
b
 b 




 r (t )dt  Rt   R(a)  R(b) where R (t )  r (t )
a
a
Position, velocity and acceleration vectors.



dr 
dv 
d 2r 
 v (t ),
 a (t ), 2  a (t )
dt
dt
dt




v (t )   a(t )dt and r (t )   v (t )dt
These integrals involve a vector constant of integration.




Note: an anti-derivative of r (t ) is of the form R(t )  C where C is a vector .
VII. Motion of projectiles.


dv
The acceleration of the projectile is a (t ) 
  gj , g is the gravity at the
dt
given point. g  32 ft / sec2 or g  9.8 meters / sec2 or g  980 cm / sec2
Equations:
The position vector of a projectile at time t is given by

1
r (t )  [v(0) cos t  x(0)]i  [ gt 2  (v(0) sin  )t  y(0)] j where  is the angle
2
of inclination from the horizontal of its initial velocity v(0).
Parametric representation of the position is given by
1
x(t )  v(0) cos  t  x(0) and y(t )  [ gt 2  (v(0) sin  )t  y(0)]
2
-3-
Definition. The range of the projectile is the horizontal distance it travels before it
returns to the ground.
Theorem: the range of a projectile is given by the formula
2
V0 )2 sin  cos , where  is the angle of inclination at which
g
the projectile is fire.
1
If g  32 ft / sec 2 , then range R  V0 2 sin  cos
16
Proof:




v (0)  V0 , r (0)  0, V (0)  V0 cos  i  V0 sin   j and a (t )   g j 



v (t )   gt j  c  v (t )   gt j  V0 cos i  V0 sin  j

 v (t )  V0 cos i  V0 sin   gt  j
Range R 
1



 r (t )  V0 cos t i  V0 sin  t  gt 2  j
2


When the projectile returns to the ground y(t)=0 and x(t)=range R
Therefore,
V0 cos t  range R and V0 sin  t  1 gt 2   0 
2


1 
1

t V0 sin   gt   0  t  0 or V0 sin   gt  0 
2 
2

2V sin 
t  0 or t  0
g
 2V sin   2
  V0 2 sin  cos
 range R  V0 cos t  V0 cos  0
g

 g
1
 range R  V0 2 sin  cos when g  32 ft / sec 2
16
-4-
Practice problems:
(when using calculators, round answers to 4 decimal places.)
1. Find the position, velocity, speed, acceleration and scalar acceleration for the given value of t.

2
3
a) r (t )  t i  t j at t=2

Position: __ r (2)  4i  8 j ________


velocity: ___ v (t )  2ti  3t j  v (2)  4i  12 j _____
2

speed:__ v (2) 
4 2  (12) 2  160  4 10 ______


acceleration: _ a (t )  2i  6tj  a (2)  2i  12 j _________
Scalar acceleration_


a (2)  2 2  (12) 2  148  2 37 ________
b) r (t )  ti  3e j  4e k at t=0

Position: __ r (0)  0i  3 j  4 jk ________
t
t


velocity: ___ v (t )  i  3e j  4e K  v (0)  i  3 j  4k _____
t

t
speed:__ v (0)  1  3  4 
2

2
26 ______

acceleration: _ a (t )  0i  3e j  4e k  a (2)  0i  3 j  4k _________
Scalar acceleration_
t
t

a (0)  32  (4) 2  25  5 ________
2. In problem 1 part a) find the distance along the curve from t=2 to t=5
Solution: distance is arc length
5
d 
2
2
2


5
5
5
 dy 
 dx 
2
2 2
2
4

dt

(
2
t
)


3
t
dt

4
t

9
t
dt

t 4  9t 2 dt
 
 



 dt 
 dt 
2
2
2
1 229 1 / 2
1 u3/ 2
Let u  4  9t , du  18tdt  d 
 u dt  18  3 / 2
18 40
229

2
229 229  80 10
 118 .978393 b
27
-5-
40
3. In problem 1 part b) find the distance along the curve from t=1 to t=3
Solution: distance is arc length
3
d 
1
2
2
   4e  dt
2
3
 dy 
 dx 
 dz 
2
t
        dt   (1)  3e
 dt 
 dt 
 dt 
1
3
  1  9e
3
2
t 2
3
 16e dt   1  25e dt   e 2t (e  2t  25)dt
2t
2t
1
3
2t
1
1
  e t (e  2t  25)dt  86.86806818 (u sin g a calculator )
1
Note:
3

3
3
1  25e dt   1  (5e ) dt   1  (5e ) 
2t
1
t 2
1
5e 3

5e

1 u2
u2
u  tan 
udu 
sec 2 
sec 2   1
t 2
1

1  tan 2 
tan 2 
sec tan  d
v sec

5e t
5e t
tan  sec d  
2
v2
 v 2  1 dv  
3
u 5e t 5e
dt 
1 u2
du
u

5e
sec 2 
tan 2 
v2 11
v2 1

sec tan d
dv   1dv  
 v  coth 1 v  sec  coth 1 (sec )  1  u 2  coth 1 1  u 2
3

1
dv
v2 1
 1  25e 2t  coth 1 ( 1  25e 2t )  1  25e 2t  coth 1 ( 1  25e 2t )
3
1
1
 1  25e

6

 coth 1 1  25e 6  1  25e 2  coth 1 1  25e 2
 
 100 .4327  coth 1 (100 .4327 )  13.6281  coth 1 (13.6281)


1 100 .4327  1 
1 13.6281  1

 100 .4327  ln
 13.6281  ln

2 100 .4327  1 
2 13.6281  1

 100 .4327  0.009957   13.6281  0.07351  100 .4227  13.5546  86.8681
-6-
4. For the scalar function f(t)= t  3t  4 and the vectors


u  t , t 2 , t 3  and v  3t  2, 4, 3t 2  find
2
 
a) Dt u  v  at t=2
answer: ___ 270 _______
Solution:


 dv du 
u

 v  t , t 2 , t 3  3, 0, 6t  1, 2t , 3t 2  3t  2, 4, 3t 2
dt dt
 2, 4, 8  3, 012  1, 4,12  8, 4,12  102  168  270
or
 
d u  v 
 
2
5
u  v  t (3t  2)  4t  3t 
 6t  2  8t  15t 4
dt
 
 at t  2, Dt u  v   6(2)  2  8(2)  15(16)  12  2  16  240  270
 
b) Dt u  v  at t=2
answer: _ 48, 84,  40 _________


   d v  d u  
Dt u  v   u 

 v  t , t 2 , t 3  3, 0, 6t  1, 2t , 3t 2  3t  2, 4, 3t 2
dt
dt
 2, 4, 8  3, 0,12  1, 4,12  8, 4,12
i
j
k
i
j
k
 2 4 8  1 4 12  48i  0 j  12k   0i  84 j  28k   48i  84 j  40 k
3 0 12 8 4 12

c) Dt  f (t )v  at t=2
Solution:
answer: __ 74, 28, 156 _

dv df (t ) 

Dt  f (t )v   f (t )

v  t 2  3t  4 3, 0, 6t  (2t  3) 3t  2, 4, 3t 2
dt
dt
 4  6  4 3, 0,12  (4  3) 8, 4, 12  6 3, 0,12  (7) 8, 4, 12


 18, 0, 72 56, 28, 84
 74, 28, 156
-7-




5. Find the position vector r (t ) if a (t )  ti  t j  t k and r (0)  10i and v (0)  10 j
Solution:
2
3


a (t )  ti  t 2 j  t 3 k and r (0)  10i and v(0)  10 j
t2
t3
t4

 
 v (t )  i 
j  k  c , c is a cons tan t vector
2
3
4
t2
t3
t4


 v (t )  i 
j  k  10 j because v (0)  10 j
2
3
4
4
 t3

t2
t

 v (t )  i    10  j  k
2
4
3

5
t4

t3
t

 
 r (t )  i    10t  j 
k  c , c is a vector
6
20
 12

t4

t3
t5




 r (t )  i    10t  j 
k  10i, because r (0)  10i
6
20
 12

 t3
 t4

t5





 r (t )    10 i    10t  j 
k
6
12
20

 

6. A projectile is fire horizontally from the top of a 100-m cliff at a target 1 km
from the base of the cliff. What should be the initial velocity of the projectile?
Give answer in yd. per second.
Solution:


v (0)  v0 i and r (0)  100 j


a (t )  9.8 j  v (t )  9.8t j  v o i  v o i  9.8t j

 r (t )  v o t i  4.9t 2 j  100 j  v o t i   4.9t 2  100 j
The projectile reaches the target when y(t) = 0 and x(t) = 1000m = 1 km
100 10 10
y (t )  0  4.9t 2  100  0  t 

4.9
7
10 10
700
x(t )  1000  v o t  1000 
v 0  1000  v 0 
 70 10
7
10
 221 .359 m / sec  495 .166 mph  726 .243 ft / s  242 .081 yd / s


-8-
7. A projectile is fired from a height of 80 ft with an initial speed of 120 ft/sec at
an angle of 60 degrees. Find the maximum height of the projectile and the time
it takes to reach the ground. Answer to the nearest hundredth.
Solution:

At time t=0 the position vector is given by r (0)  80 j and the velocity vector

is given by v (0)  120 (cos 60i  sin 60 j )  60i  60 3 j

At any time a (t )  32 j



Therefore, v (t )  32tj  c where c is a cons tan t vector



Using the given fact at t=0, v (0)  32(0) j  c  60i  60 3 j  c  60i  60 3 j

Therefore the velocity vector at any time is given by v (t )  32tj  60i  60 3 j

or v (t )  60i   32t  60 3 j
The maximum height is reached when the vertical component of velocity is zero,
that is, it stops vertically.
60 3 15 3
 32t  60 3  0  t 

32
8


The position vector at any time is given by r (t )  60ti   16t 2  60 3t j  c


But c  80 because r (0)  80 j

Therefore, r (t )  60ti   16t 2  60 3t  80 j






The maximum height is given by the vertical component when t 
Maximum height 
2
15 3
8
 15 3 
 15 3 
675 1350
707
  60 3
  80  
h  16


8

 176 .75 ft

 8 
8
4
4
4




The object hits the ground when the vertical component of position is zero.
15 3  675  320
 16t 2  60 3t  80  0  4t 2  15 3t  20  0  t 
8
 t  7.19 sec whe n it hits the ground
-9-
8. A particle moves with constant speed along a curve in space. Show that its
velocity and acceleration are always perpendicular.
Proof:
2  
Speed  v  v  v  k , where k is a cons tan t
Differentiating both sides with respect to t
 


d v  v  d k   dv dv 
   
 

v 

 v  0  v  a  a  v  0  2v  a  0
dt
dt
dt dt
 


 v  a  0  v and a are perpendicular
9. A point moves on a sphere centered at the origin. Show that its velocity vector
is always tangent to the sphere.
Solution:

Let (x, y, z) be any point on the sphere and let the vector r t   x, y, z be the
position vector.
By definition of the sphere x 2  y 2  z 2  r 2
Therefore


x  x  y  y  z  z  r 2  x, y, z  x, y, z  r 2  r (t )  r (t )  r 2  cons tan t
Differentiating both members of the equation with respect to t




d r (t )  r (t )  d r 2
d r (t )  d r (t )  


 r (t ) 

 r (t )  0
dt
dt
dt
dt








 r (t )  v (t )  v (t )  r (t )  0  2r (t )  v (t )  0  r (t )  v (t )  0
 
Therefore the velocity vector is tangent (perpendicular) to the position vector.
10. A point moves on a circle whose center is the origin. Use the dot product to
show that the position and velocity vectors of the moving point are always
perpendicular.
A point that moves on a circle whose center is the origin can be represented
parametrically by the equations x=rcost and y=rsint or by the position vector

r (t )  r cos t i  (r sin t ) j

d r (t ) 

v (t ) 
  r sin t i  (r cost ) j
dt


r (t )  v (t )  r 2 sin t cos t  r 2 sin t cos t  0
Therefore the position and velocity vectors are perpendicular to each other.
-10-
11. An artillery gun with a muzzle velocity of 1000 ft/sec is located atop a seaside
cliff 500 ft high. At what initial angle of inclination should it fire a projectile in
order to hit a ship at sea 4 miles from the base of the cliff?
Solution:



v (0)  1000 ft / s, v (0)  1000 (cos i  sin  j ) and r (0)  500 j



a (t )  32 j ft / sec 2  v (t )  32tj  c  32tj  1000 (cos i  sin  j )

v (t )  1000 cos i  (32t  1000 sin  ) j


 r (t )  1000 cos t i   16t 2  1000 sin  t j  c

 r (t )  1000 cos t i   16t 2  1000 sin  t j  500 j

 r (t )  1000 cos t i   16t 2  1000 sin  t  500 j
4 miles =4(5280 ft) = 21120 ft
When the projectile hits the ship x=21120 ft , y=0






(1000 cos )t  21120 and  16t 2  (1000 sin  )t  500  0
2
21120
21.12
 21.12 
 21.12 
t

and  16
  (1000 sin  )
  500  0
1000 cos cos
 cos 
 cos 


 16 21.12) 2 sec  2  21120 tan   500  0


 7136 .8704 1  tan 2   21120 tan   500  0
 7136 .8704 tan 2   21120 tan   6636 .8704  0
 tan 2   2.959 tan   0.929941  0    0.343303 RAD    19.6698 
-11-